Saturday, December 7, 2019

Analysis of Space Mission Sensitivity to Assumptions

For any given vehicle design,  what one assumes for mission delta-vees,  vehicle weight statements,  course corrections,  and landing burn requirements greatly affects the payload that can be carried.  The effect is exponential:  variation in required mass ratio with changes in delta-vee and exhaust velocity.

This analysis looks at trips from low Earth orbit to direct entry at Mars,  and for the return,  a direct launch from Mars to a direct entry at Earth.  The scope is min-energy Hohmann transfer plus 3 faster trajectories (see ref. 1). 

The vehicle under analysis is the 2019 version of the Spacex “Starship” design,  as described in ref. 2.  The most significant items about that vehicle model are the inert mass and the maximum propellant load.  For this study,  the vehicle is presumed fully loaded with propellant at Earth departure,  and at Mars departure.  See also Figure 1.  Evaporative losses are ignored.

 Figure 1 – Summary of Pertinent Data for 2019 Version of Spacex “Starship” Design

Since a prototype has yet to fly,  the design target inert mass of 120 metric tons is presumed as baseline.  Uncertainty demands that inert mass growth be investigated.  To that end,  the average of that design target and the 200 metric ton inert mass of the so-called “Mark 1 prototype” (that average is some 160 metric tons) is used to explore that effect.

As currently proposed,  the vehicle has six engines.  Three are the sea level version of the “Raptor” engine design,  and the other three are vacuum versions of the same engine design (basically just a larger expansion bell).  I have already reverse-engineered fairly-realistic performance for these in ref. 3.  Because of the smaller bells,  the sea level engines gimbal significantly,  while the vacuum engines cannot.  Thus it is the sea level engines that must be used to land on Mars as well as Earth:  gimballing is required for vehicle attitude control.

Analysis Process

As shown in Figure 2,  the analysis process is not a simple single-operation calculation.  The vehicle model provides a weight statement and engine performance.  The mission has delta-vee requirements for departure,  course correction,  and landing,  which must be appropriated factored (in order to get mass ratio-effective values).   There are two sets of analysis:  the outbound leg from Earth to Mars,  and the return leg from Mars to Earth. 

Each leg analyzes 3 burns.  Earth departure,  and course correction are done with the vacuum “Raptor” engines,  while the landing on Mars is done with the sea level “Raptors” to obtain the necessary gimballing.  Mars departure and course correction are done with the vacuum “Raptor” engines (Mars atmospheric pressure is essentially vacuum).  The Earth landing is done with the sea level “Raptors” to get the gimballing and to get the atmospheric backpressure capability.  

 Figure 2 – The Analysis Process and Equations,  with Data

This analysis is best done in a spreadsheet,  which then responds instantly to changes in one of the constants (like an inert mass or a delta-vee).  That is what I did here. 

Referring again to Figure 2,  for each burn,  there is an appropriate vehicle ignition mass.  At departure,  it is the ignition mass from the weight statement.  For each subsequent burn,  it is the previous burn’s burnout mass.  Each burn’s burnout mass is its ignition mass divided by the required mass ratio for that burn,  in turn figured from that burn’s delta vee and the appropriate exhaust velocity.

For each burn,  the change in vehicle mass from ignition to burnout is the propellant mass used for that burn.  For the first burn,  the propellant remaining (after the burn) is the initial propellant load minus the propellant mass used for that burn.  For the subsequent burns,  propellant remaining is the previous value of propellant remaining,  minus the propellant used for that burn. 

After the final burn,  the propellant remaining cannot be a negative number!  If it is,  one reduces the payload number originally input,  and does all the calculations again.  If this done in a spreadsheet,  this update is automatic!  Ideally,  the propellant remaining should be exactly zero,  but for estimating purposes here,  a small positive fraction of a ton (out of 1200 tons) is “close enough”.

Thus it is payload that is determined in this analysis.   This particular input (payload mass) is revised iteratively until the final burn’s remaining-propellant estimate is essentially zero.  That is the maximum payload value feasible for the mission case.

Orbits and the Associated Delta Vees

As indicated in ref. 2,  I have looked at a Hohmann min energy transfer orbit,  and 3 faster transfers with shorter flight times.  All of these are transfer ellipses with their perihelions located at Earth’s orbit.  For Hohmann transfer,  the apohelion is at Mars’s orbit.  For the faster transfers,  apohelion is increasingly far beyond Mars’s orbit.  Why this is so is explained in the reference.  See Figures 3 and 4.

Note that the overall period of the transfer orbit is important for abort purposes.  If the period is an exact integer multiple of one Earth year,  then Earth will be at the orbit perihelion point simultaneously with anything traveling along that entire transfer orbit.  This offers the possibility of aborting the direct entry and descent at Mars,  if conditions happen to be bad when the encounter happens.  Otherwise,  the spacecraft is committed to entry and descent,  no matter what.  

 Figure 3 – Hohmann and Faster Transfer Orbits,  Earth to Mars

 Figure 4 – More Details About Hohmann and Faster Transfer Obits from Earth to Mars

The cases examined in ref. 1 were all computed for Earth and Mars at their average distances from the sun.  The larger transfer ellipse with the longer period occurs when both Earth and Mars are at their farthest distances from the sun.  This leads to larger delta vees to reach transfer perihelion velocity for the trip to Mars,  and larger velocity on the transfer orbit for the trip back to Earth.

Ref 1 has the required velocities and delta-vees,  but the most pertinent data are repeated here:

Transfer              E.depdV, km/s   trip time, days   M. Vint, km/s
Hohmann            3.659                   259                       5.69
2-yr abort           4.347                   128                       7.40
No abort             4.859                   110                       7.36
3-yr abort           5.223                   102                       6.53

Transfer              M.depdV, km/s trip time, days           E.Vint, km/s
Hohmann            5.800                   259                       11.57
2-yr abort           7.548                   128                       12.26
No abort             7.509                   110                       12.77
3-yr abort           6.653                   102                       13.14

I did not examine the worst cases for all the transfer orbits in ref. 1,  but I do have the  increase in perihelion velocity for the worst case Earth departure on a Hohmann transfer for Mars:  0.20 km/s higher than average.  I also have the increase in apohelion velocity for the worst case Mars departure on a Hohmann transfer for Earth:  0.16 km/s higher than average.

I cheated here:  I used those worst-case Hohmann increases for all the faster trajectories as well.  That’s not “right”,  but it should be close enough to see the relative size of the effect of worst case over average conditions.  I also used the same additive changes on the entry velocities.

Because of the precision trajectory requirements for direct entry while moving above planetary escape speed,  some sort of course correction burn or burns will simply be required.  With this kind of analysis,  I have no way to evaluate that need.  So I just guessed:  0.5 km/s delta-vee capability in terms of propellant reserves. 

Because this is just a guess,  I did not run any sensitivity analysis on it.  However,  the delta-vee budget proposed here is factor 2.5 larger than the difference average-to-worst-case for the trip to Mars,  which suggests it is “plenty”.   It is about factor 3 larger than the difference average-to-worst-case for the return trip to Earth.  You can get a qualitative sense of this effect from examining that average-vs-worst case effect.

Propellant Budgets for Direct Landings

With this vehicle (or just about any other vehicle),  entry must be made at a shallow angle relative to local horizontal.  Down lift is required to avoid bouncing off the atmosphere,  since entry interface speed Vint exceeds planetary escape speed.  This is true at both Mars and at Earth.  Once speed has dropped to about orbit speed,  the vehicle must roll to up lift,  to keep the trajectory from too-quickly steepening downward. 

The hypersonics end at roughly local Mach 3 speeds,  which is around 0.7-1 km/s velocity,  near 5 km altitude on Mars,  and near 45 km altitude on Earth  which has about the same air pressure.  Up to that point,  entry at Mars and Earth look very much alike,  excepting the altitude.  After that point they diverge sharply,  as illustrated in Figure 5.

 Figure 5 – Entry Trajectory Data for “Starship” at Mars and at Earth

The descent and landing at Earth require the ship to decelerate to transonic speed,  then pull up to a 90-degree angle of attack (AOA,  measured relative to the wind vector).  Thus,  as the trajectory rapidly steepens to vertical,  the ship executes a broadside “belly-flop” rather like a skydiver. 

At low altitude where the air is much denser,  the terminal speed in the “belly-flop” will be well subsonic.  I assumed 0.5 Mach,  but that might be a little conservative.  This is the point where AOA increases to 180 degrees (tail-first),  and the landing engines get ignited.  From there,  touchdown is retropropulsive.

The landing on Mars is quite different.  The ship comes out of hypersonics very close to the surface,  still at high AOA and still very supersonic.  From there,  the ship must pitch to higher AOA and pull up,  actually ascending back toward 5 km altitude.  This ascent is energy management:  speed drops rapidly as altitude increases.  It’s not quite a “tail slide” maneuver,  but it is similar to one. 

At the local peak altitude,  the ship is moving at about local Mach 1,  and pitches to tail first attitude,  igniting the landing engines.  From there,  touchdown is retropropulsive.  The Martian “air” at the surface is very thin indeed,  as the figure indicates.  It may be that thrust is required to assist lift toward bending the trajectory upward:  the engines would have to be ignited earlier,  and at higher speed,  as indicated in the figure.  Whether this is necessary is just not yet known.

The low point preceding the local pull-up is at some supersonic speed;  I just assumed about local Mach 1.5,  as indicated in the figure.  That would correspond to a factor 1.5 larger landing delta-vee requirement,  implying a larger landing propellant budget. 

In either case,  I also use an “eyeball” factor of 1.5 upon the kinematic landing delta-vee,  to cover gravity loss effects,  maneuver requirements,  and any hover or near-hover to divert laterally to avoid obstacles. 

So,  for purposes of this sensitivity analysis,  the Earth landing is not of much interest,  but the Mars landing is.  The sensitivity analysis looks at the effects of Mach 1.5-sized vs Mach 1-sized touchdown delta-vee.

Analysis Results

The scope of the sensitivity analysis is illustrated in Figure 6.  As indicated earlier,  the orbital delta-vee increases worst-case-vs-average,  for Hohmann transfer,  were applied additively to the departure delta-vees for the faster trajectories.  No attempt was made to vary the course correction budgets.   Growth in vehicle design inert mass was examined.  An increase in the Mars touchdown delta-vee was examined.  Nothing else was considered.  

 Figure 6 – Scope of Sensitivities Analyzed

The results start with the worst vs average orbital delta-vee sensitivity.  These results are given in Figure 7.  These are the plots from the spreadsheet,  copied and pasted into the figure.  There are 4 such plots in the figure:  the top two are for the outbound journey Earth to Mars.  The bottom two are for the return journey Mars to Earth.  Results for all 4 transfer orbit cases are shown simultaneously by using trip time as the abscissa. 

Each has 4 data points:  these are for the Hohmann transfer at 259 days flight time,  the 2-year abort orbit at 128 days,  the non-abort orbit at 110 days,  and the 3-year abort at 102 days.  Be aware that the curves are probably not really straight between the Hohmann orbit and the 2-year abort orbit.  I did not run enough fast transfer cases in ref. 1 to get a smooth curve here.

The most significant thing in the left hand figure for the outbound trip is the about-40 ton loss of max payload between average and worst case for the Hohmann transfer.  This is a lot less than the about-130 ton payload loss using the 2-year abort orbit instead of Hohmann transfer,  or the about-210 ton payload loss for using the 3-year abort orbit. 

The average-vs-worst-case deficits are somewhat similar on the faster orbits.  The Mars entry interface velocity trend in the right-hand figure is obviously very nonlinear.  Yet,  all the calculated values fall below the entry velocity from low Earth orbit (LEO).  Any heat shield capable of serving for return from LEO will serve this Mars entry purpose,  which would be the governing case if the trip were one-way only.  There’s only a small change in entry speeds for average-vs-worst orbit case in this estimated analysis. 

