2020 Estimates for Spacex’s “Starship” to the Moon

Unlike the projected Mars missions,  Spacex has not revealed very much about how its “Starship” craft might go about travel to and from the moon. By implication,  this seems to be flights from low Earth orbit (LEO) to direct landings upon the moon,  and unrefueled flights from there back to Earth.  Those returns could be direct launches,  with or without stops in low lunar orbit (LLO),  to a direct aerobraking entry at Earth,  and the associated retropropulsive touchdown.

About the only complicating detail available for this is the possibility of using an elliptical LEO as the departure point,  to reduce the departure delta-vee (dV) somewhat.  This is based mostly upon informal comments made by Mr. Musk in his various presentations.  That gain would be at the cost of increasing the difficulty of reaching that elliptic LEO with the Starship/Superheavy vehicle.  In effect,  payload to such an orbit is reduced relative to that deliverable to circular LEO.  That is precisely because the elliptic perigee velocity is higher than circular. 

There is a very practical constraint on how far this option could be pushed:  beyond a certain eccentricity,  the apogee altitude of the elliptic LEO orbit falls within the dangerous radiation environment of the Van Allen radiation belts.  Running for the radiation shelter (whatever that is) every hour and a half is just not a practical option for a crewed mission.  That limits the elliptic LEO to about 900 miles (1400 km) apogee altitude,  and 300 km perigee altitude would be rather typical. 

The perigee velocity of a 300x1400 km elliptic LEO is 0.25 km/s faster than 300 km circular.  This reduces the departure dV by that same 0.25 km/s,  and it increases the second-stage burn dV by 0.25 km/s to reach that elliptic LEO vs circular. That reduces the payload carriable by “Starship” and its tankers to LEO,  while increasing the payload carriable by the refilled “Starship” to the moon.  It means you have to pay a price to send more payload to the moon:  lower payload to LEO and more tanker flights to refill.

What I do here in this article is investigate the one mission type (direct lunar landings,  with unrefueled return to a direct landing on Earth),  with the latest estimates that I have for Spacex’s vehicle.  This is comparable to the Mars mission study I did earlier,  in terms of methods and data.  The source documents for this are:

5-25-2020 “2020 Reverse Engineering Estimates for Starship/Superheavy”

7-3-2020 “Cis-Lunar Orbits and Requirements”

6-21-2020 “2020 Starship/Superheavy Estimates for Mars”

10-22-2019 “Reverse Engineering the 2019 Version of the Spacex Starship / Super Heavy Design”

What Is Covered In This Article

This article concerns only “Mission C” as described in the Cis-Lunar Orbits article.  There are other missions this vehicle might fly,  one of which is a dedicated lander per the recent NASA contract.  Mission C is the departure from LEO to a direct landing on the moon,  and an unrefueled return.  That return is presumed a direct escape from the moon,  aerobraking direct entry at Earth,  and a retropropulsive touchdown.  Whether one stops in low lunar orbit (LLO) or not makes no real difference.

There are multiple choices that affect the outcomes greatly,  all of which get considered herein.  One is elliptical vs circular LEO departure.  Another is the achieved inert structural mass of the Starship vehicle.  And the third is just how much of the descent dV onto the moon needs to be factored-up greatly,  to cover hover and divert requirements.  I cannot get the final answer,  but I can illustrate just how critical these issues are.  And they are.

Elliptic vs Circular LEO Departure

The orbital mechanics for this were already covered in detail in the Cis-Lunar Orbits article.  The net effects are:  (1) reduced departure dV by 0.25 km/s,  and (2) reduced payload to elliptic LEO because the final stage-two burnout velocity is 0.25 km/s faster. Handling the rocket equation estimates for the lunar trip with this effect is easy:  just change the departure dV.  Evaluating the elliptic LEO payload reduction requires determining the effect of the higher stage-two burnout velocity,  all else but payload being equal.  In effect,  you have to reduce payload to achieve the higher mass ratio,  and still be able to land.

I probably should have re-run my entire 2020-version spreadsheet model of the Starship/Superheavy vehicle to determine this,  but I chose a simpler estimate.  I think it gets about the same answer.  The higher dV at the same effective exhaust velocity results in a higher mass ratio for that burn,  which is a lower end-of-burn mass.  The difference in the two end-of-burn masses is the extra propellant burn required at full payload,  which must come out of the landing allowance.  I then assumed that same number also has to come out of payload carried,  in order not to burn up the landing allowance. 

My crude estimate says that change is about 19 metric tons.  Which reduces the 149 metric ton circular LEO payload estimate in the 2020 reverse-engineering article to some 130 metric tons to elliptic LEO.  It also uses up all the on-board propellant,  precluding any abort-to-surface landing,  before I reduce payload!  That rocket equation estimate is shown in Figure 1,  as a spreadsheet image.

