In this link and on other occasions, SpaceX has indicated that the BFR would land on Mars, "top off the tanks" as the article says, and then launch back to Martian orbit.

SSTO doesn't work on Earth for the BFR for a number of reasons, including the density of the atmosphere and the higher gravity, but presumably, the (vastly) thinner atmosphere and .38G make SSTO possible.

Can someone put some math to my pop science speculation here? Also, can someone estimate what the fuel situation for the BFR is once it's in Martian orbit? Does it need to refuel for a TEI, or is it ready to head home on its own?


1 Answer 1


SSTO on Earth requires about 9400 m/s of ∆v depending on the exact acceleration profile and other factors.

Ascent from Mars' surface to low Mars orbit requires only about 3800 m/s, and from there it's another 2500 m/s to break out of Mars orbit and get on a trajectory to intercept Earth -- a total of 6300 m/s. Allocate another 300 m/s or so for touchdown on Earth, for 6600 m/s total.

With a methane engine like Raptor, delivering exhaust velocity around 3615 m/s, the rocket equation tells us that a mass ratio of about 6.2 is needed to manage that, which is relatively modest. The current numbers SpaceX is claiming give a better than 9:1 ratio with 50 tons of return payload, suggesting closer to 7900 m/s ∆v available. The additional 1300 m/s velocity budget could be used for a faster Earth return trajectory or to reduce reentry speed at Earth intercept or both. It appears the BFS can lift off from Mars, reach orbit, and do a TEI burn on a single tank of propellant with a very large return payload.

With no payload at all, the BFS should deliver about 9500 m/s of ∆v, just enough to do SSTO from Earth.

The vast majority of the difference between ∆v required for Earth and Mars SSTO is due to gravity, not atmosphere. For a rocket of Saturn V or BFR scale, the aerodynamic losses on ascent to LEO are on the order of half of one percent of the total ascent ∆v budget; in ascent from Mars, the loss is practically zero. Low Mars orbit velocity is less than half of LEO velocity: 3360 m/s instead of 7770 m/s, and the remaining difference is the "gravity loss" term.

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    $\begingroup$ If I remember correctly, I was under the impression that the BFS part of the BFR rocket is supposed to be able to barely make it to LEO on Earth as a SSTO. Perhaps this is with a lower payload? $\endgroup$
    – Dragongeek
    Commented Feb 27, 2018 at 10:43
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    $\begingroup$ Here is a Reddit AMA where Musk makes the claim: reddit.com/r/space/comments/76e79c/…. Apparently, even a Falcon 9 has the same capability. $\endgroup$ Commented Feb 27, 2018 at 16:01
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    $\begingroup$ SSTO is (relatively) easy. SSTO with useful payload is hard. BFS is 1185t fueled, 85t dry = 13.94 mass ratio, yielding ~9500 m/s, enough to reach LEO. $\endgroup$ Commented Feb 27, 2018 at 16:04
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    $\begingroup$ I do not know, but I suspect that 2500 m/s trans-Earth injection ∆v is for a minimum energy (long) return. I expect some of that extra fuel is spent on a faster passage (2-3 month passages seems like SpaceX's solution to microgravity and radiation concerns), and the higher entry speed is managed by the heatshield and very high drag possible from the vehicle entry attitude. $\endgroup$
    – Saiboogu
    Commented Feb 27, 2018 at 17:51
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    $\begingroup$ @Saiboogu That seems likely, will incorporate. $\endgroup$ Commented Feb 27, 2018 at 17:58

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