The return voyage has trends shaped quite differently.  For Hohmann transfer,  the worst-vs-average payload loss is about 20 tons.  The deficits on the faster orbits should be similar.  The deficit for using the 2-year abort orbit instead of Hohmann is far larger at about 110 tons,  and that’s from a small return payload to begin with. 

In the right hand Earth entry interface speed plot,  the blue and orange curves in the entry interface plot fall only slightly apart.  Note that all the entry velocities are much higher than the just-below-escape speed seen with Apollo returning from the moon.  The faster transfer orbits,  and even the Hohmann transfer,  are substantially more demanding than a lunar return entry.  It is clearly the direct-entry Earth return that will size the heat shield design!  

Figure 7 – Sensitivity to Worst-Case Orbital Distances vs Averages

Results for the effects of inert mass growth sensitivity are given in Figure 8.  This is the same 4-plot format as Figure 7.  For the outbound trip to Mars,  the Hohmann mass penalty for inert mass growth is about the same 40 ton deficit as for worst-case orbit distances.  It is similar for the faster trajectories.  It is the return trip that most suffers from vehicle inert mass growth.  We lose about 40 tons from an already small return payload on the Hohmann transfer.  However the 2-year abort trajectory and the no-abort trajectory are entirely infeasible,  with their max payloads calculated as negative.  Note that both the Mars and Earth entry interface velocities are unaffected by this sensitivity.  The orange and blue curves fall right on top of each other.

 Figure 8 – Sensitivity to Vehicle Inert Mass Growth

The sensitivities to the need for a thrusted pull-up on Mars are given in Figure 9.  This follows the same format as Figures 7 and 8.  Bear in mind that the nominal design lights the engines for touchdown at about Mach 1 speed.  For this analysis,  the engines are ignited earlier,  at about Mach 1.5 flight speed,  to assist lift in pulling up to the Mach 1 “flip”,  to tail-first attitude.  That makes the landing delta vee about 1.5 times larger.  (Note that each case is also factored up by 1.5 further,  to cover any maneuver / hover needs for the touchdown.)

What the figure shows is about the same 40-ton payload loss on the voyage to Mars to cover the increased landing propellant requirement for the Hohmann transfer.  Effects on the faster transfers are similar.  This trend is comparable to the worst-case orbit losses.  The return payload is entirely unaffected,  as the landing occurs prior to refueling and loading for the trip home.

Both the Earth and Mars entry interface velocities are unaffected by this Mars thrusted pull-up scenario.  The orange and blue curves fall right on top of each other. 

 Figure 9 – Sensitivity to The Need for a Thrusted Pull-Up on Mars

Final Remarks

#1.  These results are only approximate!  Real 3-body orbital analysis,  and real entry-trajectory lifting flight dynamics models,  must be used to get better answers.  Nevertheless,  the trends are quite clear from this approximate analysis.

#2.  Flying on faster transfer orbits will cost a lot of payload capability,  on both the outbound voyage,  and the return voyage.  This effect is much worse on the return voyage,  where the allowable payload is just inherently smaller.

#3.  The effects of worst-case orbital positions-relative-to-average,  of Mars and Earth,  have a significant effect on payload,  but it is only half or less the effect of choosing faster transfer orbits.

#4.  The effect of vehicle inert mass growth from the design target of 120 metric tons to an arbitrary but realistic 160 metric tons is comparable to the effect of worst-case vs average orbits on the outbound voyage.  However it has catastrophic effects on the return voyage!  This is enough to prevent faster-than-Hohmann transfers on the voyage home,  for this vehicle model.

#5.  The effects of needing a thrusted pull-up for the Mars landing is comparable to the effects of worst-case orbit distances on the outbound voyage.  This has no effects upon the return voyage.

#6.  It is the direct Earth entry velocity that will design the vehicle heat shield for any vehicle capable of making the return.  This is substantially more challenging than was the return from the moon.  For deliberately-designed one-way vehicles to Mars,  the heat shield design requirements are comparable to entry from low Earth orbit.

#7.  My personal opinions are that thrusted pull-up will be needed,  along with the need to fly when Earth and Mars orbital distances are worst-case,  plus there will be a little inert mass growth (say by 20 metric tons to 140 metric tons vehicle inert mass).  That kind of thing is the proper design point for this vehicle,  not the most rosy projections!  Estimated performance data for this design case (at 140 metric ton inert mass) are in Figure 10 (same basic format as Figures 7,  8,  and 9).   Note that two of the faster transfers home are precluded.  The feasible one has a very small max payload value compared to Hohmann transfer. 

Figure 10 – Performance for Worst Orbits,  Thrusted Pull-Up,  and Some Inert Mass Growth

#8.  Bear in mind that the rather high max allowable payload figures feasible to Mars for Hohmann transfer are incompatible with what can be aboard “Starship” for launch to low Earth orbit.  The payloads for the faster transfers to Mars look more like what can be ferried up to LEO.  That suggests that a faster transfer to Mars is most compatible with the projected “Starship” / “Super Heavy” system design characteristics,  as these were evaluated in references 2 and 3.

#9.  Bear also in mind that a faster transfer orbit to Mars ought to include abort capability,  in case conditions at arrival prove too bad to attempt the landing.  There is simply not the propellant available to enter orbit and wait for better conditions.  Thus life support supplies must be carried to last the entire period of the transfer orbit,  and a full-capability heat shield for direct Earth entry must be used.

#10.  The fast transfer home need not be limited by abort capability.  It can be a different transfer orbit than the outbound trip.  Surprisingly,  the shapes of the plotted curves suggest that something faster than the “3-year abort” orbit could be used for the return home.

#11.  Given a way to combine two payloads to LEO into one “Starship” by cargo transfer operations on orbit,  then (and only then) the very large payloads to Mars indicated for Hohmann transfer become feasible.  Like on-orbit cryogenic refueling,  this on-orbit cargo transfer capability does not yet exist,  not even as a concept (on-orbit refueling at least exists as a concept).


#1. G. W. Johnson,  “Interplanetary Trajectories and Requirements”,  posted 21 November 2019,  this site.

#2. G. W. Johnson,  “Reverse-Engineering the 2019 Version of the Spacex “Starship”/ ”Super Heavy” Design”,  posted 22 October 2019,  this site.

#3. G. W. Johnson,  “Reverse-Engineered “Raptor” Engine Performance”,  posted 26 September 2019,  this site.

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Thursday, November 21, 2019

Interplanetary Trajectories and Requirements

Update 11-24-19:  revised delta-vees for Phobos trip,  appended below


The planning of interplanetary flights,  such as from Earth to Mars,  uses basic orbital mechanics.  As long as the spacecraft speed does not reach solar escape speed,  the form of the orbital trajectory about the sun will be an ellipse.  There are “min energy” trajectories,  and there are faster trajectories,  but all these trajectories will be ellipses.

The basics of elliptical orbits are given in Figure 1,  including just about all the relevant analysis equations,  plus a little more.  An ellipse is a symmetrical closed curve containing two foci.  The central body occupies one focus,  the other is unoccupied.  These foci are located farther from the center in the more eccentric ellipse.  A circular orbit is a sort of “degenerate” ellipse with zero eccentricity,  so that the foci come together at the center.   I wrote this for a diverse audience of both technical and non-technical people.

Bear in mind that these analytical solutions apply only to a 2-body problem (one object orbiting a central body).  The 3-body problem (object and two bodies) requires integration of the equations of motion for an exact solution.  

 Figure 1 – Basics of Elliptical Orbits (2-Body Problem)

For interplanetary trips,  we are talking about trajectories where the central body is the sun.  However,  this same analysis applies to orbits about the Earth or any other body.  The key features of elliptical orbits are the speeds,  which vary around the path,  and the min and max distances from the central body. 

“Perihelion” is the end apex of the ellipse closest to the sun,  where the speeds are highest.  “Aphelion” is the end apex of the ellipse farthest from the sun,  where speeds are lowest.  The corresponding terms for an orbit about the Earth are “perigee” and “apogee”.   For the moon,  they are “pericynthion” and “apocynthion”.  Thus the abbreviation “per” applies to the closest approach,  and “apo” the farthest approach,  regardless of the central body.

When looking up tables of characteristics of astronomical bodies,  the description of their orbits is usually cast as min and max distances from the central body.  Those would be rper and rapo,  respectively.   Their average is the semi-major axis length “a”.  The eccentricity “e” and the period “P” are easily computed from these distances,  all as given in Figure 1.  Further,  the distance of the foci from the center of the ellipse “c”,  and the length of the semi-minor axis “b”,  are also easily computed,  and given in the figure,  as is the equation for x-y coordinates all along the ellipse,  where these x-y coordinate are centered at the center of the ellipse.

The equation giving velocity at any radius “r” from the central body is actually quite simple,  as shown in the figure above.  Note that “r” is bounded:  rper < r < rapo.  There is no simple equation giving time at any particular point around the elliptical path.  Centuries ago,  it was  said “equal areas are swept out by the radius vector from the central body in equal amounts of time”. 

Today,  we would say that the time along a segment is proportional to the area swept out by the radius vector from the central body to the object,  moving along that segment.  You can obtain that time by integrating the area under the segment from one point to another,  and adding or subtracting the appropriate triangle area.  All of this should be evident upon inspection of Figure 1 above,  particularly noting the shaded area.

The “extras” in Figure 1 are the equations for calculating the escape velocity and low circular orbit velocity of any celestial body,  plus a form of mechanical energy conservation as you approach (or depart from) a body in an unpropelled state.

Hohmann Min-Energy Transfer

These trajectories are the ones with the minimum velocity requirements for travel.  The perihelion of the transfer ellipse is located at the orbit of the Earth on one side of the sun.  The apohelion of the transfer ellipse is located at the orbit of Mars on the other side of the sun.  See Figure 2.  Because the orbit of Earth is slightly elliptical,  and Mars more so,  these distances vary somewhat.  Using the average values gets you into the ballpark,  but you should really use the worst case to size your spacecraft’s propulsion capability! 

Note that because the Hohmann transfer ellipse is tangent to a very-nearly-circular planetary orbit at each end,  then the planetary velocity and transfer orbit velocity vectors are essentially parallel at each end.  What is ordinarily a vector subtraction devolves to a simple scalar subtraction,  for determining the velocities with respect to the planet,  at each end.  That is only true when the perihelion or the apohelion of the orbit are located at the appropriate planetary orbit distance from the sun.  

Specifically,  the velocity vector of the spacecraft with respect to the planet is the velocity vector of the spacecraft with respect to the sun,  minus the velocity vector of the planet with respect to the sun:

               spacecraft Vwrt planet = Vwrt sun – Vplanet wrt sun    where V is a vector velocity

 Figure 2 – The Hohmann Min-Energy Transfer Ellipse Earth-to-Mars and Back

An Approximation to the 3-Body Problem

The trouble the above evaluations right at perihelion and apohelion is that these are 2-body (spacecraft and sun) analyses,  and the close-vicinity dynamics of departure and arrival are fundamentally a 3-body problem (spacecraft-sun-planet).  As already stated,  it takes computerized 3-body analysis to get velocity requirements and detailed localized trajectories exactly right.  Yet,  you can get very,  very close to the velocity requirements with the following approximation technique. 

As in the above discussion,  you do the vector velocity subtraction to find the velocity of the spacecraft with respect to the planet,  at the perihelion and apohelion conditions.  For only Hohmann min energy transfer,  this calculation devolves to a simple scalar subtraction,  because the vectors are parallel.  Either way,  there is a velocity magnitude involved. 

