Some recent comments have got me wondering about this

Many things have landed on Mars. I'm not sure what fraction have entered orbit first versus direct deceleration from interplanetary trajectory to touch down.

In each case, what fraction of their velocity do they remove propulsively versus using atmospheric drag

I'm guessing it's 50% for interplanetary to landing and of order 10% for orbit to landing, but I could be way off.

Obviously answers will have to be rough since they may need to summarize several missions, and velocity increases once Mars' gravity kicks in.


1 Answer 1


Taking Mars Pathfinder and Viking 1 as examples:

Mars Pathfinder was a direct entry at 7600 m/s and removed about 0.7-0.8% of that propulsively. Parachute deployment was at 360-450 m/s, and landing rocket ignition at 52-64 m/s which slowed the vehicle to 0-25 m/s before cutting the bridle: https://mars.nasa.gov/MPF/mpf/edl/edl1.html

Viking 1 landed from orbit, removing about 5% of its velocity propulsively. The deorbit burn was 180 m/s, entry velocity was 4580 m/s, aerodynamic braking was used down to 60 m/s, and from there it decelerated propulsively and touched down at 2.4 m/s: https://nssdc.gsfc.nasa.gov/nmc/spacecraft/display.action?id=1975-075C, https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19760018033.pdf

  • $\begingroup$ I'm surprised the aerodynamic braking velocities are that low. $\endgroup$
    – ikrase
    Commented Apr 23, 2020 at 4:53
  • $\begingroup$ To the question "... what fraction of their velocity do they remove propulsively?" can you explicitly mention some fractions? Thanks! $\endgroup$
    – uhoh
    Commented Apr 23, 2020 at 4:55
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    $\begingroup$ Of course, given the tiny size of the final landing burn compared to the velocity shed by aerobraking, a perfectly valid answer to @uhoh's question would be that they remove all of their entry velocity by aerobraking — but then gain back some extra speed from falling under gravity, the last 60 m/s or so of which which they have to cancel using rockets because, even with a parachute, that's how high their terminal velocity on Mars is. $\endgroup$ Commented Apr 23, 2020 at 13:41
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    $\begingroup$ @IlmariKaronen: That's true for these, but generally larger payloads will take more, since it becomes harder to keep the ballistic coefficient down. Red Dragon would have used an order of magnitude more propulsive delta-v, with no parachutes and its engines starting up while still supersonic, something NASA avoided due to lack of knowledge of how it'd perform. $\endgroup$ Commented Apr 23, 2020 at 15:23
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    $\begingroup$ @Klaycon, The problem here is the OP was weak, in that it didn't explain how the fraction should be calculated (I'm not sure there is a meaningful way of expressing it as a fraction) $\endgroup$
    – user20636
    Commented Apr 23, 2020 at 20:30

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