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.
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, as indicated in Figure 1. 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.
Once in the vicinity of the moon, the easiest entry into low lunar orbit (LLO) is the figure eight trajectory into retrograde motion, as indicated in Figure 1. 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.
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, as indicated in Figure 1. But it does need to be co-planar with LLO.
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. This is also indicated in Figure 1, along with the appropriate factors to convert kinematic delta-vees to mass ratio-effective values.
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 indicated in Figure 2. 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.
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. 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. That was the value from 2019 and earlier, before the switch to stainless steel construction and the start of prototype manufacture and testing. 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. So I used 120 metric tons as “baseline”, with 100 tons as a “goal value”, in my calculations.
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.
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. None of it reaches the surface. 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 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 shown in Figure 3. 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. 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 59 tons of payload can go 1-way to the lunar surface from circular LEO, and 80 tons from elliptic LEO. 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.
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. From circular LEO, Starship can carry 149 metric tons of payload, starting less than full of propellant from LEO. It will need some 486 metric tons of propellant from 3 tankers 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 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. Only two tankers are needed in LLO to deliver 472 metric tons of propellant to the Starship there. 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 part of Figure 5.
That summary is:
Table 1 – Overall Results
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 85 tons inert thought possible in earlier years has proven to be infeasible. The 2020 thinking, per Musk himself, is 120 tons similar to the early prototypes, with 100 tons as a desired goal. This change in inert mass makes a huge difference in the achievable mass ratio out of the design, much more so than the change in propellant load.
Images of the actual spreadsheets are given in the following figures. Starship to LLO without any landings is covered in Figure 6, for both the circular and elliptic LEO cases. There are no tankers sent to LLO for this scenario.
The refill after landing scenario is covered in Figures 7 and 8. 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.
The refill before landing scenario is covered in Figures 9 and 10. 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.
Figure 10 – Spreadsheet Image: Starship Refilled Before Landing, From Elliptic LEO
Update 7-18-2020: 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. 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. 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.
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.
The tankers are presumed to be cargo/passenger Starships flown at zero payload, just as before. 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.
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. 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. If you are carrying 149 tons of payload, you 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. One fly-along tanker is not feasible. 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, 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 illustrated in summary in Figure A. The image of the spreadsheet is given in Figure B. 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 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