The approximation is to treat this relative velocity with respect to the planet as a velocity “very far from the planet”,  and to approximate the pull of the planet’s gravity during the close encounter as an unpowered gravitational acceleration toward the planet,  from “very far” to “very close”.  This is done with conservation of mechanical energy,  based on the “far from planet” velocity magnitude (with respect to the planet),  which makes the vector direction more-or-less irrelevant,  except to the trajectory details.   Conservation of mechanical energy says:

               0.5 m Vfar2 = 0.5 m Vnear2 - change-in-PE far-to-near,  where m = spacecraft mass

This approximation then makes use of the very convenient fact that the change in potential energy from very far to very near is,  in point of fact,  numerically equal to the spacecraft kinetic energy associated with the escape velocity of the planet:

               0.5 m Vesc2 = change-in-PE far-to-near

Thus,  after dividing off the “0.5 m” factors common to all three terms,  we have a very simple way to estimate the spacecraft velocity magnitude with respect to the planet,  once it is “very close”.  This velocity would apply for either entry into planetary orbit,  or for the initial direct entry into the local atmosphere for aerobraking (of any type).  That simple equation is:

               Vfar2 = Vnear2 – Vesc2   or  Vnear = (Vfar2 + Vesc2)0.5

Doing the full 3-body problem on the computer refines the actual trajectory to be flown,  but does not refine the spacecraft propulsion velocity requirements very much at all,  beyond these simple estimates.  These estimates are really quite good,  and apply to departure as well as arrival.

This is illustrated in Figure 3,  for both arrival and departure at both Earth and Mars.  For arrival,  Vnear is denoted as Vint for “interface velocity”,  and Vnear = Vbo for the “burnout velocity” at departure.  

 Figure 3 – The Approximation for Close Encounter Estimates Without Doing 3-Body Analysis

Please note that Mars arrival and departure trajectories are directed retrograde,  with respect to Mars,  because the planet’s orbital velocity about the sun exceeds the transfer ellipse apohelion velocity with respect to the sun.  In effect,  the planet literally runs over the spacecraft from behind upon arrival.  You want to time your arrival at Mars orbital distance very slightly ahead of Mars’s arrival,  so that you don’t miss closing gravitationally with the planet,  or get left behind for not “leading the target” enough. Similarly,  you must accelerate in the retrograde direction escaping from Mars,  so that you end up at the apohelion of your return ellipse,  with the appropriate slower velocity about the sun.

The situation at Earth is different,  because the transfer ellipse perihelion velocity with respect to the sun exceeds the orbital velocity of Earth about the sun.  Thus departures and arrivals are in the posigrade direction with respect to Earth.  Upon arrival,  the spacecraft is literally running into the Earth from behind.  It literally runs away from the Earth in a posigrade direction upon departure. 

Actual Departure and Arrival “Close-In” Operations

Arrival at,  and departure from,  Mars is depicted in Figure 4.  These are similar at Earth,  only the numbers are different.   You are “close-in” at speed Vnear,  which is Vint for arrival.  If Mars (or Earth) were airless,  this is the theoretical delta vee you have to “kill” in order to land direct.  They are not airless,  so your choices are entry into orbit,  or some sort of direct aerobraking entry.  

 Figure 4 – Recommended Approximate Analyses for Close-In Operations

If you are entering Mars orbit,  you want to enter it on the side of the planet and specific location where the orbit velocity vector is fairly parallel with your own spacecraft velocity vector.  That gives the smallest delta-vee requirement:

               dV = Vint – Vorbit  on the “correct” side

If you enter orbit on the wrong side,  the delta vee is much larger:

               dV = Vint + Vorbit  on the “wrong” side

Orbital entry dV values need no factoring,  because these are brief impulsive burns in space.  It is not feasible to use low-thrust long burn electric propulsion for this,  at least not for manned craft.  The spiral-in times are months long.

If instead you are aerobraking (whether one pass or multi-pass),  your initial entry interface speed is essentially Vnear = Vint.  From there deceleration is by drag,  not propulsion,  until touchdown.  On Mars,  retropropulsive touchdown is required at one level or another,  since terminal parachute speeds are high subsonic at best.  Whatever the terminal velocity is,  that is the theoretical delta-vee you need to “kill” for touchdown.  I recommend that theoretical value be increased by factor 1.5 to cover maneuver and hover allowances.

Departure from Mars is the exact reverse,  but in the same direction as arrival.  You want to end up at Vnear = Vbo still close to the planet,  so that your Vfar is the aphelion velocity of the transfer ellipse back to Earth.  If departing from orbit,  you burn on the side where the orbital motion is locally retrograde,  so that the delta-vee required is lower:

               dV = Vbo – Vorbit

Departing Mars direct from the surface,  your theoretical dV is Vbo,  and it must be directed retrograde.  This dV needs to be factored-up for small drag and gravity losses.  Recommendations are given in the figure. 

While not shown in the figure,  departing Earth is the same process and analysis,  just with everything oriented in the posigrade direction.  If you leave orbit,  you do it on the side where orbital motion is posigrade,  and end up at Vbo in a posigrade direction.  That delta vee is:

               dV = Vbo – Vorbit

If you depart directly from the surface,  your theoretical delta vee is Vbo,  which must be factored-up for gravity and drag losses.  Recommendations are in the figure.

Arriving at Earth has exactly the same values.  If you enter into orbit,  you do it on the side where orbit motion is posigrade,  so that the delta vee is:

               dV = Vint – Vorbit

If you enter the atmosphere for aerobraking,  your entry interface speed is Vint.  Depending upon the design of your spacecraft,  there may or may not be a touchdown burn.  If there is,  it is some terminal speed to “kill”.  I recommend factoring that up by 1.5 for hover and maneuver allowances.

Faster Trajectories

There is no such thing as a “direct flight to Mars”.  All fast trajectories are still elliptical about the sun,  unless your spacecraft propulsion is capable of far greater than solar escape speed.  None are,  at this time in history.

Faster trajectories reduce the travel time at the expense of higher required delta-vees.  There is no way around that bit of physics.  What you want to do is employ the higher-speed end of your transfer ellipse as your trajectory,  and arrive at your destination before you reach the lower speed portion of your ellipse.  Thus for Earth-Mars,  your transfer ellipse perihelion will still be at Earth’s orbital distance,  while your transfer ellipse apohelion will be well beyond the orbit of Mars.

There is no point to putting the transfer ellipse perihelion inward of Earth’s orbit,  because that moves your trip segment towards the slower end of your transfer ellipse.  You would thus average lower speeds over about the same path length. 

For any such faster transfer ellipse,  the situation is as depicted in Figure 5.  Your perihelion velocity is higher,  so your departure delta vee is higher.  But the same departure Vbo and Vfar calculations apply,  as for the Hohmann ellipse. 

You will “get off” the transfer ellipse when your distance from the sun is at Mars’s distance.  You need to time your arrival to be just as Mars gets there.  It will be a real vector subtraction to determine your velocity with respect to Mars at arrival.  Its magnitude is your Vfar.  The geometry of closure with the planet is more complicated because of the nonparallel angle between the vectors to be subtracted.  Even so,  just use the “kinetic energy thing” on Vfar and Vesc to get Vint.  From there,  it’s the same basic choices for orbital entry or direct landing,  even though the detailed geometries are changed.  

 Figure 5 – Faster Ellipse Trajectories

Departing Mars is the reverse.  Whether from orbit or direct from the surface,  you will end up at Vbo near the planet.  The “kinetic energy thing” gets you Vfar.  That speed and an appropriate direction must add vectorially with the planet’s orbital velocity vector,  to obtain the velocity vector you want for the return trajectory (correct magnitude,  and direction tangent to the transfer ellipse path).  You have to time this such that Earth will be at your perihelion point when you get there.

The easiest way to get the angles for the vector additions is to just plot the transfer ellipse,  and a circle at the Mars distance.  Draw the tangents where they cross,  and measure the angle “a” between them with a protractor.  Otherwise,  when evaluating the points on the trajectory,  compute the slope at the encounter point numerically.  The tan-1(slope) is numerically its angle “a1” below reference,  where reference is a line parallel to the semi-major axis (a negative angle on the perihelion half of the ellipse,  and positive on the apohelion side). 

The encounter coordinates give you the angle a3 of the radius vector at encounter as the value tan-1(y/(c-x)).  The circle approximation for Mars’s orbit through the encounter point has a tangent normal to that radius vector.  Its angle below reference “a2” is 90o-radius vector angle a3.  The difference in the angles is the angle “a” between the velocity vectors.  See Figure 6.  

 Figure 6 – Vector Geometries at Encounter

The easiest way to get the time from perihelion to the Mars encounter point is (again) to plot the transfer ellipse and a circle at the Mars orbit distance.  Bound this with the semi-major axis,  and with the radius vector from the sun to the Mars encounter point.  Then use a planimeter to measure the swept area. 

The area of the entire ellipse corresponds to the period of the whole transfer orbit.  Or,  if desired,  half the area of the ellipse corresponds to the travel time from perihelion to apohelion.  The ratio of your planimeter area swept for the trip,  to the ellipse area,  is the same as the ratio of 1-way trip time to orbital period.  Or if ratioed to half the ellipse area,  the ratio of travel time to one-way trip time.

If you don’t have a planimeter with which to measure areas on your plot,  then integrate numerically the area under the ellipse curve (relative to semi-major axis) from the perihelion point to the encounter point.  Add or subtract as appropriate the area of the right triangle formed by the encounter vector from the sun to Mars as its hypotenuse.  A spreadsheet would work for this.  (If you have values for a and b,  then you have the equation for the ellipse,  by which to generate coordinate values for x and y.)

Trajectories to Venus or Mercury

Trips to Venus or Mercury work almost the same way as trips outbound to Mars or further.  The difference is that the transfer perihelion is at the destination,  not at Earth.  This is shown in Figure 7.  If Hohmann min energy,  then the apohelion is at Earth’s orbit.  If a faster trajectory,  the apohelion is outward from Earth’s orbit.  

 Figure 7 – Hohmann-and-Faster Ellipses to Venus or Mercury

Quite frankly,  as fast as these inward-from-Earth trips are with Hohmann min energy ellipses,  there seems little point to the added complications of a faster trajectory.  Venus is only 143 to 149 days away,  and Mercury is only 95 to 117 days,  using min energy Hohmann trajectories.  Compare that with Mars:  235 to 283 days away,  and Ceres-as-typical-of-the-asteroid-belt at 428 to 517 days away.   Times outward of Earth are longer simply because the distances are larger,  and the velocities are lower.

Reference Data for Solar System Bodies

The universal gravitation constant for Newtonian gravity is G = 6.6732E-11 N-m2/kg2.  The masses and radius data for some selected bodies are as follows:

Body     mass, kg     eq.R, km    avg.R, km
Sun        1.991E30   695950      695950
Earth     5.979E24   6378.5       6371.3
Mars      6.418E23   3386          3380
moon     7.354E22   1738.7      1738.3
Phobos   2.72E16     11.3           10.4

Basic orbital data for selected bodies are as follows:

Body      rper, km                rapo, km
Earth     1.4707E8            1.5207E8
Mars     2.0656E8            2.4912E8
Moon    363,299               405,506
Phobos  a = 9408 km --------------

Specific Case Study Numbers for Trips from Earth to Mars

For purposes of typical results,  I presume that both Earth and Mars are at their average distances from the sun,  meaning the radii to their orbits are their “a” values.  I also set the transfer orbit perihelion distance at the Earth orbit distance (in this case its value of “a”) for this study.  What changes is the transfer orbit apohelion distance

I ran 4 values of transfer orbit apohelion distance:  (1) Hohmann transfer at Mars “a” = 2.28E8 km for comparison,  (2) 3.21E8 km to get a 2-year orbit period so that a free return to Earth is possible,  (3) 4.00E8 km near the inner edge of the asteroid belt,  and (4) 4.671E8 km to get a 3-year orbital period so that a free return to Earth is possible from an orbit whose apohelion is well within the asteroid belt.

For all cases,  departure is from low Earth orbit (LEO).  Arrival at Mars could be any of 3 cases:  (1) direct aerobraking entry leading to a landing,  (2) entry into low Mars orbit (LMO),  or (3) rendezvous with and touchdown upon Phobos,  Mars’s inner moon.  Once Vfar is determined for each of the transfer orbit cases,  then each arrival sub-case must be analyzed separately. 

