This is a report of my follow-on efforts after the Moon
Direct evaluations of “2020 Estimates For Spacex’s Starship to the Moon”, 7-5-2020,
this site. That effort showed
very clearly that the unrefueled delta vee for a round trip direct to the lunar
surface is beyond the mass ratio capability of this vehicle design, as it is currently understood in terms of
posted data and public presentations.
This effort looks at refueling in lunar orbit, either before, or after,
the surface landing.

**Source Data:**

I got most my data from the Spacex website, regarding their 2020 thinking and their performance
projections for Starship and its Raptor engines. Inert structural mass data is missing from
that website, so I got that from Elon
Musk’s presentation in front of one of the prototypes at Boca Chica. All that is documented in articles posted on
this site. That list is now as follows:

7-5-2020 “2020
Estimates For Spacex’s Starship to the Moon”
(direct to the moon)

7-5-2020 “How the
Spreadsheet Works”

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 Was Analyzed:**

Trips to the moon can start from either circular low Earth
orbit (LEO) or an elliptical LEO. That
elliptical option increases perigee speed,
so that departure delta-vee is reduced,
at the cost of increasing the delta-vee required of the second stage to
reach LEO from launch,

**. The eccentricity of this elliptical option is limited to apogee altitudes that are not far into the inner edge of the Van Allen radiation belts. For this analysis, I took that limit to be 1400 km (900 statute miles). The circular LEO I took to be 300 km altitude, and used that same altitude as the perigee of the elliptic option.***as indicated in Figure 1*
Once in the vicinity of the moon, the easiest entry into low lunar orbit (LLO)
is the figure eight trajectory into retrograde motion,

**. Circular LLO altitude I took to be 100 km, approximately the same as for Apollo. The entry into this orbit has a delta vee no more than 0.804 km/s, unfactored, as indicated. Direct entry to a landing from this trajectory has at most 2.533 km/s delta-vee, unfactored. The unfactored landing from LLO has a delta-vee of 1.680 km/s. Note that I used a bit more than surface escape, and exactly surface circular orbit, speeds for these transitions. The surface values provide an approximation to the potential energy due to altitude, as well as the kinetic energy, which potential energy then gets included in the effective velocity.***as indicated in Figure 1*
If entry into LLO involves a rendezvous with something
else, then the vehicle making the rendezvous
maneuvers must get into a slightly different orbit with a different
period, until the orbital positions line
up, then recircularize into LLO. Already being so low, the clear choice is elliptical with perilune
at 100 km, and apolune higher still.

I picked an apolune altitude of 1262 km (center-to-center
3000 km) as a slight overkill, with a
period still short enough to be practical.
The axis of this orbit can be any desired orientation relative to the
Earth-moon axis of the transfer trajectory,

**. But it does need to be co-planar with LLO.***as indicated in Figure 1*
The return in all cases examined leads to a direct
aerobraking entry at Earth, followed by
Spacex’s “belly-flop” maneuver, and a
final deceleration to touchdown from a low altitude terminal fall velocity of
70 m/s, per the simulations on the
Spacex website.

**, along with the appropriate factors to convert kinematic delta-vees to mass ratio-effective values.***This is also indicated in Figure 1*
All the in-space/in free fall factors are unity. The three possible landings are factored
differently, reflecting factor 1.008
applied to the bulk of the burn to cover lunar gravity losses (no drag losses
in vacuum), with factor 2 applied to the
last 0.25 km/s of the burn to cover hover and divert needs. At Earthly touchdown, factor 1.5 is applied to that burn to cover the
hover and divert needs.

Figure 1 – Lunar Orbits and Delta-Vee Characteristics

The characteristics of the vehicle are

**. This is the Spacex “Starship” vehicle, which is both the second stage for launch, and if refueled in orbit, the spacecraft for journeys outside LEO. The figure depicts the cargo/passenger version most often depicted on the Spacex website. It has pressurized accommodations in the nose for crew and passengers (blue in the figure), followed by a depressurizable cargo hold for deliverable dead-head cargo (left clear in the figure). The bulk of the vehicle volume is propellant tankage (orange in the figure), depicted here without any details.***indicated in Figure 2*
In the tail are 6 Raptor engines that burn liquid oxygen and
liquid methane as the propellants. 3 of
these are the sea level design with the shorter, smaller expansion bell that works both at sea
level and out in vacuum, and the other 3
are the vacuum design with the larger,
longer expansion bell, usable
only in vacuum. The 3 vacuum engines are
essentially fixed in place, while the 3
sea level engines are set to gimbal significantly, allowing for considerable attitude control as
they burn.

