2020 Starship/Superheavy Estimates for Mars

Update 7-2-2020:  I found a mistake in my performance-estimating spreadsheet.  I used the unfactored delta-vee requirements for figuring mass ratios,  instead of the factored delta-vee requirements.  I corrected this in the spreadsheet,  and used those revised images to correct the spreadsheet results depicted here as Figures 5,  7,  and 9.  Then I corrected the results shown in Figure 10 by re-editing that figure. Where numbers changed in the text of the article,  I edited them with line-out and red text.

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These results are based upon the earlier article posted here,  titled “2020 Reverse-Engineering Estimates for Starship/Superheavy”,  dated 25 May 2020,  and also upon the earlier article titled “Interplanetary Trajectories and Requirements”,  dated 21 November 2019. 

The first examined the latest data available for the Starship/Superheavy designs,  and projected its performance into low Earth orbit (LEO),  complete with a look at what kind of tanker flights might be required to leave LEO.  That same set of weight statements and engine performances is assumed here.

The second examined trajectories to Mars,  including both Hohmann min-energy,  and faster trajectories,  and including direct entries,  and entries into low Mars orbit,  and to Phobos.  Of most interest here was the faster trajectory of an exactly 2-year period,  which provides a 2-year abort path if the direct landing on Mars cannot be made,  for any reason whatsoever. 

This article also examines direct entry and landing upon Mars,  starting from LEO,  via either Hohmann or the faster 2-year abort trajectory.  Further,  it looks at stopping in low Mars orbit (LMO),  followed by deorbit and landing on Mars.  It does not examine going to Phobos,  which would be broadly similar to stopping in LMO,  except a bit worse,  energetically.

Returns are launched directly from the Martian surface,  assuming a full refill using propellants made locally on Mars.  Whether one stops in LMO or just escapes directly,  the total departure velocity requirement is the same,  for the same return trajectory.  Stopping in LMO relieves a tight launch window from the surface,  replacing it with a fairly tight window for trans-Earth injection from LMO,  which is actually an easier target to hit,  in actual practice.

A summary illustration (not to scale) of the two basic transfer trajectories,  with the LMO option,  is given in Figure 1.  All figures are at the end of this article.

It must be understood at the outset that the kinematic velocity requirements for this-or-that trajectory are not what you design your vehicle with,  using the rocket equation!  Those velocity requirements must be factored-up appropriately to “mass ratio-effective velocity requirements”,  in order to size a credible design,  or make credible performance calculations.  That is because the rocket equation was mathematically derived for zero-gravity,  zero-drag conditions. The real world often is not.  See Figure 2.

I have been to the Spacex website many times over the last several years.  Among the other things posted there have been atmospheric entry simulations for their Starship vehicle.  There are two such:  one for entry at Earth,  and one for entry at Mars.  These are rarely posted at the same time,  and are quite different,  because the Martian atmosphere resembles the Earthly atmosphere above 105,000 feet (about 33 km).  See Figure 3. 

The Earthly entry trajectory ends hypersonics at about Mach 3,  above 40 km,  followed by a bending downward while pitching upward to dead-broadside descent angle-of-attack.  This dead-broadside fall is termed the “belly-flop” or “skydiver” maneuver.  It is supersonic up in the thin air,  and very subsonic down in the thick air near the surface.  Spacex lists this as 68 meters/second under about 5 km.  There, they pitch to tail-first while igniting the engines to land.  That is a very modest delta-vee on the order of 0.070 km/s,  but it has to be factored-up significantly to cover hover and divert allowances.

The Mars entry ends hypersonics at about Mach 3 very close to the surface:  about 5 km altitude.  Spacex’s simulation shows a lifting pull-up to decrease speed while gaining altitude,  to around Mach 2-ish at 10 km/s.  Right after that is when they pitch to tail-first and fire-up the engines to brake for landing,  at something like 900 meters/sec = 0.9 km/s.  Again,  that needs a significant factoring-up to cover hover and divert needs.

It may well prove to be,  that Spacex will need to fire up the engines earlier at Mars entry,  in order to augment lift with thrust to achieve the pull-up maneuver.  If so,  the landing burn delta-vee is even higher,  but I did not assume that to be the case for this analysis!

The Hohmann min-energy trajectory is depicted in Figure 4.  This is an ellipse with its perihelion at Earth’s orbit about the sun,  and its aphelion at Mars’s orbit about the sun.  Half of it covers the voyage to Mars,  the other half covers the return voyage.  For purposes of this analysis,  I only looked at Earth and Mars at their average distances from the sun.  Those variations do affect the results,  Mars more than Earth because of its more eccentric orbit.  But,  the averages are close enough to find out “what ballpark you are playing in”,  which is the point here.