Finding the encounter point requires solving the equations of the ellipse and the circle models simultaneously,  unless one does this graphically.  The ellipse is centered at the origin of the x-y coordinates,  the circle is not (being centered at the positive-x focus where the central body is located,  in all the figures above depicting this).

Now,  for the transfer ellipse,  the semi-major (a) and semi-minor (b) axis distances,  and the distance to the foci (c),  are all known.  Its center is the origin (0,0).  Its equation is:

               x2/a2 + y2/b2 = 1

The circle is offset from the origin,  centered at (c,0) to the right of that origin,  with a radius equal to the Mars orbit distance (call it R).  Its equation is therefore:

               (x – c)2 + y2 = R2

Solving the circle equation for the y2 term gets us something we can substitute into the ellipse equation,  getting us one equation in one variable (x) that we can solve:

               x2/a2 + [R2 – (x – c)2]/b2 = 1               substitution into ellipse to eliminate y2

We have to expand the squared binomial in the second term,  and then distribute the 1/b2 coefficient,  followed by collection of like terms in x2 and x:

               x2/a2 + [R2 – (x2 - 2xc + c2)]/b2 = 1
               x2/a2 + [R2 – x2 + 2xc – c2]/b2 = 1
               x2/a2 + R2/b2 – x2/b2 + 2xc/b2 – c2/b2 = 1
               (1/a2 – 1/b2)x2 + (2c/b2)x + (R2/b2 – c2/b2 -1) = 0

That result is a quadratic equation in standard form Ax2 + Bx + C = 0,  where:

               A = 1/a2 – 1/b2
               B = 2c/b2
               C = R2/b2 – c2/b2 -1

for which the most convenient solution is by means of the quadratic formula:

 x = -B/2A +/- (D^0.5)/2A,  where D is the discriminant D = B2 – 4AC. 

For there to be one and only one x solution,  the discriminant must be zero,  so that x = -B/2A.  If the discriminant is positive,  there are two real solutions for x per the formula.  If the discriminant is negative,  there are no real-number solutions at all. 

Once we have values for solution x,  the corresponding y coordinates can be determined from either the ellipse equation or the circle equation (we are interested in the positive-y roots for the arrival encounter;  the negative-y roots correspond to the departure point):

               y = +/- [R2 – (x – c)2]0.5                    from the circle equation
               y = +/- [b2(1 - x2/a2)]0.5                   from the ellipse equation    

Once the x-location (along the semi-major axis) is known,  we can integrate numerically under the ellipse curve (relative to the semi-major axis) from the solution x to the perihelion x value.  The area of half the ellipse (to one side of the semi-major axis) is 0.5*pi*a*b.  There is a triangle formed by the radius vector to encounter:  its height is the y coordinate at encounter.  Its base is the focus length c minus the x coordinate.  Thus the triangle area is 0.5*y*(c-x).  To create the area swept by the radius vector,  the area of this triangle subtracts from the integral area,  as long as the encounter x is less than c.  The swept area factor SAF is the swept area divided by the ellipse half area.  This area ratio applies to half the orbit period,  for the one-way trip time.

I used a spreadsheet to do this analysis,  supplemented by hand plots of the orbits to ensure the calculations were getting the right answers.  This process required multiple iterations before I got it “right”.  Figure 8 shows the basic transfer orbit-related data that are independent of the exact nature of Mars arrival.  The basic orbital parameters a,  b,  c are given (blue-highlighted values expressed as Mkm (millions of km) have the most significant figures).  Average,  perihelion,  and apohelion velocities (with respect to the sun) are given in km/s.  The period and half-period values are shown,  with the half-period shown in seconds,  days,  and months.  

 Figure 8 – Transfer Orbits Descriptive Data

Also included in the figure are the basics of Earth departure velocities and Mars arrival velocities.  All the Earth departure velocities are tangent to both the transfer orbit and Earth’s orbit,  since the transfer perihelion is always at Earth’s orbit.  The scalar difference between perihelion velocity and Earth’s average orbital velocity is the Vinf value,  typical of “far from Earth” velocity needs,  and measured with respect to Earth.   Adjusted for the effects of Earth’s gravity to a “near Earth” value,  this produces the Vbo values with respect to Earth. 

Mars arrival is a little more complicated,  since the velocities must add vectorially for all but the Hohmann min energy transfer case.  In the middle group in the figure,  the angle “a” is that between the velocity of the spacecraft “V” in its transfer orbit at Mars encounter,  and the velocity vector of Mars that is tangent to its orbit.  Only for the Hohmann case is this angle zero. 

The velocity “far from Mars” with respect to Mars is the velocity vector V at angle a relative to Mars’s vector,  minus Mars’s velocity vector.  This is the Mars arrival Vinf in the figure.  I did not include all the spreadsheet details of computing that angle,  but my hand plots showed that I was indeed computing the correct values.  Adjusted for Mars’s gravitational attraction,  this corresponds to a higher Vint “near Mars”. 

The bottom group in the figure relates to the swept-area estimated 1-way trip time.  Again,  I didn’t include all the details of the numerical integration,  but my area estimates were indeed confirmed by the hand-plotted orbits.  The blue highlighted data are the transfer orbit parameters expressed as Mkm,  for the most significant figures. 

The principal results are plotted versus transfer orbit apohelion distance in Figure 9.  These include the half-period of the orbit,  the “near Earth” departure requirement Vbo with respect to Earth,  the “near Mars” arrival speed Vint with respect to Mars,  and the 1-way trip time.  

 Figure 9 – Principal Results Plots

Note that the 1-way trip time of 8.62 months and the half-period of the transfer orbit (8.62 months) are identical for the Hohmann transfer case,  where the apohelion distance is the average orbital distance of Mars.  This requires 11.57 km/s achieved burnout speed to depart,  and arrives close to Mars at some 5.69 km/s,  more-or less lined up tangent to Mars’s orbit. 

The next faster case investigated has a half-period of 12.00 months,  for a full round trip of exactly 2 years.  If the Mars encounter were to fail,  the spacecraft would arrive back at Earth’s orbit just as Earth got there.  This offers the possibility of a free-return abort,  if 2 years in space is tolerable.  Otherwise,  the 1-way trip to Mars is much faster at 4.26 months.  It costs more:  the Earth departure burnout requirement is 12.26 km/s,  and the near-Mars encounter velocity is some 7.40 km/s,  skewed off-tangent at about 34-35 degrees.

The third case used an even 400 million km apohelion distance.  Its total transfer orbit period is 30.3 months,  which is nonresonant with Earth’s period about the sun.  There is no free-return abort using this orbit!  The 1-way trip time is really fast at 3.67 months,  but this costs quite a bit.  The required departure burnout speed is 12.77 km/s,  and the near-Mars encounter velocity is some 7.36 km/s,  skewed about 38 degrees off tangential.  This apohelion is actually into the inner edge of the main asteroid belt.

The final case has a half-period of 18.0 months,  or a full round-trip period of 3 years,  resonant with Earth.  This offers the possibility of a free-return abort,  if 3 years in space is tolerable.  The 1-way trip time is the shortest of the cases investigated,  at 3.40 months.  The cost is high:  13.14 km/s at departure burnout,  and a near-Mars encounter speed of 6.53 km/s to deal with.  That last is skewed about 33-34 degrees off tangential. 

Most Mars mission designs will be leaving from orbit about the Earth.  The Spacex Starship is one of those.  Low circular Earth orbit (LEO) has a speed about the Earth of about 7.9 km/s,  departure starts from there and must reach “Vbo”.  At Mars,  the choices are (1) direct entry and retropropulsive touchdown,  (2) entry into low Mars orbit (LMO) with possibly a separate vehicle to deorbit and enter for a retropropulsive landing,  and (3) entry into Phobos’s orbit and touching down propulsively on Phobos.  For that last,  I simply applied a 1.5 factor for maneuver and hover to the escape speed from Phobos,  to estimate a mass ratio-effective delta-vee requirement.

A few of my estimation items are simply assumptions.  I assumed that any of the transits (both outbound and inbound) have a course correction allowance of dV = 0.5 km/s.  I also assumed that any vehicle returning to LMO must rendezvous with another vehicle,  necessitating a rendezvous allowance.  I assumed that allowance to be dV = 0.1 km/s.  Anything deorbiting from LMO to land upon Mars requires an allowance for a deorbit burn.  For this I used dV = 0.05 km/s.  Finally,  I assumed the terminal speed after Mars entry to be about a Mach number,  leading to an estimate of the speed to be “killed” (and about 0.5 Mach on Earth).

A summary of the detailed transit and terminus estimates is given in Figure 10.  Note that the transfer ellipse influences departures and arrivals,  but nothing else.  Delta-vee information is highlighted blue,  and entry speeds for aerobraking are highlighted green.

 Figure 10 – Details of the Various Transit and Terminus Estimates (see Update 11-24-19)

This same information is rearranged into groups representing each kind of overall mission,  with the data parametric upon the transfer trajectory used.  The presumption is that the return to Earth uses the same transfer ellipse trajectory as the outbound trip to Mars. 

Figure 11 shows the velocity requirements (blue) and entry interface speeds (green) for a mission that departs Earth orbit,  makes a direct entry and landing upon Mars,  then makes a direct escape from Mars,  leading to a direct entry and landing at Earth.  Touchdowns are assumed retropropulsive.  There are many mission designs which might use this architecture.  The most notable recent example is Spacex’s proposed “Starship”. 

Figure 12 shows the velocity requirements (blue) and entry speeds (green) for a typical orbit-to-orbit mission to Mars.  This is broken down into a transit velocity requirement,  and a separate landing velocity requirement,  since it is likely the lander is a separate vehicle.  Similarly,  the takeoff is separate from the transit home.  Earth return is to LEO,  not a landing.  That is presumed to be a different vehicle.

 Figure 11 – Summary for the Direct Landing and Direct Earth Return Mission

 Figure 12 – Summary for the Orbit-to-Orbit Mission

The Phobos (only) visit is shown in Figure 13.  Because the landing allowance is so small,  it is presumed the transit vehicle also makes the landing.  Transit is from LEO to Phobos orbit,  and from Phobos orbit to LEO for the return.  There is no aerobraking in this scenario.  Upon return to LEO,  it is presumed that some other vehicle is used for the final return-to-Earth landing.

 Figure 13 – Summary for the Phobos (Only) Mission (see Update 11-24-19 instead!)

As mentioned above,  I utilized by-hand plots (pencil and paper) to verify correct calculation of many items.  I did not need such a plot for the Hohmann min-energy transit,  but I did for the 3 faster trajectories.  The following two figures are photographs made of those working hand plots,  two plots per page.  Note that I mis-plotted one of them and had to try again.  

 Figure 14 – Photo of Hand-Plotted Orbits Made for Verification,  part 1

Figure 15 – Photo of Hand-Plotted Orbits Made for Verification,  part 2


Update 11-24-19:  Revised Delta-Vee Estimates for Phobos Mission

I have corrected an error in estimating Vinf for the Phobos misson.  I used the surface escape speed 5 km/s for the conservation of mechanical energy estimate,  when I should have used the escape velocity out at Phobos's orbital distance,  some 3 km/s.  This reduces the Vnear values substantially,  thus reducing the estimates for delta-vee required.  This changes Figure 13 entirely,  and the Phobos Departure/Arrival details in Figure 10.  The revised data are given in Figure 16 below. 