Engine performance figures from the Spacex website are
indicated in the figure. These data
match very well with earlier ballistic reverse engineering that I did on these
engines. For these analyses, I presumed any landings required the sea
level Raptors in order to get the gimballing for fine control. I presumed the vacuum engines plus attitude
thrusters would provide adequate control for takeoffs.

The mass characteristics are also indicated in the
figure. The Spacex website lists the
maximum propellant load as 1200 metric tons,
and the payload as “100+ metric tons”,
without mention of the inert structural mass.

**That was the value from 2019 and earlier, before the switch to stainless steel construction and the start of prototype manufacture and testing.***Musk’s Boca Chica presentation in front of one of the prototypes specifically calls out the 85 metric tons inert on one of his slides as an error.***So I used 120 metric tons as “baseline”, with 100 tons as a “goal value”, in my calculations.***Musk now says the prototypes are running about 120 metric tons inert, and he would be very happy if they could get that down to 100 tons.*Figure 2 – Vehicle Characteristics

Trajectory-wise, I
looked at 3 scenarios: (1) just going to
LLO without landing, (2) refilling in
LLO from tankers sent to LLO, after
landing on the moon, and (3) refilling
in LLO from tankers sent to LLO, before
landing on the moon. Each of these was
investigated for departures from circular LEO,
and from elliptic LEO.

**Results Obtained:**

For the trip to LLO,
I did add-in the delta-vee for rendezvous with “something” (perhaps a
space station?) located in LLO.
However, the cargo has to be
off-loaded there in LLO and transferred to that “something” by means not
contemplated yet.

**I did this for max propellant load at departure in LEO, and determined the max payload masses that could be delivered and still have non-negative propellant remaining after landing on Earth. Those results***None of it reaches the surface.**are depicted in Figure 3.*
The deliverable payloads to LLO (from circular and elliptic
LEO)

__both exceed__what Starship/Superheavy can deliver to circular or elliptic LEO. If the inert is 20 tons lighter at the goal value, then the payloads to LLO are 20 tons heavier still. All of these numbers are**. What one would really do is reduce the payloads to those deliverable to LEO, then reduce the refill tonnages so that propellant remaining is just still non-negative after landing on Earth. This helps reduce the tanker flight requirement in LEO.***shown in Figure 3**For this scenario, no tanker need be sent to LLO, but no cargo is delivered to the lunar surface, either.*

The second scenario was to land upon the moon, then ascend to LLO to refuel from a tanker
(or tankers) waiting there. If the
payload delivered was low enough, the
Starship could ascend to LLO with non-negative propellant remaining. As it turns out, only one tanker was needed in LLO, and the onus of rendezvous was placed on the
tanker. This tanker was an identical
Starship, flown at zero payload, and with just enough propellant on board at
departure to reach LLO, make the
rendezvous, offload propellant, and still return and land on Earth with
non-negative propellant remaining.

The results shown in Figure 4 indicate that

**. Disappointingly, both numbers are less than the website “100+ tons” figure. In both cases 36 metric tons of propellant is required to get the Starship back to Earth with non-negative propellant remaining after landing, because both payloads back to Earth are zero. Only one tanker need be sent to LLO, and it need not be fully loaded with propellant at departure from LEO. The propellant load is a little higher for the elliptic case than the circular case, but neither is a full load.***59 tons of payload can go 1-way to the lunar surface from circular LEO, and 80 tons from elliptic LEO*

*This mission is feasible as indicated, with two vessels sent to the moon (Starship to the surface and a tanker for it in LLO), and a bit over 1700 to 1900 metric tons of propellant needed in orbit, to send them there.*
The third scenario sends the Starship to LLO, where the tankers must rendezvous with it
there, to refill it, before it descends to the surface. It offloads payload before escaping from the
moon direct to Earth for entry and landing.
The tankers return from LLO to Earth entry and landing. For this scenario, I limited the payload to that which
Starship/Superheavy can send to LEO (a bit less to elliptic LEO than to
circular LEO). I limited the refilled
propellant loads in the cargo Starship to that required to reach LLO for
refill, with non-negative propellant
remaining.

After refilling in LLO,
Starship descends to the moon and offloads its payload. Then it escapes directly to Earth. The tankers return to Earth from LLO.

**, starting less than full of propellant from LEO.***From circular LEO, Starship can carry 149 metric tons of payload***that must rendezvous with it, in LLO. That’s a total of 4 vehicles sent to the moon, fueled with some 3313 metric tons of propellant sent to LEO to refill them there. These numbers are***It will need some 486 metric tons of propellant from 3 tankers**given as part of Figure 5.*
From elliptic LEO,
the plan is almost identical,
only the numbers change.

