To be more specific: Suppose a space station is orbiting in a cislunar orbit (about 1000 miles away from the Moon). Now you keep adding more mass (huge cargo) to the cislunar space station. The space station departs for Mars with that huge cargo (much more mass). There will be fuel needed for putting the ship on the trajectory to Mars and initial change of velocity. But will the fuel required be much higher or the same if the mass is added, say, 1000 times more? Thank you for asking for clarification.

  • $\begingroup$ ...I really don't see what the station being cislunar have to do with your question. $\endgroup$
    – SF.
    May 25, 2017 at 9:40

2 Answers 2


To a first approximation, the amount of fuel required to send a spacecraft on a given trajectory (i.e. at a given velocity) is proportional to the mass of that spacecraft. So if the craft loaded with cargo is 1000 times the mass of the craft without cargo, it will need 1000 times as much fuel to make the same flight to Mars.

The reason for this is the Tsiolkovsky rocket equation,

$$ \Delta v = v_\text{e} \ln \frac {m_0} {m_f} $$

Where $\Delta v$ is the required change in velocity, $v_\text{e}$ is the exhaust velocity of the rocket engine, $\ln$ is the natural logarithm operation, ${m_0}$ the initial mass of the spacecraft, and ${m_f}$ the final mass of the spacecraft after the fuel has been burned.

Shifting a spacecraft's trajectory from lunar orbit to Mars (or from Earth orbit to lunar orbit, or from Jupiter orbit to Saturn orbit, etc) requires a particular total change in the spacecraft's velocity which is independent of the spacecraft's mass; this is $\Delta v$. For a flight from lunar orbit to orbit around Mars the total required value is about 3400 m/s.

For a given engine design, in vacuum, the exhaust velocity is more or less constant, which is $v_\text{e}$. For high-thrust chemical rockets this will likely be either about 3100 m/s (hypergolic fuels) or 4500 m/s (hydrogen fuel).

That leaves the mass ratio ${m_0} / {m_f}$. Rearranging the equation, we get $$ \frac {m_0} {m_f} = e^{\Delta V\ / v_\text{e}} $$ or, for moon-Mars on hydrogen, $$ \frac {m_0} {m_f} = e^{3400 / 4500} = \frac {m_0} {m_f} = 2.128 $$

If the final mass -- spacecraft plus cargo -- is 100 tons, then the initial mass -- spacecraft, cargo, and fuel -- must be 212.8 tons, i.e. 112.8 tons of fuel must be carried. If the final mass is 1000 tons, then the initial mass must be 2128 tons; 1128 tons of fuel must be carried.

In practice, it won't be perfectly proportional. If this is the same spacecraft in both cases, probably 90 tons of the final dry mass consists of fuel tanks and structure, so in the first case you'll have about 10 tons of payload, and 910 tons of payload in the second case, but if you're expecting to haul different amounts of payload, you'll design a ship with modular fuel tankage, so you don't have to carry the penalty of huge empty fuel tanks when sending a light load to Mars.

  • $\begingroup$ That's a very rough approximation of the rocket equation :) $\endgroup$
    – Antzi
    May 25, 2017 at 3:20
  • $\begingroup$ Not that rough, if we assume that the cargo hold and fuel tankage is modular, so that you don't carry the full weight of the 1000 X tanks when not carrying cargo. In such a case the $m_0 / m_f$ ratio, and thus the delta-v, remains approximately the same. $\endgroup$ May 25, 2017 at 3:31
  • $\begingroup$ Thank you Russell. But even from a cislunar orbit (no drag from atmosphere or gravity), is the fuel required to set the spacecraft to Mars still proportional to the added mass? Why? $\endgroup$
    – DRSD
    May 25, 2017 at 3:41
  • $\begingroup$ I'll expand my answer with my reasoning. $\endgroup$ May 25, 2017 at 12:39

Here is a version of the rocket equation:

$Mf = 1 - e^{-delta v/v_{exhaust}}$

Mf is the mass fraction that is propellent. The rest of the mass fraction will be rocket and payload.

From EML2, delta V for Trans Mars Insertion (TMI) is about 1 km/s. Exhaust velocity of hydrogen/oxygen bi-propellent is about 4.4 km/s. Let's plug those numbers into the rocket equation:

$Mf = 1 - e^{-1/4.4}$
$Mf = 1 - .8$
$Mf = .2$

In this case propellent would be 1/5 the mass and the remaining 4/5 would be rocket and payload.

If rocket and payload is 4 tonnes, you'd need 1 tonne of propellent.
If rocket and payload is 400 tonnes, you'd need 100 tonnes of propellent.

Where mass savings might be realized is what needs to be lifted off from earth's surface. Let's take a look at what it takes to get from earth to low earth orbit. Earth to LEO is about 9 km/s. To minimize gravity loss higher thrust propellents like oxygen/rp-1 is used. This has about 3 km/s exhaust velocity.
$Mf = 1 - e^{9/3}$
$Mf = .95$

So to get to LEO a rocket and payload of 1 tonne would take 19 tonnes propellent.
A rocket and payload of 100 tonnes would need 1900 tonnes propellent

Earth's surface to EML2 is about 12.5 km/s, only a tad less than earth's surface to TMI. But if we could load up on propellent and life support consumables at EML2, a much lighter vehicle could be launched to EML2 vs luanching a fully stocked and fueled vehicle from earth's surface to TMI.

EML2 is about 2.5 km/s from the moon's surface. There may be rich volatile deposits at the lunar poles. A volatile rich asteroid parked in a loose lunar orbit would be even closer to EML2.


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