Figure 16 -- Revised Estimates for the Phobos Mission


Tuesday, October 22, 2019

Reverse-Engineering the 2019 Version of the Spacex “Starship” / ”Super Heavy” Design

Multiple Updates planned (dates not defined),  appended to bottom as completed (dates defined):

Update 10-23-19:  Change the Stagepoint
Update 10-24-19:  Going to Mars and the moon
Update 10-26-19:  Return from Mars
Update 10-28-19:  To the Moon from Elliptical Orbit
Update 10-29-19:  Effects of Price Upon Elliptical Departure
Update 10-30-19:  Soil Bearing Loads Estimates
Update 10-31-19:  Static Overturn Stability
Update 11-21-19:  Choice of Landing Engines


This design concept has changed dramatically since Elon Musk’s first official “reveal” at a presentation in 2017.  It will still dramatically change as flight testing gets underway with the various “prototypes” that have been,  and will be,  built.  As of this writing,  none but a very generic test vehicle of limited scope called “Starhopper” has yet actually flown,  in some very limited flight tests.

That being said,  Musk has recently revealed the most realistic design concept yet for this vehicle in 2019.  It actually resembles in its overall shape the test vehicles being constructed for flight tests out of Boca Chica,  Texas,  and Cape Canaveral,  Florida.  The entry heat protection isn’t included yet in the test vehicles.   Presumably,  the inboard profiles of these test vehicles are reasonably representative in terms of propellant tankage,  but likely very little else.  Such is the nature of experimental flight test.

The design is first and foremost a 100%-reusable two stage system for launching payload to low Earth orbit (LEO).  Instead of a disposable second stage and separate payload shroud,  the innovation here is to make the entire second stage a reusable spacecraft named “Starship” that is capable of re-entry and landing,  and to contain the actual payload within it,  much the same way as cargo is contained within ships at sea,  or inside air freighters.  This vehicle uses engines that burn methane and liquid oxygen. 

The big first stage “Super Heavy” booster is a scale-up of the Falcon booster core idea,  refitted with higher-performing engines that burn methane and liquid oxygen instead of the kerosene and liquid oxygen that the Falcon stages used.  These are in fact the same basic engines the second stage uses,  a design named “Raptor”.  This first stage booster conducts a post-staging burn to reverse course,  a second burn to control supersonic entry into the atmosphere,  and a third and final burn to land at the site from which it was launched,  exactly like the smaller Falcon cores.

This concept has been through many fundamental design changes.  Initially,  it was a 10 meter diameter system,  but in the last couple of presentations,  has converged upon a 9 meter diameter system.   Initially,  its basic structure was supposed to be carbon composite,  but this last iteration has settled upon stainless steel construction instead,  which is a little heavier,  but far more robust in terms of resisting a multiplicity of cryogenic propellant effects,  as well as the high-temperature effects during re-entry.  Although heavier,  it resists heat far better than the aluminum construction used in the Falcons,  in turn better than organic carbon composite at resisting heat,  and by far.

A two-layer liquid-cooled heat protection scheme with this stainless steel construction,  has been abandoned in favor of a simple single shell with an ablative heat shield on the windward surfaces,  and bare metal radiative cooling on lee-side surfaces.  Landing pads at the tips of 3 or 4 fins have been abandoned in favor of 6 landing legs that extend right from the base of the vehicle,  not from any fins or wings at all.  Whether that is the “right” choice remains to be seen.

The first stage now features 6 tip-mounted landing pads on the tips of 6 small fins at the rear of the stage.  It also features the same 4 grid fins at the forward end of the stage,  just larger,  that provided effective aerodynamic control forces after entry,  with the Falcon stages. 

Whether these landing leg and small size landing-pad designs are truly appropriate remains to be seen.  They are certainly appropriate for landing upon a thick reinforced-concrete pad.  Or possibly upon a level,  flat,  thick,  solid slab of rock.  It is unlikely in the extreme that this kind of landing pad design is appropriate for landing upon soft soils of various types.  Such soft soils are inherent for off-site launch abort landings on Earth,  as well as the great bulk of possible landing sites on the moon or Mars. 

Data Sources

I got the latest data from two primary sources:  (1) Spacex’s own website,  which now for the very first time documents some of the expected values for the “Starship”/”Super Heavy” design,  and (2) Wikipedia articles for “Starship”,  “Super Heavy”,  and the “Raptor” engines that are supposed to power them. 

Beyond this,  I have already used Spacex and Wikipedia data to reverse-engineer the expected performance parameters for the methane-liquid oxygen “Raptor” engines.  This includes both the sea level version,  and the vacuum-only version,  which uses the same powerhead,  just a larger expansion bell. 

“Raptor” Engines

My engine performance reverse engineering is documented here on my “exrocketman” site, as the article titled “Reverse-Engineered “Raptor” Engine Performance” and dated “26 September 2019”.    Readers should be aware that my reverse engineering of the “Raptor” engine was done for lower values of r (the oxidizer-to-fuel mass flow rate ratio) than Spacex currently claims. 

What that means is that actual “Raptor” performance may be very slightly different than what I estimated from my old 1969-vintage data.  That being said,  as near as I can tell,  they (Spacex) have yet to demonstrate the full claimed chamber pressure (of 4400 psia) in actual tests so far!  So there are some very real uncertainties remaining in “Raptor” performance numbers,  as of this writing! 

Nevertheless,  my estimated data pertain to engines that really do achieve the 4400 psia chamber pressure.  I actually got about a second or two higher specific impulse than Spacex is claiming.  Be aware that the data on the Spacex website showing 330 sec Isp at sea level and 380 sec Isp in vacuum,  is for two different versions of the design:  the sea level short bell,  and the vacuum long bell.   These share the same powerhead.  All the pump drive gas eventually goes through the nozzle.  I assumed 5:1 throttle-down ratio on chamber pressure,  so that the min Pc = 880 psia.  Be also aware that Isp is reduced when throttled down (my reverse-engineering of “Raptor” article gives those numbers).

Vehicle Weights and Loadings

What is shown on the Spacex website are payload masses and propellant capacities,  but not inert masses or launch masses.  Thrust values are shown,  but these are rounded to only two significant figures.  The Wikipedia articles actually give data usable on weight statements,  particularly the inert and launch masses.  Spacex now shows a 9 meter diameter,  with 50 meter length for the “Starship” second stage,  and 68 meter length for the “Super Heavy” first stage booster. 

The inert mass and propellant capacity of the Starship second stage spacecraft have increased from the values publicized in 2018.  Inert mass of a carbon composite structure was widely said to be 85 metric tons.  With the switch to stainless steel hull construction,  this is now unsurprisingly larger at 120 metric tons.  Plus,  propellant capacity has grown from 1100 metric tons to 1200 metric tons.  Payload is quoted as “100+ metric tons”,  with the goal being 150 metric tons,  or perhaps even more.  The ignition mass quoted on Wikipedia is for reduced propellant and only 100 tons of payload. 

Ignition mass (“Starship” alone) given as 1320 m.tons,  which implies this weight statement:

The inert mass and propellant capacity of the “Super Heavy” first stage is quoted as 230 metric tons and 3300 metric tons,  respectively.  The ignition mass quoted on Wikipedia is for the stage alone,  without the Starship upper stage.  Thus the weight statement is different when you include the upper stage.

These numbers were the startpoint for a spreadsheet-based study of launch to LEO and return.  In that study,  payload up was allowed to be different from payload down. 

Spreadsheet Study for LEO Operations

The basic assumptions and analysis conditions used in this study are summarized in Figure 1,  with numerical values shown.  The most critical thing was picking the actual staging point.  Nominally,  most two stage vehicles stage at roughly 3 km/s and about 50+ km altitude.  Trajectory has largely bent over to not quite horizontal at staging.  For this study,  I used a staging velocity of 2.8 km/s,  at “exoatmospheric altitude”,  and 27 degrees up from horizontal,  which sets the vertical and horizontal velocity components at staging.  All the aerodynamic drag loss,  and most but not all of the gravity loss,  is applied to the first stage “Super Heavy”.  

Figure 1 – Analysis Conditions and Values

Post-staging,  the “Super Heavy” booster immediately flips end-for-end and does the first return burn dead-horizontal,  which kills the downrange velocity but not the vertical velocity.  This burn actually reverses the direction of flight to a modest horizontal return velocity.  

From there the stage arcs high back toward its launch point,  requiring an entry burn to reduce pathwise velocity to about 0.5 km/s as shown,  and extend the grid fins for stability and control.

For touchdown,  the assumed speed is just about Mach 1,  to be “killed” by the final landing burn. That burn delta-vee requirement has the factor 1.5 applied to it,  to cover course and speed adjustments to make the pinpoint landing.

These return burns require a certain amount of propellant,  computed from dry tanks mass without the upper stage payload,  first.   The ascent burn requires a large amount of propellant,  computed from ignition mass with the second stage payload,  second.  For the trajectory (and assumptions) shown,  I was able to increase the “Starship” ignition mass slightly (from its nominal 1320 metric tons to 1322.5 metric tons),  and reconcile the amounts of propellant with a full capacity propellant load.  This is shown in the spreadsheet image of Figure 2.  Inputs are highlighted yellow.  Significant outputs are highlighted blue (or green).  Note that masses got converged to within a few dozen kilograms,  out of hundreds of tons. 

Figure 2 – “Super Heavy” Booster Operation Analysis

The “Starship” second stage cannot use all its propellant just to reach orbit from the staging condition.  It must do a deorbit burn,  and it must do a retropropulsive landing burn.  The landing must kill a “belly-flop” velocity of about half a Mach number,  down at low altitude in the denser air.  I used factor 1.5 on this delta-vee requirement to cover course and speed adjustments to make a pinpoint landing. 

The spreadsheet is set up to use different tonnage values for descent and ascent payload,  but for this analysis,  I made them the same.  I was able to increase the payload figure significantly,  while decreasing the propellant load figure,  and still reconcile the propellant quantities.  The variation in payload and propellant amounts was made at the fixed ignition mass determined from the “Super Heavy” analysis.  You do the landing first,  then the ascent. 

These numbers are the spreadsheet image shown in Figure 3.  

Figure 3 – “Starship” Operation Analysis

What this analysis shows is that there are limits on how heavy “Starship” can be,  and still successfully reach orbit from the staging conditions that “Super Heavy” can reach.  Yet for LEO operations,  “Starship” propellant load can be reduced,  for more-than-minimum cargo.  The number I got was actually larger than the “goal” value for payload. 

Note in the spreadsheet images that I looked at the figures for mass and thrust at some significant points in the flights of the two vehicles.  The resulting kinematic net acceleration levels,  along with thrust/weight values,  are shown in Figure 4

The “Super Heavy” uses all 37 sea level “Raptor” engines during its ascent,   with max gees just under 4 at staging.  Takeoff thrust to weight is a healthy 1.5 factor,  for a net upward kinematic acceleration of half a gee. 

The “Starship” uses the 3 vacuum engines for its ascent,  and for its deorbit burn.  The arrival on orbit is at just over 2 gees.  It uses the 3 sea level “Raptor” engines for its touchdown.  Two could do the job,  but if three are running throttled,  loss of one can be compensated most quickly by throttling up the remaining two.  Throttle levels at thrust equal to weight are in the 50% range,  with 20% the presumed min. 

Figure 4 – Selected Summary Values from the LEO Analyses

Setup for Going Outside LEO

The propellant needed for “Starship” deorbit and touchdown is about 29.89 metric tons.  This would be in the header tanks nested inside the main tanks.  The possibility of abort-to-surface must be considered for any “Starship” sent to LEO,  whether it says there or is refueled and sent elsewhere.  That is why I made the descent payload equal to the ascent payload in my spreadsheet analysis.  So,  one still cannot use all the propellant just to reach LEO for a “Starship” intended to be sent elsewhere.  It’s a flight safety thing.

The limitations of the “Super Heavy” dictate a less-than-capacity propellant load for “Starship” in routine LEO operations.  This was identified in the spreadsheet analysis.  It might be possible to relieve this limitation somewhat by lowering the staging velocity.  That is another analysis out-of-scope here.