Figure 1 – Effects of Elliptic vs Circular LEO,  and Sources of Inert Vehicle Mass

Inert Mass Trends

For the 2019 and earlier versions of Starship/Superheavy (under various names),  the Starship inert structural mass was given by Spacex as 85 metric tons.  That was before they switched to stainless steel construction,  and it was before they began to manufacture prototypes.  The propellant load for the stage was also given as 1100 metric tons. 

With the 2020 estimates,  both the stage inert is different,  and its propellant load is different.  The website now shows 1200 metric tons propellant load.  Musk’s presentation in front of a prototype at Boca Chica says the inert mass is 120 metric tons,  that his slide showing 85 tons is in error,  and that he would be delighted if they ever get the inert down to 100 tons.  See again Figure 1.

Because in vehicle developments,  there is usually growth,  not loss,  of inert mass,  I chose the current prototype inert mass of 120 metric tons as “baseline”,  with Musk’s preferred goal of 100 metric tons as a bounding optimistic estimate.

Retropropulsive Landings in Vacuum vs Atmospheres

The worst-case direct kinematic dV for landing direct upon the moon from the transfer ellipse is some 2.533 km/s,  and never much less than that.  Unlike aerobrake landings on Earth or at Mars,  all of this must come from the on-board propulsion.  In the past,  I have factored up the kinematic landing dV by 1.500 to cover hover and divert needs.  With aerobrake landings,  the propulsive kinematic dV is quite small,  since most of the deceleration occurs by hypersonic entry drag.  Such are typical of Earth and Mars,  although the trajectory details are wildly different,  due to the wildly-different surface densities.

For a vacuum landing (as on the moon),  the kinematic dV is quite large,  and is also likely one long,  continuous burn.  I suspect that factoring all of it by 1.5 is very likely “overkill”.  Factoring most of it by the lunar gravity loss factor of 1.008 seems reasonable.  Then factoring only a small,  terminal portion of it by 1.50 (or even 2.00) for hover and divert,  would thus result in a lower overall average factor applied to the kinematic dV.  How much of the dV to factor up by the higher factor is nothing but a guess.  I arbitrarily chose the last 0.25 km/s,  that being about the last 10% of the burn.  I chose to factor that portion by 2.00,  with the other 90% factored only by 1.008 for lunar gravity losses.  See Figure 2.

Figure 2 – Revising the Vacuum Landing Factor Applied to Kinematic Delta-Vee

That gives us two dV-factor values to explore,  the overkill 1.50 applied to the whole 2.533 km/s,  and the arbitrary 1.106 applied to the whole 2.533 km/s,  reflecting factor 2.00 on the last 0.25 km/s,  and factor 1.008 on the rest.  This turns out to be a very major effect.

Return Payload

There are two possibilities which sort-of-bound the results.  One is to carry the same payload both ways.  The other is to carry zero payload on the return flight.  Failing feasibility,  one can look at zero return payload with all propellant used in the landing,  thus stranding the vehicle upon the moon.  I did so.

Methods

I did this with a spreadsheet,  very similar to that used for the 2020 Mars estimates.  In point of fact,  it is another worksheet in that same spreadsheet file.  The difference is refueled versus unrefueled,  for the return flight.  The worksheet has two weight statements,  one for departure,  the other for return.  They have inputs for different payloads each way.  The difference is that for Mars,  the return flight is presumed to start with a full propellant load.  The lunar model does not:  it used the propellant still on board at landing as the propellant supply for the return trip.  This ignores evaporative losses.

You will note that for each burn in the list,  there is a selection of which engines to use,  along with their effective specific impulse.  That sets the exhaust velocity for the burn,  which leads to mass ratio from the rocket equation,  given an appropriately-factored kinematic dV.  In order to achieve thrust gimballing for landing control,  it is the three sea level Raptor engines that must be used for the lunar landing,  at their vacuum specific impulse levels. 

The same was true for the Mars landing in that study,  and it is true for all the Earth landings in all the studies,  just at sea level specific impulse in Earth’s atmosphere.  All the in-space burns are done with the vacuum Raptor engines.  Those would include LEO departure,  course corrections,  and lunar (or Mars) liftoff.  There is no Earth liftoff with only three Raptors (sea level or vacuum).  There is only a second-stage burn with three vacuum Raptors,  and at dV virtually perpendicular to the gravity vector.

There is one additional calculation done here for the lunar mission that was not done for the Mars mission.  That is a little block to the right of the main calculation block,  showing the change in payload for a revised inert mass.  The fundamental assumption here is that every ton of inert mass saved is an extra ton of carriable payload.  So each spreadsheet image reported here has the inert mass variation included. There is one image for circular LEO departure,  and another for elliptic LEO departure.  That leaves re-running the whole set to evaluate the effects of the factor applied to lunar landing dV,  and re-running another set to evaluate the effects of zero return payload.