**.***130 metric tons of payload gets delivered***. The starship lands, offloads, then escapes directly back to Earth. The two tankers return to Earth from LLO. There are 3 vehicles sent to the moon from elliptic LEO, and some 2623 metric tons of propellant must be sent to elliptic LEO to refill them there. These numbers are also***Only two tankers are needed in LLO to deliver 472 metric tons of propellant to the Starship there***.***part of Figure 5*
That summary is:

Table 1
– Overall Results

**Conclusions:**

It will be difficult indeed to get Starship, as we currently understand its likely
characteristics in 2020, to land on the
moon without refilling somewhere near the moon before returning to Earth. It is possible to deliver over-100 ton lots
of payload to low lunar orbit,

__but not to the surface__, without any refilling (except that in LEO before departure).
It is possible to deliver under-100 ton lots of payload to
the lunar surface, with refueling after
the landing, in LLO from a single tanker
Starship sent to LLO. That will require
less than 2000 tons of propellant ferried by tankers up to LEO, for the two
vehicles sent to the moon.

It is possible to deliver over-100 ton lots of payload to
the lunar surface, with refueling before
the landing, in LLO from 2 or 3 tanker
Starships sent to LLO. That will require
over 2500, to over 3000, tons of propellant ferried by tankers up to
LEO, for the 3 or 4 vehicles sent to the
moon.

It is the change in inert structural mass for 2020 that
drives this result, despite the increase
in max propellant load from 1100 to 1200 metric tons for 2020.

**The 2020 thinking, per Musk himself, is 120 tons similar to the early prototypes, with 100 tons as a desired goal.***The 85 tons inert thought possible in earlier years has proven to be infeasible.***much more so than the change in propellant load.***This change in inert mass makes a huge difference in the achievable mass ratio out of the design,***Details:**

Images of the actual spreadsheets are given in the following
figures. Starship to LLO without any
landings is

**, for both the circular and elliptic LEO cases. There are no tankers sent to LLO for this scenario.***covered in Figure 6*
The refill after landing scenario is

**. Figure 7 shows the circular LEO departure option, and covers both the cargo Starship and the 1 tanker Starship sent to LLO. Figure 8 covers the elliptic LEO departure option, for the cargo Starship and the 1 tanker.***covered in Figures 7 and 8*
The refill before landing scenario is

**. Figure 9 shows the circular LEO departure option, and covers the cargo Starship and the 3 tanker Starships sent to LLO. Figure 10 covers the elliptic LEO option, for the cargo Starship and the 2 tankers.***covered in Figures 9 and 10*
Figure 10 – Spreadsheet Image: Starship Refilled Before Landing, From Elliptic LEO

Figure A – Essential Characteristics of the Fly-Along
Refueling Scenario from Circular LEO

**Based on comments by reader Rok, I looked at refueling the Starship cargo vehicle with tanker(s) after departing LEO, during the transit to the moon.**

__Update 7-18-2020__:**There is no major rendezvous maneuver allowable, when two vehicles must be on the same trajectory. They must essentially be within direct unaided human eyesight of each other.**

*For that to work, both the cargo Starship and its tanker(s) must depart LEO simultaneously, or else do a free-return abort back to Earth.*
In this scenario, the
cargo Starship leaves orbit with only enough propellant to depart, plus ~ 20 ton abort landing allowance of
propellant. It refuels on the way to the moon,
incurring some of the rendezvous delta vee (I arbitrarily selected 20
m/s). The tanker(s) must leave LEO
rather full of propellant, so that it
(they) may offload a significant amount to the cargo Starship on the way to the
moon. They also must provide 20 m/s
rendezvous delta-vee. How much a tanker
can deliver is set by the ship propellant capacity, and the need to make a free return and
landing.

The cargo starship then makes a direct landing on the
moon, offloads all its payload, and then makes a direct escape from the moon
onto the return trajectory for Earth, at
zero payload. It makes a direct entry
and landing on Earth. If it were to
carry return payload, then the refueling
propellant quantity from the tankers must increase. There is a limit to what they can hold, which will set max return payload rather low.

**All Starships are presumed to have 3 sea level Raptor engines that gimbal, and 3 vacuum Raptor engines that are essentially fixed. Whether nominal or abort, all landings must be made at a feasible weight, that being set by engine thrust and a factor 1.5 thrust to weight ratio.**

*The tankers are presumed to be cargo/passenger Starships flown at zero payload, just as before.*
For flight safety purposes,
it must be assumed that there is one engine “out”, so the landings must be made on the thrust of
two landing engines, not all three. The sea level Raptor is listed as “2 MN
thrust” on the Spacex website.

**If you are carrying 149 tons of payload, you***That sets max landing mass of any of the Starships at 271 metric tons, which is 120 tons inert plus 151 tons of propellant-plus-payload.*__cannot have more than 2 tons of propellant remaining__after landing, and still meet that thrust/weight criterion.
I ran this scenario for the case of departing from circular
LEO. I did not investigate elliptic LEO
departure, although it would have a
similar effect as before. I ran the
nominal 149 metric ton payload that is deliverable to circular LEO, and determined a two-tanker solution.