That limitation no longer applies,  when refueling a “Starship” on-orbit for a trip elsewhere.  The full propellant load can be used.  That would be the full 1200 metric tons,  less the about 30 tons of return-and-land propellant,  for about 1170 metric tons to be delivered by the tanker flights.  From there,  destinations could be the moon or Mars.  Those analyses are extras to be added later,  as updates.


Update 10-23-19:  Change the Stagepoint

I re-ran the LEO analysis with the intent to match stagepoint vertical velocity with the return velocity such that a max-range 45 degree arc is obtained.  This required reducing both path angle at staging,  and staging velocity.  Doing the trades showed the result was very sensitive to path angle at staging,  and less sensitive to staging velocity.

Therefore,  I reduced the path angle to “get in the ballpark” of a 0.5 km/s vertical velocity to match the intended horizontal 0.5 km/s return velocity.  Then I reduced staging velocity to get the desired result.  I iterated this procedure about 3 times before settling on staging at 2.63 km/s at an 11 degree path angle.  This gives me a booster returning 45 degrees upward,  for a max-range arc trajectory.

Looking at “Super Heavy” performance capability then uses payload mass (ignition mass of the “Starship” upper stage) as my free variable to balance out the propellant demand against the 3300 metric ton propellant capacity from the Spacex website.  What I saw was an increase in the allowable max mass of “Starship,  from 1322.5 metric tons,  to some 1488 metric tons,  favorable indeed!  These results are shown in Figure 5,  an image of the revised Super Heavy spreadsheet analysis.  It still lands using 3 engines throttled back to no less than about 38% thrust. 

Figure 5 – “Super Heavy” Results for Revised Stage Point

This increase in the “Starship” ignition mass is partly used in the larger demanded delta-vee from stagepoint to orbit,  but it also allows a larger payload,  and a larger propellant load relative to the Spacex-stated capacity.  In the spreadsheet,  the balancing variable is payload,  with ignition mass set by the “Super Heavy” result.  This revised stagepoint is quite beneficial,  as can be seen in the “Starship” results of Figure 6.  

Figure 6 – “Starship” Results for Revised Stage Point

In addition to carrying more payload (now slightly over 200 metric tons),  the kinematic acceleration levels just after staging are significantly better,  being about 0.27 gee instead of the earlier 0.07 gee.  That is the effect of reducing trajectory path angle at staging!  We still land on 3 sea level “Raptors” throttled to about 50% thrust as a min value. 

For the next update,  traveling outside LEO after refueling on-orbit,  the analysis can be theoretically be simplified considerably,  by carrying the same payload value both ways.  That way,  all the delta-vees are simply factored appropriately and summed to a total mass ratio-effective delta-vee value.  Reducing return payload is the way to try to address shortfalls in available mass ratio.

A caveat here:  these numbers that get depend very fundamentally upon characteristics of the “Starship” and the “Super Heavy” that simply are not yet known with certainty.  I base this stuff on projections,  not facts!  Small changes in things like inert mass have a huge effect.  These things will NOT be known with certainly until a lot of test flights and vehicle changes have been made! 


Update 10-24-19:  Going to Mars and the moon

For Mars,  the trip is one-way,  unless propellant can be produced locally on Mars,  in quantities and at rates,  that are sufficient to support the return trip.  For the moon,  the mission is presumed to be two-way and unrefueled,  and only by direct landing on,  and departure from,  the moon. 

This lunar scenario proved too limited to carry significant payload to the moon,  or to carry the same payload two ways,  which explains why Musk’s recent presentations have referenced launching moon missions from an elliptical orbit instead of a circular LEO.  The higher perigee velocity of the elliptical orbit reduces the delta-vee onto the transfer orbit to the moon.  But this makes getting the “Starship” and its payload all the way to orbit harder (reducing max feasible payload),  because the perigee velocity is a higher velocity to reach from the stage point velocity. 

For both cases reported here (Mars and the moon),  the vehicle payload deliverable to LEO is based on what the “Super Heavy” and “Starship” can successfully send to a circular LEO,  with the revised stage-point data that increased allowable ignition masses for the “Starship”.  Any elliptical orbit effects will have to be the subject of a future update. 


It turned out to be possible to send a very small payload to low Mars orbit (LMO),  with enough propellant still onboard to deorbit,  and to make the landing on Mars.  The only feasible option for heavy payloads to Mars was the direct entry trajectory straight off the interplanetary trajectory.  This avoids the large burn otherwise required to enter LMO.   But it also means you CANNOT abort the landing,  no matter what!  And that includes the surprise occurrence of giant planetary dust storms!

An image of the spreadsheet I used for Mars is given in Figure 7.  A cartoon of what could be done is given in Figure 8.  Scope includes both direct landing and stopping in low Mars orbit before landing. 

Bear in mind that this analysis is based upon a “Starship” fully-refueled by tankers on-orbit to the full 1200 metric tons of capacity.  It is also based upon a spread of payloads up to and including the max payload “Starship” can carry to a circular LEO,  as boosted by “Super Heavy”,  with the reduced stage point velocity and path angle already investigated above.

I have not investigated what those tankers might really be able to accomplish.  That would be the subject of another update.  Bear in mind that Spacex has published next-to-nothing about the tankers!  If the cargo/crew “Starship” is used for a tanker at a nominal 200 metric tons of payload in extra tanks instead of the regular cargo/cabin section,  then 6 such flights could deliver 1200 metric tons of propellant to a circular LEO. 

Note also that I have included a (wild guess !!!) course correction payload allowance,  as well as factoring-up the landing delta-vee for pinpoint landing needs,  up to and including brief hover. This was an attempt to be as realistic as possible in these estimates.

 Figure 7 – Both Mars Options,  as Analyzed in the Spreadsheet

What I got was all the payloads up to the max-orbited 203 metric tons are indeed feasible for direct landing on Mars,  as long as the vehicle was filled to max propellant capacity in LEO.  Bear in mind that my transfer trajectory was a minimum-energy Hohmann transfer orbit,  not something faster!  Depending upon where Earth and Mars are in their orbits vs perihelion or aphelion,  the one-way flight times are never less than 8 months,  and usually just short of 9 full months.  The max is about 9.5 months.

The smaller payloads arrive with significant unburned propellant (seen in the spreadsheet as unused delta-vee capability).  This could be used for some degree of faster transfer trajectory.  The big payloads really cannot be flown faster this way.  Too little is left over.

 Figure 8 – Mars Results in Cartoon Form

I was surprised to find that a fully-fueled “Starship” with the smallest payload I considered (50 metric tons) actually had enough delta-vee capability to make the burn to enter LMO,  and still have enough propellant to deorbit,  and land.  Bear in mind this is ONLY for a min-energy Hohmann transfer to Mars.  I did not investigate whether this could be done at 55 tons,  because the delta-vee margin was already quite low at 50 tons.  This orbit entry would be the only way to abort a landing in a dust storm,  but the payload capability available when doing this,  is just not very attractive at all!

Once again,  be fully aware that going to Mars with “Starship” is a one-way trip,  without adequate propellant production capability on Mars.  That is both adequate produced quantities,  and production at appropriate rates,  of methane and oxygen,  all in the required ratio. 

Bear in mind that the power and machinery requirements for this have considerable scope.  Not only must there be methane production,  but there must also be local water electrolysis to produce the oxygen,  and to produce the hydrogen needed to make the methane.  There must also be some way to compress carbon dioxide from that near-vacuum-of-an-atmosphere,  to make that methane. 

Beyond that,  there must be the machinery to liquify the methane and the oxygen.  Plus,  there must be the machinery to sub-cool it to the density required to put up to 1200 metric tons inside the tankage volume of “Starship”.  On Earth,  that stuff is big and heavy.  And very hungry for electric power!

And finally,  there must be some more machinery to keep such propellant cool enough not to boil away,  over very extended periods of time.  Approximately 2,  to maybe 6 years,  depending upon how often return flights are to be made.

It's not “just a Sabatier reactor”!

               The Moon

Going to the moon is different,  that being an unrefueled return requirement.  Again,  be aware that I analyzed the case of departing from a circular LEO.  Spacex has already said their “Starship” must use a more elliptical orbit about the Earth  in order to do the moon mission.

I could not use the same payload on the return voyage,  as on the voyage to the moon.  That proved infeasible.  I investigated a list of payloads to the moon,  resetting vehicle mass statements appropriately in response to an input return payload in the spreadsheet.  I set that return payload to approximately 10 metric tons,  corresponding to 10 persons,  with a not-just-life support allowance of 1 metric ton per person.  That would be the person,  a couple of space suits,  and expendable life support supplies for several days. 

What I found was disappointing,  as the spreadsheet image in Figure 9 shows.  

 Figure 9 – Moon Mission Analysis Results From the Spreadsheet

From a circular LEO departure orbit,  my numbers say that a fully-fueled “Starship” cannot deliver more than about 27 metric tons of payload to the moon,  starting from that circular LEO departure.  The 10 ton return payload is transporting 10 people home,  all in the one mission.  The “eob-drytnk” results highlighted blue in the figure are delta-vee-difference numbers in km/s units.  Negative numbers are conclusively indicative of an infeasible shortfall.

The cartoon in Figure 10 shows the highlights of these results,  including the trajectory concept,  which is NOT the figure-eight trajectory Apollo used to reach lunar orbit!  Bear in mind that transfer orbit apogee velocity is less than the moon’s orbital speed about the Earth.  The moon literally “runs over” the spacecraft from behind.

If your intent is lunar orbit,  it will be retrograde.  The Apollo “figure-eight” trajectory reflects this,  in the form of a computer-analyzed 3-body problem.  If you are landing direct,  retrograde orbit is irrelevant. 

Figure 10 – Results for “Starship” Lunar Landings

Possible future updates would include both (1) the elliptic orbit departure for lunar missions,  and (2) the return trip home from Mars to a direct entry and landing at Earth.  Elliptic orbit departure implies a much higher apogee.  If this falls within the Van Allen belts,  then either no crew can be aboard,  or else there must be considerable radiation shielding available. 

There is also the unaddressed issue of making rough-field landings on the moon or Mars.  This gets complicated by the far larger weights on Mars of the fully-fueled vehicle ready to launch on the return.  That issue is not a problem for the moon,  since the vehicle is not refueled there.  Even so,  it seems unlikely that sufficient attention has been paid to exerted bearing pressures being within the allowable safe bearing pressures of soils thought to be representative of the bulk of both Mars and the moon.  Excess bearing pressures lead to landing pads sink into the soil,  which will happen unevenly.  That is a problem for takeoff due to the extra friction holding the ship down.  That is worthy of another update.

There is also the issue of static overturn stability of a landed “Starship” on the moon or Mars.  This presumes the landing pads do NOT sink into the soil,  which if it happened,  would happen unevenly.  This is more of a must-fall-within-the-footprint problem for the weight vector,  something long-established in classical freshman engineering mechanics.  This is also worthy of an update,  addressing both general land slope,  and localized roughness. 


For this analysis,  I assumed (1) full capacity propellant load of 1200 metric tons,  (2) Hohmann min-energy trajectory home,  (3) no prior constraint on return payload,  and (4) 3 sea level and 3 vacuum engines.  The ascent from Mars is a direct escape straight onto the transfer orbit,  using the vacuum engines.  I assumed a 0.5 km/s course correction delta-vee,  same as for the trip to Mars.  This is also done with the vacuum engines.

At Earth,  entry is direct from the interplanetary trajectory,  with aerobraking to about Mach 3 at about 45 km altitude.  From there,  it belly-flops broadside to the relative wind,  finally decelerating to about half a Mach number at low altitude.  Just prior to the end,  the  vehicle turns tail-first for the retro-propulsive touchdown.  It must use the sea level engines for this!

For the Mars ascent,  gravity losses are the Earthly 5% ratioed down by Mars’s surface gravity (0.384 gee).  Aerodynamic drag losses are the Earthly 5% ratioed by the surface density ratio to Earthly standard (0.007).  The net ascent dV factor is then just under 1.02.  The touchdown velocity requirement is factored up by 1.5,  as is my usual practice,  to cover pinpoint landing needs.