For the baseline outbound voyage,  entry at Mars is direct from the interplanetary trajectory.  There is only a course correction budget needed along the way.  The three burns are LEO departure,  course correction,  and direct landing upon Mars.  The value of the course correction budget (0.5 km/s) is nothing but a guess,  but it is a good ballpark guess,  being about 2-3% of the typical velocity with respect to (wrt) the sun during the voyage.  Note that in this baseline there is no possibility of aborting the direct landing for reasons of adverse conditions at Mars.

For the baseline return voyage,  entry at Earth is direct from the interplanetary trajectory.  The same 0.5 km/s course correction budget is presumed for this leg.  The three burns are Mars departure,  course correction,  and direct Earth landing.  There is no possibility of aborting the direct Earth landing.  The difference is Mars departure,  which could be broken-up into two burns:  one to achieve LMO,  the other to escape onto the return interplanetary trajectory.  The total is the same,  either way.

What I got for this mission is the spreadsheet image shown in Figure 5.  The weight statements,  engine selections for each burn,  and engine performances,  are all listed in the spreadsheet image.  It covers both the voyage to Mars,  and the return to Earth,  as separate sections,  with different weights (payload values) for each leg.  Payload both ways exceeds what the “2020 Reverse-Engineering Estimates” article found to be the maximum payload deliverable to LEO.  The payloads shown are the max values to the nearest metric ton,  that gave positive fractional-ton remaining propellant,  after all burns were fully expended (the spreadsheet really is iterative).

Therefore,  to meet these weight statements,  not only must the Starship’s propellant be refilled to 100% of capacity on LEO,  but also some extra payload mass must be brought up to LEO and loaded on board.  Those means are not explored here.  However,  284 220 metric tons of total payload (people plus cargo) is a generous allotment that this design can deliver to Mars,  if an 8.62 month one-way voyage is acceptable. 

The Hohmann return payload of 153 144 metric tons is less,  because the net total delta-vee requirement to return is higher.  This mission spends about 13 months at Mars waiting for the orbits to be “right” to come home,  so you have 13 months to make 1200 tons of propellant!  The next opportunity is 26 months after that. The min round trip total is about 30.2 months.

The faster transfer trajectory that I chose to analyze is the 2-year abort return ellipse.  Its perihelion is at Earth’s orbit,  but its aphelion is far beyond Mars,  nearly to the main asteroid belt!  Its period is exactly two years,  so that if you follow it,  the Earth will be at its perihelion just as you arrive there.  In order to have Earth there when you are there,  the period of any such orbit must be an integer multiple of 1 year.  It is simply not a free-return abort orbit,  if the Earth is not there when you get there.  See Figure 6.

I used the same landing allowances and course correction allowances for the faster transfer case.  The LEO departure and Mars surface departure velocities are significantly higher,  because the energy of this orbit is also significantly higher.  That raises required mass ratio,  and lowers max payload.  Again,  the payloads I found were maximum to the nearest ton,  for a positive fractional-ton remaining propellant,  after all burns were fully expended.  What I found is the spreadsheet image in Figure 7. 

Max payload to Mars still exceeds equals what can be delivered to LEO without any separate freight deliveries!  Max payload returnable to the Earth is actually quite limited at only 39 34 tons!  This mission has about 15-1/2 months on Mars in which to manufacture the return propellant.  It is 2 years long overall.

As for stopping in LMO instead of direct Mars entry,  I only looked at the outbound voyage to Mars.  I did look at both the Hohmann min-energy transfer,  and the faster transfer using the 2-year abort orbit.  This added more burns to be made.  Now,  there are LEO departure and course correction as before,  plus an entry into LMO,  plus a deorbit from LMO,  and the same landing burn as before.  The faster transfer has the higher LEO departure and LMO entry requirements.  The final landing burn and the deorbit and course correction burns,  are all the same.  This is shown in Figure 8.

What I found for this,  is the spreadsheet image given in Figure 9.  I was quite surprised to find that it is indeed possible for a Starship to enter LMO and still have the propellant remaining to land.  It can only do this from the Hohmann min-energy transfer,  and only with very drastically-reduced payload (only 39 34 metric tons).  This is just not a possible landing abort mode,  unless you deliberately use Starship to transfer very small payloads to Mars.  From the faster transfer trajectory,  we are 21.5 33.6 tons short of the min landing propellant needed to avoid a fatal crash,  even at zero (!!!) payload.  That is just not feasible.

