The question What is the terminal velocity on Mars? and its answer has got me rethinking the BFR's reentry and propulsive landing simulation/animation shown in the recent SpaceX video First Private Passenger on Lunar BFR Mission starting at about 41:19 and again at 43:07.

I've made a GIF of the descent between about 5 and 1 kilometers below.

Does this represent a landing on the Moon, or Mars, or Earth, or are they all about the same? If they are substantially different (as "I suspect") then what would a landing on Mars be like? Unlike BFR or even Falcon-9 first stages, without much atmosphere on Mars, would it have to be propulsive all the way down? Or at least the last few km?

animated GIF from screenshots.

enter image description here

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    $\begingroup$ It's Musk, so any information is only as good as the next tweet, but previously it has been stated that most of the Mars speed reduction will be aerodynamic. Obviously for the moon it'll be all propulsive, and for Mars the transition from aerodynamc to propulsive will occur earlier than on Earth. $\endgroup$
    – user20636
    Commented Sep 24, 2018 at 7:19
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    $\begingroup$ It's obviously not the moon, since they are aerobraking most of the way through the descent, and the moon has no air. I kind of remember Elon Musk actually saying what it was, but I can't find it right now. My bet would be on Earth, though, since the whole press conference was about the "fly around the moon and land back on Earth" trip. $\endgroup$ Commented Sep 24, 2018 at 7:29
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    $\begingroup$ He does say "Earth" at one time when he (presumably) demonstrates the attitude of the ship off-screen using his own body, but it is not clear if he just uses that as a synonym for "ground" while going off-script: youtu.be/zu7WJD8vpAQ?t=43m11s $\endgroup$ Commented Sep 24, 2018 at 7:37
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    $\begingroup$ @JörgWMittag indeed I think so too. So a Mars landing would need to be quite different, but I don't have any feeling for how much deceleration would be available to this craft starting from low orbit, if it's much/most of it or only a small fraction and most would be propulsive. $\endgroup$
    – uhoh
    Commented Sep 24, 2018 at 8:33
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    $\begingroup$ @JörgWMittag: The Moon does, technically, have air. Not enough to aerobrake off of, though. $\endgroup$
    – Vikki
    Commented Feb 13, 2019 at 23:48

2 Answers 2


Mars entry follows a different series of maneuvers (a simulation using the previous fin layout: spacex.com/sites/spacex/files/mars-entry.mp4) that involves entering inverted and using downlift to stay within the atmosphere. They don't use parachutes, as they scale up poorly in both size and velocity (and development cost, for that matter).

As the video points out, over 99% of the energy was removed aerodynamically, but whereas the final retropropulsive landing burn is sub-sonic on Earth (circa mach 0.3) it will be supersonic (circa mach 2.4) on Mars.

For more on that, see:

             Start of "Landing Burn" (roughly) 

                   Earth        Mars
altitude (km):      0.9          2.4
speed (mach):       0.3          2.3

animated GIF from screenshots from the linked SpaceX mp4:

BFS landing SpaceX from screenshots

  • $\begingroup$ +n! This is precisely the answer to my question, thank you very much! $\endgroup$
    – uhoh
    Commented Sep 25, 2018 at 2:09
  • $\begingroup$ I hope you don't mind the edit, feel free to roll back (click "edited" to view the edit history) or ping me, or edit further. $\endgroup$
    – uhoh
    Commented Sep 25, 2018 at 2:55
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    $\begingroup$ Nah, it wasn't originally intended as an answer, and needed more work I didn't have time to give it. $\endgroup$ Commented Sep 25, 2018 at 3:17
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    $\begingroup$ So how much of it is slowed down by the atmosphere from the start? $\endgroup$
    – Muze
    Commented Sep 25, 2018 at 19:35
  • $\begingroup$ @Muze 2nd paragraph in the answer says "As the video points out, over 99% of the energy was removed aerodynamically..." Since kinetic energy is proportional to v^2 that means about 90% of the velocity. As the video starts the velocity is about 7,000 km/s and when the propulsion starts it's at about mach 2.3 which is about 10% of of 7,000 which checks out nicely. So the answer is 99% of the energy, and 90% of the velocity. $\endgroup$
    – uhoh
    Commented Sep 26, 2018 at 1:51

Elon made that simulation about BFR returning from the Lunar trip and Landing on Earth as clearly seen in the video at about 43 mins 11s(as mentioned in the comments on the question ). Aerodeceleration is possible and necessary only on bodies with an atmosphere i.e., Earth or Mars and very unlike the Moon. Moon would require very little propulsive deceleration as well as little fuel to get off from, unlike these massive planets with their large gravities. mars re-entry is riddled with challenges, especially one because of it's existent but thin atmosphere which constantly has severe weather conditions dust storms as studied in https://www.sciencedirect.com/science/article/pii/S0273117797002780. Robust control, propulsive along with large surfaces like parachutes for deceleration is the way to go. The fins will also have poorer performance in an atmosphere like mars and may need retro-thrusters to change attitude and enter Mars Atmosphere.

https://www.nap.edu/read/11220/chapter/5#27 explains all of this in much mroe detail if you want to know more. Hope to have answered sufficiently

  • $\begingroup$ The comment is by far the best answer so far to the main question. Can you make it into one? $\endgroup$ Commented Sep 24, 2018 at 11:05
  • $\begingroup$ I would ask that user to make it into one @steve linton $\endgroup$
    – Rajath Pai
    Commented Sep 25, 2018 at 0:20

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