For example, for the Apollo lunar modules, for landing on the Moon the weight of fuel was approximately 60% of the total weight (on Wiki 8 tons of the 15 tons). For departure from the surface, the weight of fuel was about 50% of the total weight (2.5 tons from 5 tons).

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    $\begingroup$ The amount of fuel you'll need to take off depends on the mass of your payload. Do you just want the percentage? $\endgroup$
    – Phiteros
    Apr 3, 2017 at 22:09
  • $\begingroup$ the fuel needed to land depends from the orbit velocity and the search for the right place to land. The LM had a 720 seconds burn. $\endgroup$ Apr 3, 2017 at 22:54
  • $\begingroup$ Just percentage. I would like to know at least approximately how much the rocket needs "refuel" to leave the surface and flew at low Mars orbit. $\endgroup$
    – Majkl
    Apr 4, 2017 at 7:33
  • $\begingroup$ It depends on the the engine efficiency, which will mostly depend on the fuel chosen. It would need around 3800 m/s $\endgroup$
    – user20636
    Nov 6, 2017 at 13:06

1 Answer 1


Assuming that you are talking about a propulsive Landing only, without parachutes, you would need around 3.8 kilometers per second of Delta V (from the map below). It should be noted that the amount of Delta V required to land on a body from orbit roughly equals the amount required to reach orbit. The Solar System - A subway map Source of image

To land: There is some air at Mars which will slow you down somewhat, so 3.8km/s of Delta V is a ballpark number which is higher than what it is in reality. As the drag due to air resistance will slow the vehicle down and therefore require less Delta V to land.

One major factor in knowing how much fuel you will actually need to perform the landing is the efficiency of the fuels. Taking into account the Apollo LEM which used Aerozine 50 and N2O4 and got an isp of 311 (3047m/s), it had around a 60% fuel mass to total mass ratio. But the moon requires much less delta V to land on than Mars. So a vehicle running the with the same fuels (N2O4 and Aerozine) would need a ratio of .72 fuel mass to total mass. However this ratio really depends on the efficiency of the engine, as a more efficient engine can have a lower ratio, but a less efficient engine will need a higher ratio. .72 which was calculated via the rocket equation does not take into account air resistance during landing (which will help in this case). So a lander would need at most a ratio of .72 for a propulsive landing only, neglecting air resistance.

To get to orbit from the surface: Since the amount of Delta V required to land from orbit is roughly the same to ascend from the surface to orbit, we will assume 3.8km/s of Delta V needed. It is helpful to not that this number is actually a lot closer to reality as air resistance will play much less of a role in ascent than descent. Let’s use the same fuels as the Apollo LEM once again and assume an efficiency of 311s (3047m/s).


The mass final in this case is around 2.8, giving a percentage of 72% (as mentioned earlier) for the fuel mass to total mass to reach Mars orbit. This is definitely lower than what would actually be used as this is the bare minimum to reach orbit, once in orbit the vehicle may need to rendezvous and use more fuel.

Once again, this ratio of .72 is only for the LEM descent engine. Use a more efficient engine and get lower ratios and vice versa.

  • $\begingroup$ What's the source of the delta-v "subway map" that you posted? (It looks like there's an attribution on it, but it's too blurry for me to read.) $\endgroup$ Nov 8, 2017 at 21:36
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    $\begingroup$ @MikeHarris ah yes I must have forgot to link this sorry. google.com/amp/s/amp.reddit.com/r/space/comments/29cxi6/… $\endgroup$ Nov 8, 2017 at 21:38
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    $\begingroup$ @MikeHarris I updated with a better copy of the subway map so that the attribution is more visible. $\endgroup$
    – called2voyage
    Nov 8, 2017 at 22:07
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    $\begingroup$ In their recent "making life interplanetary" presentation SpaceX claims that their reentry/landing trajectory at mars sheds 99% of the energy aerodynamically. So I don't think dV for landing roughly equals dV for starting is a very good approximation. $\endgroup$ Nov 11, 2017 at 7:56

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