I have summarized all these results in Figure 10.  Using the “2020 reverse-engineering” numbers,  payload transferable to Mars looks to be more,  or at least equal to,  than what can be transferred to LEO,  for both Hohmann min-energy transfer (284 220 metric tons for 8.62 months one-way),  and for fast transfer on the 2-year abort orbit (197 149 metric tons for 4.26 months one-way).  The earlier article indicated a max of 149 tons could reach circular LEO,  with full recovery of both stages.

I looked at the LEO tanker issue in the “2020 Reverse Engineering” article.  Flying a cargo/passenger Starship at zero payload results in 133 metric tons of deliverable and transferable propellant,  to LEO.  That’s 9 flights to fully refill the one vehicle going to Mars,  assuming it arrives in LEO with sufficient propellant remaining to deorbit and land.  Flying a reconfigured Starship with extra tanks,  as a dedicated tanker design,  can deliver the full 149 metric tons to LEO.  That’s 8 flights to fully refill the one vehicle going to Mars.  Sorry,  8 or 9,  not 4 to 6.  The numbers do not support the optimism!

None of these tankers has the payload capacity to deliver what would increase the payload of the vehicle going to Mars from the minimum 149 tons.  Such increases would have to come from separate “freighter” flights.  There has been no way identified yet,  for transferring cargo from one Starship to another while in LEO.   No one has even mentioned this.

It is possible to stop in LMO before landing if you sharply reduce your payload before you ever depart LEO,  if you are on a Hohmann min-energy trajectory (78 50 metric tons for 8.62 months one-way).  But it is impossible,  even at zero payload,  to land without crashing,  if you fly faster!  The sharply restricted payload (78 50 versus 149 metric tons),  makes this option very undesirable,  but discarding it also eliminates a possible landing abort method,  if landing conditions unexpectedly prove adverse at Mars,  such as a giant dust storm with high winds.

Going to Phobos will not help this LMO-impracticality picture.   That slightly reduces the delta-vee requirements to rendezvous with,  and then land on,  the moon,  but it significantly increases the delta-vee requirements needed to reach the surface of Mars.  This has already been thoroughly explored in the article titled “Interplanetary Trajectories and Requirements“,  dated 21 November 2019,  on this site.  Other work I have done and not posted verifies that the least-energy path from Phobos to Mars is a burn to escape from Phobos,  a burn onto a transfer ellipse to LMO altitude,  a circularization burn at LMO altitude,  followed by a deorbit burn,  entry,  and a landing burn.

There are two critical items necessary to make this Starship/Superheavy design concept more practical:  (1) the ability to put closer to 300 220 metric tons of payload in LEO while maintaining an abort deorbit and landing capability,  and (2) coming up with a real tanker design.  Those are in addition to the rough-field landing capability problem,  and the inert mass growth problem that always occurs in experimental flight testing.  You have to increase the size of SuperHeavy to do that LEO payload increase,  so as to increase the staging velocity,  while still maintaining flyback capability for recovery. 

 Figure 1 – Types of Missions Considered

 Figure 2 – Factors Applied to Kinematic Delta-Vees to Obtain Mass Ratio-Effective Delta-Vees

Figure 3 – Trajectory Characteristics,  Final Speeds,  and Factors for Entry,  Descent,  and Landing

 Figure 4 – Characteristics and Velocity Requirements for Hohmann Min-Energy Transfer


 Figure 5 – Spreadsheet Image of Results for Hohmann Min-Energy Transfer AS EDITED 7-2-20

 Figure 6 – Characteristics and Velocity Requirements for Fast Transfer with 2-Year Abort

 Figure 7 – Spreadsheet Image of Results for Fast Transfer with 2-Year Abort AS EDITED 7-2-20

 Figure 8 – Hohmann and Fast Transfers to Mars Stopping in Low Mars Orbit Before Landing

 Figure 9 – Spreadsheet Image of Results for Stopping in Low Mars Orbit Before Landing AS EDITED 7-2-20

Figure 10 – Summary of Results for Transfers To and From Mars AS EDITED 7-2-20

Repeat-Pass Aerobraking at Mars?

This is a topic that often comes up,  as a potential way to save propellant for arrival into Mars orbit from the interplanetary trajectories.  There are two problems:  (1) Mars’s atmosphere is very thin,  so that the densities effective for drag deceleration at orbital speeds are quite close to the surface,  and (2) trusted sources such as the Justus and Braun tome on EDL (Entry,  Descent,  and Landing) indicate that the density of Mars’s atmosphere at braking altitudes is erratically variable by a factor of 2,  totally unlike the case at Earth.