I split the spreadsheet image into two figures (see Figures 11 and 12),  followed by a summary cartoon of the trip home from Mars (see Figure 13).  What I found was that under these assumptions,  the max payload sendable home is 162.6 metric tons (assuming a full-capacity 1200 metric tons of propellant at launch from Mars).  This also assumes course changes up to the full 0.5 km/s delta-vee get made.  

 Figure 11 – Spreadsheet Analysis for the Return Voyage,  Part 1

 Figure 12 -- Spreadsheet Analysis for the Return Voyage,  Part 2

Figure 13 – Illustration of the Salient Features of the Return Voyage

For the feasible payload sizes,  the 3 vacuum Raptors operated at full thrust are adequate to launch the fully loaded and fueled “Starship” from the surface of Mars (thrust/weight at or exceeding 1.2).  It is not possible to make the ascent with one vacuum engine “out” as the thrust is less than the Mars ignition weight of the vehicle.  One of the sea level engines must be fired up to replace the lost vacuum engine.  The sea level engine will operate in vacuum,  just at slightly reduced vacuum thrust and Isp relative to the vacuum engine designs. 

This way to handle an engine-out situation is included in the Figure 11 data.  The slightly-reduced Isp does eat into the propellant margins (0.5 km/s worth of course correction,  and the factor 1.5 applied to the touchdown burn).  It would be wise to reduce the max payload limit to compensate (but I did not determine that here).

Figure 12 takes this through the aerobraking and retro-propulsive landing on Earth.  Here the sea level engines must be used,  with no backup from the vacuum engines.  The reverse-engineering I did on the vacuum Raptor indicated backpressure-induced flow separation in the nozzle,  at about 20,000 feet altitude at full thrust,  and at about 55,000 feet altitude at min thrust (20%).  However,  at this point in the voyage,  the remaining propellant load is small enough that even two sea level engines can do the landing,  with three such running at 40-something percent of thrust.

The illustration in Figure 13 summarizes these results,  but without the details of the thrust/weight and engine-out analyses.  The max payload listed corresponds to a min-energy (slow) trip,  a big course correction,  and landing with essentially-dry tanks on Earth.  It would be wiser to reduce this max payload figure to something in the 150-160 metric ton range.

The remaining-propellant figures in Figure 12 at payloads less than max are what could be used to support a faster trajectory home.  These numbers are significant at 100 and especially 50 metric ton payload values.  (A similar thing applies to the outbound voyage.)

Not analyzed here is the effect of not having a full 1200 metric tons of return propellant available.  That would be catastrophically-significant.  The remaining-propellant figures correspond roughly to how big the shortfall might be,  and still make the voyage. They are not large numbers,  compared to the nominal capacity.  This indicates just how crucial it is to have a propellant production facility on Mars that exceeds requirements in terms of quantities and rates.


The idea behind this effort was to depart from an elliptic orbit with perigee at the usual circular orbit altitude (about 400 km),  and an apogee skirting the inner “edge” of the Van Allen belts (at about 1400 km).  The higher perigee velocity relative to circular orbit velocity is a delta that subtracts directly from the delta-vee required to depart Earth orbit.  The apogee restriction is to eliminate the need for radiation protection during these flight operations.

This is a more difficult orbit for the “Super Heavy” / “Starship” combination to reach from surface launch.  For this analysis,  I presumed the reduced staging velocity and path angle already covered,  which had given the “Super Heavy” a larger throw weight.  For “Starship” to reach the higher perigee velocity of the elliptic orbit,  its payload must reduce at constant “thrown weight”.  What that really means is that there is a tradeoff between the larger payload deliverable to the moon,  and the number of tanker flights needed,  since each can carry somewhat less,  flying to that more-demanding elliptical orbit.

Basic orbital calculations gave me Figure 14,  from which the change in velocity at 400 km was at the very most 0.259 km/s,  given a limited apogee at 1400 km,  right at the inner “edge” of the Van Allen belts.  (That boundary is not a sharp line.)  From this,  I picked a nominal increase of 0.25 km/s perigee velocity for the vehicle rocket equation analyses.   This variation of perigee velocity increase over circular,  versus the resulting apogee altitude,  is depicted in Figure 15.

 Figure 14 – Trading Apogee Altitude Against Increase in Perigee Velocity

 Figure 15 – Depicting the Selected Elliptical Orbit for Lunar Departure

The vehicle spreadsheets look exactly like those used for the LEO operations and the moon mission from circular orbit.  Only the detail numbers change.  Those are not depicted here to save space.  The max payload delivered by “Super Heavy” / ”Starship” to circular LEO was 203 metric tons,  in a “Starship” limited to a 1488 metric ton ignition mass and a less-than-capacity propellant load. 

For this selected elliptic orbit,  the ignition mass limit is the same 1488 metric tons,  but the payload reduces to 179.8 metric tons,  while the propellant load increases closer to capacity at 1188.2 metric tons.  (That capacity is 1200 metric tons.)  Bear in mind that one cannot use 100% of the “Starship” propellant to reach the desired orbit,  no matter what.  The deorbit and landing allowances must be reserved,  to cover the unplanned event of an abort back to Earth,  for any reason at all. 

For circular LEO,  I had estimated 6 tanker flights each able to carry a nominal 200 metric tons of propellant as payload.  Those 6 flights could fully refuel the “Starship” for lunar departure,  all the way to a capacity load of 1200 metric tons.   For the elliptic case,  I figured 7 tanker flights at 171.4 metric tons (falling within the 179.8 metric ton maximum) that could deliver the full 1200 metric tons to refuel the “Starship” for lunar departure.
The flight to the lunar landing has a departure delta-vee reduced by the 0.25 km/s change to perigee velocity.  Mid-course correction and lunar landing requirements are the same as the circular orbit case.  The return flight and Earth landing has exactly the same characteristics as the circular-orbit case,  right down to the reduced 10 ton return payload.  One must chase the required mass ratios through the entire mission,  while “book-keeping” the remaining propellant all the way to final Earth landing.  Negative remaining propellant is a conclusive indication of infeasibility.

For the circular orbit departure case,  my earlier analysis indicated the max payload deliverable to the moon was only 27 metric tons.  Launching instead from the elliptic orbit,  the payload increased to just about 52 metric tons,  although at the “cost” of 7 tanker flights,  not 6.  See Figure 16 for the comparison of the two departure orbits and their effects on payload and number of tanker flights. 

Figure 16 – The Lunar Payload Tradeoff Results

I did not analyze still-higher apogee elliptic orbits precisely because those apogees fall within the Van Allen belts,  making the radiation exposure risk very real indeed.  Should that risk be worth taking,  the trend says that lunar payload increases further,  while the number of tanker flights also increases,  raising cost per ton of payload delivered.   The analysis responds sensitively to this,  since the available perigee velocity change (up to 1 km/s at 6000 km apogee) is a fair fraction of the original departure delta-vee from circular orbit (3.29 km/s).  That would require venturing way into the inner Van Allen belt,  where radiation levels get very high indeed.

As to whether that radiation risk is worth taking,  that would depend upon there being adequate radiation shielding built into the “Starship”.  This is something Mr. Musk claims will be available,  but absolutely no details have been shown publicly.  I cannot yet analyze for that issue. 

But,  assuming that the refueling “Starship” in elliptic orbit needs to stay clear of the Van Allen belts,  then the payloads deliverable to the moon (~50 tons) are rather disappointing compared to what might be delivered to Mars (~200 tons).  This is because both the lunar mission is not refueled on the moon,  and that unrefueled return adds to the effects of the higher overall velocity requirement for a two-way lunar mission (8.475 mass ratio-effective km/s) versus a one-way direct-landing Mars mission (5.4 mass ratio-effective km/s).

In comparison,  the one-way Mars mission via LMO (with only 50 ton payload) has a mass ratio-effective  velocity requirement of 7.25 km/s.  The mass ratio-effective velocity requirement for a direct return from Mars is 6.075 km/s.  All of these figures are for Hohmann min-energy transfer orbits.  Faster trajectories have significantly higher velocity requirements! That topic is not under study here.


Update 10-29-19:  Effects of Price Upon Elliptical Departure

In response to a question posed by reader Rob Davidoff (see comments below),  I checked the numbers as a function of assumed price per launch of “Starship” / “Super Heavy”.  That launch price is a complete unknown at this time.  So I just tried a low guess and a high guess.  Note that the number of tankers gets added to 1 ship carrying the lunar payload.  While launch price affects price per delivered ton greatly,  I was surprised to find that the ratio of price per delivered ton from elliptical orbit to the price per delivered ton from circular orbit,  seems to be quite independent of the launch price for any one flight.  See Figure 17.

Figure 17 – Launch Price Effects on Cost per Delivered Ton to the Moon

I also briefly relaxed the constraint of not entering the Van Allen belts,  for a slightly taller elliptical orbit that was 400 x 1900 km,  with the difference in perigee to circular velocity 0.383 vs 0.25 km/s.  At the fixed stagepoint thrown mass of 1488 metric tons,  this “Starship” requires a full propellant load of 1200 tons,  with at most 170 tons max payload getting there.  Departing from there to the moon fully fueled,  the max deliverable payload to the moon is 67 metric tons.  Note that the apogee associated with this departure orbit penetrates about 400-500 km into the lower Van Allen belt,  presenting potentially a very significant radiation hazard to the crew.

Orbit                    payld, m.ton
400 km circ.                      27
400x1400 ellip.                 52
400x1900 ellip.                 67


Applied surface bearing loads depend upon the local weight of the mass (1 gee Earth,  0.384 gee Mars,  0.165 gee moon),  and the total area of landing pad footprints actually in contact with the surface.  Force/area means these get expressed as applied bearing pressures.  The allowable bearing pressures depend upon the nature of the surface:  hard rock safely supports more pressure than soft sand. 

Exceed the maximum safe applied pressure,  and the landing pads start to “dig in” like tent stakes,  regardless of what the surface really is.  And this will happen very unevenly,  from leg to leg!  That is just an ugly little fact of life. 

We have no local surface bearing strength test data from any of the probes sent to Mars,  or from the probes and Apollo landings upon the moon.  What we have is only a sense of equivalent Earthly soil or surface types,  as seen by these probes and landings. 

Pretty much the entire surface of the moon resembles “fine loose sand” on Earth.  The presence of various sizes of rocks scattered within the sand has little or nothing to do with its bearing strength!  It’s neither cemented nor is it compacted,  it is just meteor-smashed dry regolith!

The vast bulk of the surface of Mars resembles the regolith on the moon,  meaning it is more-or-less equivalent to Earthly “fine,  loose dry sand” in strength.  The presence of rocks of various sizes scattered within this regolith has no impact on its safe bearing strength!

There are places on Mars where this is not true,  but they are a minority of its surface!  The main thing that comes to mind is ancient alluvial deposits of sand and rocks (small rocks somewhat like coarse gravel) on Mars,  which in those ancient wet times would have compacted and cemented like the natural beds of sand and gravel here on Earth.

A very tiny minority indeed of Mars’s surface would be exposed outcrops of “solid” rock.  But,  there’s “solid rock”,  and then there is “solid rock”,  in terms of just how strong that rock really is!  This makes quite a difference here on Earth!  Unfractured igneous rocks on Mars would be much stronger than fractured sedimentary rocks,  same as here on Earth.  The variation in effective strength is enormous! 

Yet all of these terrains would show about the same density from remote sensing,  simply because most common minerals show specific gravities in the 2.5 to 3.5 range!  Crystal density has nothing to do with gross rock self-cohesion.  Therefore,  the remote sensing that we can do,  is just not yet reliable at all for this safe bearing load purpose!  It does not matter how inconvenient or unpopular that statement might be,  it is still true.

