I've tried to get my head around this and have decided that, for the moment, I don't want to learn to solve logarithmic equations. Too much else going on. I was reading Mars Direct and got the impression from the very abbreviated info there that about 80% would be fuel. That surprised me because the gravity is so much weaker and the air is so thin on Mars. Is that right?


1 Answer 1


Here's a Map to the Solar System. It details, roughly how much $\Delta v$ you need to get from one place to the next.

Subway Map

You can take the rocket equation to quickly calculate your Fuel fraction for any given $\Delta v$. We take the basic form ($m_0$ starting mass; $m_1$ final mass; $v_\text{e}$ effective propellant exit velocity) $$ \Delta v = v_\text{e} \ln \frac {m_0} {m_1} $$ And introduce the fuel fraction $M_f=\frac{m_0-m_1}{m_0}$ $$ M_f = 1-\frac {m_1} {m_0}=1-e^{-\frac{\Delta V}{ v_\text{e}}} $$

(I was too tired to derive that by myself, so I took it out of Wikipedia. Not that it's hard to do...)

Mars Direct wants to use in-situ-produced methane and oxygen for the return. The theoretical ISP of this fuel combination is 368.9s. That equals an effective exhaust velocity of $368.9\text{s}\cdot \text{g}=3618\frac{\text{m}}{\text{s}}$.

To return to earth we need at least an intercept. Dennis Tito wants to return from mars and plunge right into the earth's atmosphere (yes, I know it's more complicated), so if you fly that way, you only need to intercept earth, not go into an orbit before you land.

That way your $\Delta v$ would be $(3800+1400+1060)\frac{\text{m}}{\text{s}}=6260\frac{\text{m}}{\text{s}}$.

Pluck that into the equation above and you get:

$$ M_f = 1-e^{-\frac{6260\frac{\text{m}}{\text{s}}}{3618\frac{\text{m}}{\text{s}}}}=0.82 $$

Since this a very ideal calculation, with a flatrate model for Mars air drag and gravity losses, perfect engine, etc, 80% is a bit optimistic. But then again, maybe the maker of that solar system map took a worse constellation for reference, than Mars Direct did. I don't know.

Source for the ISP: Modern Engineering for Design of Liquid-Propellant Rocket Engines (AKA The Huzel)

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    $\begingroup$ Man, that delta v map is excellent. I didn't see that one when i searched for them. $\endgroup$
    – kim holder
    Sep 10, 2014 at 23:32
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    $\begingroup$ The $g$ has nothing to do with Mars. It is simply a conversion factor between English units (pounds-force thrust per [pounds-mass propellant per second]) and Metric units (Newtons per [kg/s] == m/s) for Isp. $\endgroup$
    – Mark Adler
    Sep 11, 2014 at 1:19
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    $\begingroup$ I did factor in aerobraking. Moon assists give you nothing. Mars' moons are too small and I already assume that you can kill the whole $v_\infty$ by aerobraking. Sure there are assumptions to a subway map like that (as I pointed out in my answer). Some return maneuvers use Venus for a fly-by. That may work. Looking forward to your answer. $\endgroup$ Sep 11, 2014 at 10:43
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    $\begingroup$ No, it's not a typo. If you land, you will not need any impulse, because you can use the Atmosphere. It's so dense that in 1978 one of the pioneer Pioneer 13 probes survived an impact although it was not designed to and had had no parachute. If you land, you just need a tiny parachute. But the enormous atmosphere makes launching from Venus' surface almost impossible. $\endgroup$ Sep 12, 2014 at 11:24
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    $\begingroup$ Yes, just launching. Not only does the drag slow you down, but chemical engines are much less efficient in a high-pressure environment. The only reasonable way to launch from Venus in my view is to use a balloon to float to the upper atmosphere, and then start an engine there. $\endgroup$ Sep 12, 2014 at 12:14

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