Results Obtained

The baseline case is 120 metric tons inert,  with the lunar landing kinematic dV =2.533 km/s factored by 1.50.  There are calculation blocks for both circular and elliptic LEO departure on the worksheet page.  These are given separately in Figures 3 and 4.  The effects of reduced inert mass are also given,  as described above.  These are for the same payload carried both ways. 

This was disappointing in the extreme:  the vehicle simply cannot make the two-way trip from circular LEO,  or from elliptic LEO.  That is shown by the negative remaining propellant numbers for the return trip,  even at zero payload both ways.  This makes the inert mass reduction results shown in both figures irrelevant.  It is futile to show zero return propellant,  because the payloads are already zero.  That leaves only the landing dV factor as something practical to pursue.

Figure 3 – Circular LEO Departure,  Two-Way Payload,  “Overkill” Landing

Figure 4 – Elliptic LEO Departure,  Two-Way Payload,  “Overkill” Landing

I did look at a one-way flight to the moon,  stranded there with essentially no propellant remaining.  Under those circumstances,  the effects of inert mass reduction do apply,  and are given in the figures.  Again,  this is for the “overkill” factor upon the kinematic dV for lunar landing.  Figure 5 shows the spreadsheet images for the circular departure case,  and Figure 6 for the elliptic case.

Figure 5 – Circular LEO Departure,  One-Way Stranding,  “Overkill” Landing

Figure 6 – Elliptic LEO Departure,  One-Way Stranding,  “Overkill” Landing

Reducing the overall-average factor applied to the kinematic landing dV does dramatically improve these results,  but not enough to be truly attractive.  Circular departure is shown in Figure 7,  for only the one-way stranding on the moon.  It includes showing the reduction in inert mass.  No two-way trip is feasible,  even at zero return payload and zero payload to the moon. 

Figure 7 – One-Way Stranding is Feasible for Circular Departure at Reduced Landing Factor

The same data for elliptic departure at the same two-way payload is shown in Figure 8.  These results are not very attractive,  even at only 100 metric tons inert mass.  Deliverable payload is just pathetically low,  at either inert mass.

Figure 8 – Two-Way Payload for Elliptic LEO at Reduced Landing Factor

Figure 9 shows the elliptic departure results for reduced landing factor and for zero return payload.  These are still quite pathetic. 

 

Figure 9 – Results for Elliptic Departure,  Zero Return Payload,  and Reduced-Factor Landing

Evaluation of These Results

In no case was the payload deliverable to the moon in the least attractive,  except for the one-way strandings,  which are extremely unattractive because the vehicles cannot be re-used!  This is because the lunar landing dV is large,  no matter how we factor it,  quite unlike the Mars landing dV’s.  The sum of departure,  course correction,  and landing dV is just too large,  no matter how exactly it is figured.  That plus the departure,  course correction,  and Earth landing dV’s for the return are just far beyond any realistic values for mass ratio and exhaust velocity performance out of this vehicle.  What that really says is that the direct landing scenario is just the wrong mission for this vehicle to fly to the vicinity of the moon!  The “kicker” for this is the unrefueled return,  which is quite unlike the nominal Mars mission. 

Refueling this vehicle on the moon is simply not a practical option,  as there is no practical source of carbon for making methane out of local materials.  Further,  the best information we have (poor though it is) says that water is only available near the south pole of the moon,  and we do not yet understand how concentrated,  or pure,  this resource is.  Making Raptor engine propellants on the moon looks to be unlikely at this time!  For any reusability,  that means any lunar fights must return to Earth unrefueled.  And this analysis shows that cannot happen,  at any practical payload levels.

This vehicle,  as it is currently understood,  is just not well-suited for direct landings upon the moon.  The rocket equation and publicly-available numbers prove that beyond a shadow of a doubt,  no matter the assumptions made factoring landing kinematic dV values.  And,  while elliptic LEO departure offers some small improvement,  it is nowhere near enough to make a significant difference,  as long as crewed excursions into the Van Allen radiation belts are excluded for crew safety reasons (which they should be).

The only attractive deliverable payload numbers are for non-reusable one-way strandings upon the moon.  Thus as we currently understand it,  this design is only feasible for some other mission,  than direct landings upon the moon,  excepting only one-way strandings there.

Update 7-10-2020:  Here is a sketch (Figure 10) that illustrates at a glance the results that I found.  The total mission dV is about 8.2 to 8.5 km/s,  which is just too high for this single chemically-fueled stage to accomplish unless it is just stranded in a one-way trip to the moon.  This vehicle (as we currently understand it) has to be used in a different way,  in order to send large payloads to the moon and still be reusable,  without any refueling on the moon. 

 Figure 10 – Results at a Glance

Update 7-12-2020:  see "How the Spreadsheet Works" dated 7-5-2020 this same site,  to see exactly how I calculated these things,  and how the spreadsheet images got generated.