# How much fuel is needed for the motor landing and taking off on mars?

For example, the Apollo lunar module landing on the moon was worth approximately 60% fuel of the total weight (on wiki 8 tons of the 15 tons). Departure from the surface was about 50% of fuel (2.5 tons from 5 tons).

• The amount of fuel you'll need to take off depends on the mass of your payload. Do you just want the percentage? – Phiteros Apr 3 '17 at 22:09
• 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. – Antonio Cipolla Apr 3 '17 at 22:54
• 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. – Majkl Apr 4 '17 at 7:33
• It depends on the the engine efficiency, which will mostly depend on the fuel chosen. It would need around 3800 m/s – JCRM Nov 6 '17 at 13:06

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. 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).

3800=3047ln(10/Mf)

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.

• 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.) – Mike Harris Nov 8 '17 at 21:36
• @MikeHarris ah yes I must have forgot to link this sorry. google.com/amp/s/amp.reddit.com/r/space/comments/29cxi6/… – Jake Blocker Nov 8 '17 at 21:38
• @MikeHarris I updated with a better copy of the subway map so that the attribution is more visible. – called2voyage Nov 8 '17 at 22:07
• 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. – Roman Reiner Nov 11 '17 at 7:56