Discussion of the Problems

The first problem brings up the risk of accidentally impacting the surface of the planet,  since that surface is so closely adjacent to any practical aerobraking trajectories.  Very precise trajectory control is required to make this work,  much more so than here at Earth.  It is not impossible,  but it is very demanding,  when this technique is to be used at Mars,  which lacks the navigation satellites we have here at Earth.

The second problem has two impacts:  (1) factor two density variation is factor 2 drag variation,  all else being equal,  which may be too large to assure aerobraking into orbit,  and (2) since the variation is erratic,  you cannot know the density that exists at aerobraking altitude until you are actually in the atmosphere attempting your aerobraking.  It will show up in your reduced speed and the peak gees decelerating to it.

Conditions of the Approximate Analysis

The altitude for “entry interface” at Mars is generally considered to be 140 km,  per Justus and Braun,  who recommend using 135 km for Earth entry interface altitude.   On Earth,  peak decelerations are up nearer 80-90 km,  while these occur lower on Mars,  primarily because Martian surface densities look like Earth densities at 30-35 km altitude.  I picked an arbitrary 50 km for my ellipse periapsis altitude,  to fall approximately in the max braking range on Mars. 

For speed at entry interface,  I used the 7.5 km/s that Spacex quotes for its peak entry speed with the Starship design.  That is intended for direct entry from the interplanetary trajectory.  Here I am investigating a braking alternative from those same conditions.  That entry speed corresponds to a velocity “far” from Mars near 5.6 km/s relative to Mars,  which implies a trajectory somewhat faster than a min energy Hohmann transfer orbit.

The remaining variable is the initial braking ellipse orbit apoapsis distance.  To be a ellipse at all,  this must fall between circular at the braking altitude,  and very eccentric with periapsis velocity very close to Mars escape.  I used a 10 m/s difference between escape and periapsis velocity as my largely-arbitrary criterion for the most elongated ellipse that I would consider “stable”.

Anything falling outside those limits cannot lead to repeat-pass aerobraking.  If the density and drag are too high,  the braking pass will inherently become an unplanned direct entry.   If the density and drag are too low,  braked speed exceeds escape,  and the vehicle bounces off the atmosphere into deep space.  Both outcomes are likely fatal,  especially bouncing off into deep space.

How the Repeat-Pass Aerobraking Works

The way this works is the vehicle approaches on a hyperbolic path to pass past Mars at the braking altitude.  As it approaches Mars,  gravitational pull increases its speed to the entry interface value,  approximated as:  Vint² = Vinf² + Vesc²,  where Vesc is computed at the braking pass altitude.  Vint is the interface speed,  and Vinf is the speed relative to Mars “far” from Mars.  For this analysis,  I just set Vinf such that I got Vint = 7.5 km/s.

During the pass,  density and velocity vary,  so drag varies.  There is a drag work integral with path length along the braking pass path.  That integrated drag work subtracts from the vehicle kinetic energy at entry interface,  so that kinetic energy (and vehicle velocity) are lower at the exit from the braking pass.  That final velocity must fall in the range of feasible periapsis velocities for the initial ellipse,  or else you enter direct if too low,  or bounce off into deep space if too high.

Once on a feasible ellipse,  there is drag braking at each periapsis pass,  which reduces vehicle mechanical energy and apoapsis altitude.  After multiple braking passes,  the orbit circularizes at the braking altitude,  and the vehicle then deorbits.  That is the concept behind repeat pass aerobraking.   See Figure 1.

Figure 1 – Repeat-Pass Aerobraking Conditions and Analysis Results

How the Analysis Was Conducted

I did not attempt to evaluate the actual drag work integrals.  Instead,  I looked at kinetic energy differences between entry interface speed and periapsis speed conditions for the range of feasible braking ellipses.  Those kinetic energy differences are the required drag work integrals. 

The max kinetic energy reduction (drag work integral) is for a circular orbit (degenerate ellipse of eccentricity zero) at braking altitude.  That orbit has about a 1.7 hour period.  The min credible kinetic energy reduction (drag work integral) is for an extremely-elongated ellipse with an apoapsis radius near 850,000 km.  Its period is nearly 3.3 months.  The ratio of those kinetic energy reductions (ratio of the drag work integrals) is about 1.39.  