**I then bounded the problem at zero payload in the cargo Starship, to see if a one-tanker solution was feasible in that limiting case. It was not,***One fly-along tanker is not feasible.**therefore there are*__only two-tanker solutions__for this refueling-along-the-way scenario.
The results I got for nominal circular-LEO payload delivery
are

**. The image of the spreadsheet is***illustrated in summary in Figure A***. I don’t really see a vast improvement over refueling in LLO before the descent to the moon’s surface, although a little less propellant need be ferried up to LEO (2700-2800 tons vs 3300-3400), if you do it as a fly-along refueling. You do incur the simultaneous departure requirement, which you do not incur, if you refuel in LLO. Refueling in LLO***given in Figure B*__after__the surface excursion requires less propellant be sent to LEO, but carries far less payload (by roughly a factor of 3).
In LLO, one vehicle stays
in circular LLO with the shorter period,
the other must go elliptical at the same perilune, and a slightly longer period. After a few orbits, the two will be in the same place at the same
time at perilune, and the vehicle in
elliptic orbit can circularize. And THAT
is the rendezvous technique! This cannot
work unless one of the vehicles is on a different trajectory that repeatedly
converges with the trajectory of the other vehicle in a short time. You simply cannot do that on the transfer
trajectory from Earth to moon, because
you will arrive at the moon before your orbits converge again.

In the event of an abort because of non-simultaneous
departures, both the cargo Starship and
the tankers will be overweight to land,
particularly the tankers. These
vehicles will have to vent some propellant to space (~15 tons for the cargo
Starship, and about 190 tons for the two
tankers) before entering Earth’s atmosphere.
If they do not, they will crash
due to insufficient thrust, required to
land that weight. And that is what the
271 metric ton landing mass requirement prevents.

Figure B – Spreadsheet Image of the Fly-Along Refueling
Scenario Analysis from Circular LEO

If im calculating it right.

ReplyDelete120t starship needs to be 280t on the moon, to get back to earth.

With 150t of additional payload, it needs to be 910t from earth escape to land on the moon.

With full 1200t of fuel and 150t payload from LEO, it still weighs 600 at earth escape. S

it needs 310t of fuel.

How about one tanker on a free return trajectory?

If I understand your intent correctly, you want to refuel not in LLO but on the free return to Earth. The timing of the launch to enable the rendezvous would be extremely critical. You could not even be a couple of minutes off. Being able to rendezvous reliably is why I did it in LLO, and also why I put a rendezvous budget on the tankers. It takes a difference in orbital periods to accomplish. -- GW

DeleteHow feasible/infeasible are multiple spacecraft performing the same departure burns while being say a few kilometers apart?

ReplyDeleteIts more of a software / navigation problem than hardware.

Coast phase is 3 days , Apollo program did an undocking, docking during that time.

Then again, it only reduces fuel requirements to LEO by a couple hundred tons.

Then I do not understand what it is that you propose. A tanker flying by the moon and a cargo craft launching from the moon have NOT rendezvoused. Nor can they,

Deleteexcepting by a launch timed and a burn timed to the exact second. Putting them on the same trajectory toward Earth does NOTHING to effect such a rendezvous, so the 3 day travel time is irrelevant to that picture. You have to be within a couple of miles, not hundreds of km, to just "thruster" your way together. Sorry, that's just orbital mechanics. -- GW

I mean departing from LEO.

DeleteI haven't thought about that scenario before. Both vehicles would have to depart simultaneously from LEO in order to be "pre-rendezvoused" to make it work. There would have to be some sort of abort in case this simultaneous departure failed. It's lot easier to handle the rendezvous without the simultaneous time constraint, if you do it in a closed, short-period orbit situation, one circular, one elliptical, sharing either a periapsis or an apoapsis. -- GW

DeleteTo clarify. Theres a fully fueled tanker and fully fueled starship in leo, same orbit, physically within say 10km.

DeleteThey both do trans lunar injection burns at the same time.

Wouldnt they end up in a simmilar enough trajectory for the tanker to be able to match and dock with low dv usage?

It does take 3 days to get to the moon.

Abort could be at any time before elliptical orbit reaches lunar sphere of influence, after that it has to be a free return trajectory.

Actually, refueling in LLO has less abort options.

Rok: Take a look at the 7-18-20 update appended to the article. I looked at your idea, and found it feasible, and comparable, to refueling in LLO before the descent. It is a two-tanker scenario, you cannot do it with one. Three vessels must depart simultaneously. -- GW

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