Calculating Fuel-Payload Ratio for Martian Rocket Lift-off

[jbriggs444]

Oh, that was a surprise! Does that also mean that if we assume atmospheric breaking on both ends of the travel, it will be the same fuel-payload ratios both ways? And that that is the very simple answer to my question?

No. It is asymmetric. You gain a big advantage with atmospheric braking on Earth (high escape velocity/orbital velocity and good available atmosphere). Not so much on Mars (much lower escape velocity/orbital velocity and thinner atmosphere) and none at all on the moon (even lower escape velocity/orbital velocity and no atmosphere).

So there is some advantage to be had on the outbound trip but a much greater advantage to be had on the return.

 

[Flisp]

But if we would go from orbit to orbit (outside atmosphere) it would be symmetrical?

 

[jbriggs444]

But if we would go from orbit to orbit (outside atmosphere) it would be symmetrical?

Yes, indeed.

 

[Flisp]

Yes, indeed.

Great, thank you. And if I want a more accurate comparisson I have to piece the trips together the way you discribed it before. Earth to orbit, orbit to orbit, orbit to Mars surface and then back.

 

[jbriggs444]

Great, thank you. And if I want a more accurate comparisson I have to piece the trips together the way you discribed it before. Earth to orbit, orbit to orbit, orbit to Mars surface and then back.

Credit where credit is due, I think that was @Janus, not me.

 

[Flisp]

Well, thanks to you all. Great forum! If I publish something on my blog I will mention you guys just the way I did with ny article about interstellar flight. Owe you a lot.

 

[Janus]

What? How can that be? The fuel-payload ratio was certainly not the same both ways when we flew to the moon!

In terms of leaving Earth orbit, to arriving at Moon orbit, compared to leaving Moon orbit and returning to Earth orbit, it was. The difference between going to and returning from the Moon is that with going to the Moon, part of the mass of what would be considered "payload" was due to the fuel needed for the return trip.
From LEO to the Moon or back the other way requires a delta V of ~4 km/s. The Service module had a ISP of 314s, or an exhaust velocity of ~3077 m/s. This would require a fuel to payload ratio of ~2.7 ( 2.7 kg of fuel for every kg of payload. So this mean's that for every kg you wanted to return from the Moon, you had sent an additional 2.7 kg in fuel to the moon. So leaving LEO, you had to start with ~10 kg of fuel for every kg you would want to return from the Moon. ( For the actual Moon missions, we left quite a bit of the mass in the form of the LEM back at the Moon, so this reduced the return trip fuel cost some, allowing us to trim some of the total fuel cost. )

Apollo was of course able to save on return fuel by using aero-braking while returning from LEO. If they had been forced to use a powered landing like the takeoff was, this would have required a much different approach. Possibly by docking the Command module to a fueled lander module parked in LEO (no sense in carrying fuel that you don't need to all the way to the Moon.

 

[Flisp]

From LEO to the Moon or back the other way requires a delta V of ~4 km/s. The Service module had a ISP of 314s, or an exhaust velocity of ~3077 m/s...

Thank you, Janus, great to have some real numbers to check against.