I retrieved the safe bearing pressure capacities of various Earthly surfaces from my old edition of Marks’ Mechanical Engineer’s Handbook,  and selected for each type the minimum allowable safe bearing pressure.  This minimum is the value you must use for estimates regarding that apparent type, until and unless you have actual soil bearing strength test data from the specific site you are considering!  

We have no such test data yet for Mars,  or the moon,  but we understand the general range of values for each type here on Earth.  Soil types on Mars and the moon seem equivalent to certain Earthly types.  That apparent equivalency is currently the best we can do right now. 

I have already looked at the ranges of feasible landed masses for “Starship” on Mars and the moon,  plus abort flights on Earth,  in the article and updates above.  Each of these has a local weight at local gravity.  I picked “typical” max local weight values for each of these scenarios. 

Note that an abort-to-surface anytime during the second stage ”Starship” ascent to LEO,  will lead to a fatal crash with any “Starship” fitted with 3 sea level and 3 vacuum engines.  The same is true for a failure requiring “Starship” to abort-to-surface during the “Super Heavy” booster burn.  That is because the Earth weight of the descending vehicle far exceeds the combined maximum thrust of the 3 sea level engines. 

The thrust available from the 3 vacuum engines is very limited indeed,  because full flow separation in the vacuum bells starts around 20,000 feet altitude at max thrust.  The sonic-only thrust coefficient is going to be about 1.2 at the very most!  Plus,  with separated flow,  the vacuum engine bells may fail thermally-structurally,  if operated in separated mode for any appreciable length of time. 

The moon landing as conducted from circular LEO has at most 27 metric tons of payload to the moon,  and features no more than 10 metric tons of return payload to Earth.   If instead conducted from an elliptic orbit that still avoids penetrating the lower Van Allen belt,  this payload increases to 52 metric tons,  with the same 10 ton return payload.  Moon missions are entirely unrefueled on the moon. Excepting payload,  there is no other difference between lunar landing and lunar takeoff masses. 

The Mars landing as conducted from circular LEO can carry at most 203 metric tons of payload to Mars,  and at most 162 metric tons of payload back to Earth,  all figured at min-energy Hohmann transfer.  Mars missions are entirely one-way trips to Mars,  unless the full propellant load of some 1200 metric tons can really be manufactured on Mars!  Departure weights are far higher than landing weights. 

I figured the local weights for all these missions and scenarios from the masses the missions required.  These have been incorporated into a table of applied and allowable bearing pressures that varies with Earthly surface type.  Those results are shown in Figure 18. 

This is based upon the observation that for the 2019 version of “Starship”,  there are 6 landing legs extending from the vehicle base (one each from each aft “wing” fillet,  two others distributed across the ventral surface,  and the last two distributed across the dorsal surface).  Based upon the visual characterization from published Spacex images,  the landing pads of these short-extension struts are “fractional-meter in dimension” at most.  I have used an arbitrary 0.5 m diameter as a sort of “default” value.  It is not right,  but it is “in the ballpark”. 

There are NO published figures from Spacex regarding how big these landing pads really are!  Whatever size they really are,  the local weight divided by the total pad area in contact with the surface,  is the applied bearing pressure!  This MUST be below the minimum credible value of the surface’s safe bearing capacity!  No “ifs”, “ands”,  or “buts”.  For purposes of magnitude sizing,  I consider 0.96 MPa the same number as 1 MPa.  This is just roundoff error. 

 Figure 18 – Comparison of Applied vs Allowable Pressures for “Starship” Operations

The first conclusion drawn from Figure 18 is that the moon landing of a “Starship” is just NOT feasible,  given the current landing leg and landing pad design,  by something like a factor of 4!  If you consider the larger payload of an elliptical-orbit launch,  this discrepancy increases to around factor 5! 

The second conclusion from Figure 18 is that a one-way Mars landing might be feasible with the 2019 “Starship” design,   but only for a very tiny minority of Mars’s surface:  a “solid rock” outcrop,  that is also very smooth and very level!  Failing this circumstance,  something equivalent to a thick,  reinforced,  concrete landing pad is simply required,  even for just the landing!  Much less the takeoff!

It gets worse by around a factor of 5 for a refueled takeoff from Mars,  even with a reduced return payload.  Mars takeoff at fully-fueled launch mass requires something pretty close to 5 MPa bearing capacity,  as seen in Figure 18.  There simply are no surfaces on Mars that are that known to be that strong!  Lava flows,  even if unfractured,  are just not that flat and level.  This is reinforced-concrete-equivalent landing pad stuff!  No way around that!

Larger landing pad areas are just absolutely required!  Figure 19 shows options for both square landing pads,  and 2:1 ratio length:width pads.  Thee are computed from the max local mission weights as-rounded up to the nearest 100 KN,  at fine loose sand strengths of 0.1 MPa,  or a bit higher,  as shown.   

Figure 19 – Landing Pad Sizing Data

The worst case of all is not an abort landing upon an Earthly soft sand beach (quite the common scenario otherwise),  but fully-fueled takeoff from the vast bulk of Mars,  which is similar to soft,  fine,  loose sand,  regardless of any rock content present!  We need something like 14-15 square meters of total landing pad area to make this happen reliably!  If the landing legs bury themselves into the ground as we refuel on Mars,  the resulting “tent stake” friction acts to prevent liftoff!  The ship may even topple over (and explode) as it is being refueled.

These required pad areas are about 3.8 m x 3.8 m for 4 square panels,  deployed from the upper and lower surfaces of the aft fixed-wing surfaces.  If 2:1 length:width,  with length chordwise along the fixed wing chord,  these are still four panels,  each about 4.06 m by 2.03 m in dimension. 

Either way,  such are easily designed into the wings as panels folded out by hydraulics,  much like landing gear bay doors on Earthly aircraft.  Bear in mind that “outrigger” legs perpendicular to the body are still needed to get a big,  stable “footprint” out of this. 

There is not yet one single public word from Spacex regarding this critical issue.  I doubt they will address this until an abort during a prototype test flight has to land on soft ground,  because it could not reach the intended paved launch/landing pad.  The resulting very serious problems will no doubt spur action and changes to the design. 


Update 10-31-19:  Static Overturn Stability

We start with static stability of “Starship” as a flight vehicle.   Now,  there is a shortfall of broadside projected area of the nose,  relative to the rest of the cylindrical body,  of the ”Starship” upper stage spacecraft.  But there is also a shortfall of broadside projected area of the forward steering canards relative to the aft fixed wings.  These differences tend toward balancing each other.

Thus,  the center of aerodynamic pressure is somewhere very near the geometric center of the 50 m long “Starship” upper stage.   Call it 25 m from the base,  of the 50 m long by 9 m dimeter vehicle,  per Spacex published data.

For minimal “arrow stability” purposes,  there is a need for the center of gravity of an object to be about 75% of its body width,  ahead of its center of pressure.  For the 9 m diameter “Starship”,  this would be about 6.7 m ahead of the center of pressure,  or about 31.7 m ahead of the base of the vehicle. 

The 2019 version of Spacex’s “Starship” has some 6 landing legs that extend minimally and vertically from around the base of a 9 m diameter vehicle (radius 4.5 m to the center of each pad).  Assuming these are equally spaced around the periphery in a hexagonal array,  they correspond to a 30-60-90-degree triangle with a hypotenuse of 4.5 m.  The min distance from the vehicle centerline to nearest edge of this footprint hexagon (halfway between two pads) is then (4.5 m) (cos (30 degrees)) = 3.897 m.  The max is just the radius,  4.5 m.

Overturn stability on a flat,  level surface then requires that the vehicle weight vector always fall within the polygon representing the landing pad footprint.  The weight vector “hangs” from the center of gravity,  some 31.7 m higher than the base of the vehicle.  What that really means is that tan(angle) = min distance/cg height,  as landed.  For aft base-to-cg = 31.7 m,  and centerline-to-nearest-edge 3.897 m,  that angle is 7 degrees.  (Figured at the radius to a pad,  it is 8 degrees,  but you are interested in the smaller value as your true limitation.)  See Figure 20. 

Figure 20 – The Geometry of Static Overturn Stability

What that really means is that for the 2019 “Starship” design,  any straight outcrop of sufficiently-hard rock must NOT be sloped any more than 7 degrees off horizontal,   for a successful landing to be made at all!  Local surface roughness variations make this requirement even more stringent

For 7 degrees max tilt,  the local roughness variation of an otherwise absolutely-horizontal surface may not exceed radius*tan(7 deg) = 0.55 m.  What that really means is that the max boulder size under any one landing pad may not exceed 0.55 m,  without a topple-over risk, on otherwise flat and completely level ground.  This applies to roughness in general,  not just boulders.  Landing with a couple of adjacent pads down in a runoff channel has the same tilt-over effect,  as illustrated in the figure.

If already sloped,  this limitation is far lower still!   A wider landing pad base diameter than 9 meters would greatly improve this picture.   That gets back to landing pads deployed from the surfaces of multiple fins-as-landing legs,  for which the effective polygon circumscribed diameter is larger than 9 m,  as the first concept that would come to mind.

4 such support points is significantly more stable than 3 such,  but 5 or 6 such are not a whole lot more stable than 4.  This could be done,  while simultaneously addressing the need for more pad area,  folding out panels from the two fixed wings of the 2019 version of “Starship”:  two from the upper surfaces on each side,  and two from the lower surfaces.   Adding more legs from other periphery locations just makes things even better by increasing the size of the footprint polygon,  although they need to extend outward significantly for better stability (allowing more slope angle).  See Figure 21 for a new idea.

Figure 21 – An Idea for Higher Pad Area and Static Overturn Stability Simultaneously


All of the above analyses notwithstanding,  please bear in mind that these are design analysis estimates for a design concept that has yet to “gel” into a real design undergoing actual development.  There are now some initial flight test prototypes based on this design concept (and not the earlier concepts),  but the fundamental nature of experimental flight test suggests that major design changes are yet to come.  These changes will be uncovered by those flight tests.  

I have pretty-well addressed the performance potential and possible shortfalls of the “Starship” / “Super Heavy” design concept as it currently exists,  based on the sparse data currently available.  There would seem to be little point to pushing this further,  until the design concept evolves into a real candidate design during flight test.  So,  unless a reader has a question that I might answer,  I am pretty well done with this particular article. 


Update 11-21-19:  Choice of Landing Engines

One should be aware that there are two variants of the Raptor engine to be installed on "Starship".  These differ only in the size of the exit bell (expansion ratio),  they share the same hot gas generation and throat area.  Those are the smaller-bell "sea level" Raptor,  and the larger-bell "vacuum Raptor".  The vacuum Raptor has higher vacuum Isp precisely because the expansion ratio is larger. The sea level Raptor has good vacuum performance far higher than its sea level performance,  but not as high a vacuum performance level as the vacuum Raptor.

My estimates for these engines indicated the vacuum Raptor could not be successfully operated in Earth's atmosphere below the stratosphere,  due to backpressure-induced flow separation in the nozzle.  But Mars's atmosphere is so thin as to permit operation of the vacuum Raptor at essentially its vacuum performance level.  The vacuum designs have higher Isp.  So in my analyses above,  I used the sea level Raptors for landing on Earth,  and the vacuum Raptors for landing on the moon or Mars.

I have since become aware of gimballing restrictions in the 2019 version of the Starship design.  The vacuum Raptor engines cannot gimbal very much,  if at all,  due to a lack of room to swing the bells about.  The sea level Raptor engines have plenty of room for significant gimballing,  because there is plenty of room to swing those smaller bells about.

Now,  significant gimballing is required in order to do retropropulsive landings,  anywhere.  What that really means is that the sea level Raptor engines must be used for touchdown on the moon and Mars,  as well as Earth.  The Isp during the touchdown burn is lower by nearly 20 seconds for this geometry-driven choice.  The touchdown burn is not a large delta-vee item,  but this does mean max payloads are a little lower than I estimated in the above analyses.  (I did not re-run all those estimates.)