Effects of Density Variability

For the same braking altitude (and it really won’t be,  but we are ignoring that effect here),  the drag work integral could vary max-to-min by a factor of 2,  simply because the density can vary by that factor of 2.  It varies erratically,  and you cannot know what that density really is,  until you have completed the braking pass with the wrong exit velocity,  which is far too late to adjust your braking pass altitude.  The factor 2 variation easily encompasses both unplanned direct entry and bouncing off into deep space.  Therefore you must be prepared to compensate “on-the-fly” as you exit the first braking pass by firing rockets to adjust your speed to the proper (or at least a feasible) value. 

It’s not that repeat-pass aerobraking won’t work,  because it most certainly will.  But at Mars,  the control of speed coming out of that initial braking pass requires very precise control and adjustment “on the fly”,  with a rocket burn of some significant size.  That plus the final circularization burn into a stable low Mars orbit will be less than,  but as much as some fair fraction of,  the direct orbit entry burn of almost 4 km/s.

Control is simpler,  and to more certain requirements,  with the orbit entry burn approach.  But if one has real-time velocity information,  and fast-response propulsion,  one can do a real-time-tailored delta-vee burn to exit the initial braking pass with the correct velocity.  You have to plan this for the very worst case,  even if you never use that propellant.

Conclusions

My conclusion is that the risk potential for a fatal outcome is just too high relative to the feasible braking orbit entry conditions,  unless you bring extra propellant and some very sophisticated,  precise,  and fast-acting controls.  This is not a technique that should be preferred for use at Mars!  One would have to add some serious rocket braking to prevent bouncing off,  or unintentionally landing direct but too steeply.  If you have to do that,  then you might as well just do rocket braking into orbit under much more certain control.

Details

The details of the calculations are shown in Figures 2 through 4.  These are spreadsheet images,  two images per figure.  I looked at a total of 6 ellipse conditions,  to see what the trend shapes looked like.  The approach to the bounce-off point,  just past the most elongated orbit,  is sort-of asymptotic in terms of kinetic energy,  so that energy estimate is insensitive to the exact max apoapsis criterion.

Remember,  because of unpredictable upper-atmosphere density variations,  the drag integral can vary through a factor of 2.  The ratio of drag integrals leading to a feasible ellipse is only 1.39.  Even if you do not care what ellipse you end up on,  the chances are very significant that you could see a fatal outcome (unintended direct entry or bouncing off into deep space),  without a significant propulsion allowance for adjusting speed as you leave the initial braking pass. 

If you care which ellipse you end up on,  you simply must have propulsive speed adjust capability coming out of that first braking pass. Either way,  the propulsion must be sized for the worst possible case,  which Murphy’s Law says will happen.

Figure 2 – Spreadsheet Images for Circular and Apoapsis at Phobos

Figure 3 – Spreadsheet Images for Apoapsis at Deimos and a One-Week-Period Orbit

Figure 4 – Spreadsheet Images for a One-Month-Period Orbit and the Max-Credible-Elongated Orbit

Interplanetary Trajectories and Requirements

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

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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 r_per and r_apo,  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:  r_per < r < r_apo.  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 V_{wrt planet} = V_{wrt sun} – V_{planet 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 V_far² = 0.5 m V_near² - 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 V_esc² = 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:

               V_far² = V_near² – V_esc²   or  V_near = (V_far² + V_esc²)^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,  V_near is denoted as V_int for “interface velocity”,  and V_near = V_bo 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 V_near,  which is V_int 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 = V_int – V_orbit  on the “correct” side

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

               dV = V_int + V_orbit  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 V_near = V_int.  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 V_near = V_bo still close to the planet,  so that your V_far 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 = V_bo – V_orbit

Departing Mars direct from the surface,  your theoretical dV is V_bo,  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 V_bo in a posigrade direction.  That delta vee is:

               dV = V_bo – V_orbit

If you depart directly from the surface,  your theoretical delta vee is V_bo,  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 = V_int – V_orbit

If you enter the atmosphere for aerobraking,  your entry interface speed is V_int.  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 V_bo and V_far 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 V_far.  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 V_far and V_esc to get V_int.  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 V_bo near the planet.  The “kinetic energy thing” gets you V_far.  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 90^o-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      r_per, km                r_apo, 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 V_far 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:

               y² = R² – (x – c)²                               circle solved for y²

               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 V_inf 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 V_bo 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 V_inf 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 V_int “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 V_bo with respect to Earth,  the “near Mars” arrival speed V_int 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 “V_bo”.  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

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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

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