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.

**And they are.***I cannot get the final answer, but I can illustrate just how critical these issues 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

**as a spreadsheet image.***shown in Figure 1,*
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.

**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.**

*With the 2020 estimates, both the stage inert is different, and its propellant load is different.*

*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.

**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.***How much of the dV to factor up by the higher factor is nothing but a guess.**See Figure 2.*
Figure 2 – Revising the Vacuum Landing Factor Applied to
Kinematic Delta-Vee

**, 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.**

*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***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.

**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.***The fundamental assumption here is that every ton of inert mass saved is an extra ton of carriable 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

**The effects of reduced inert mass are also given, as described above. These are for the same payload carried both ways.***given separately in Figures 3 and 4.*
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.**shows the spreadsheet images for the circular departure case, and***Figure 5***for the elliptic case.***Figure 6*
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

**, 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.***shown in Figure 7*
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

**. These results are not very attractive, even at only 100 metric tons inert mass. Deliverable payload is just pathetically low, at either inert mass.***shown in Figure 8*
Figure 8 – Two-Way Payload for Elliptic LEO at Reduced
Landing Factor

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

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

**Evaluation of These Results**

**This is because the lunar landing dV is large, no matter how we factor it, quite unlike the Mars landing dV’s.**

*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!__**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.**

*The sum of departure, course correction, and landing dV is just too large, no matter how exactly it is figured.*

*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.

**For any reusability, that means any lunar fights must return to Earth unrefueled.**

*Making Raptor engine propellants on the moon looks to be unlikely at this time!*__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.

Figure 10 – Results at a Glance

*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.*

**Here is a sketch (**

__Update 7-10-2020__:**) 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***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.**

__Update 7-12-2020__:
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