Why does a rocket have far more payload capacity when placing an object in LEO compared to when it's placing the object in farther orbits, like on the Moon? If a rocket has sufficient thrust to take an heavier object into space, why wouldn't some additional stages of thrust allow it to take the same payload to the Moon?

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    $\begingroup$ The relationship between delta-v (roughly correlated to altitude gain) and fuel required is non-linear. If you want to lift more fuel into orbit, then you need to burn a lot more fuel to get that fuel into orbit. It's called the tyranny of the rocket equation. You don't just need the fuel to get from LEO to lunar orbit, you also need the fuel to lift the fuel. That's why Lunar payload capacities are much lower than you would intuitively expect. $\endgroup$
    – UEFI
    Commented Nov 9, 2020 at 13:46
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    $\begingroup$ As far as the first stage is concerned, those "additional stages of thrust" are just payload, and directly subtract from the amount of actual payload you can get off the ground. $\endgroup$ Commented Nov 9, 2020 at 14:16
  • $\begingroup$ related, if note a dupe: Why are there such large differences in launcher payloads to higher orbits (GEO, Lunar) compared to LEO? $\endgroup$
    – Manu H
    Commented Nov 9, 2020 at 15:36
  • $\begingroup$ I think you may be interested by the Wikipedia articles using this file (mostly the article about delta V) $\endgroup$
    – Manu H
    Commented Nov 9, 2020 at 15:39
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    $\begingroup$ @UEFI Your comment does a good job of describing the problem intuitively. You should post it as an answer. $\endgroup$ Commented Nov 9, 2020 at 16:42

4 Answers 4


The other answers are correct, but might be too hard to grasp intuitively. The simplest way to understand this is to reason the opposite way.

You have a rocket that can fly to the Moon. At some point in its flight, it already has enough speed to orbit the Earth, and some fuel to propel it to the Moon. If, instead of having extra fuel for the remainder of the journey, you'd put the same mass as payload, you'd have exactly this: more payload in LEO.

  • $\begingroup$ Perhaps this is closest to what I was looking for. By the way, I have no intention to orbit earth in LEO, its just the altitude I was referring to. $\endgroup$
    – Niranjan
    Commented Nov 11, 2020 at 18:14
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    $\begingroup$ I like this one, it's the most straightforward way to state what I was going for in my answer, but avoids all the complications. $\endgroup$ Commented Nov 11, 2020 at 18:53
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    $\begingroup$ @Niranjan simply reaching the altitude of the lower Earth orbit (say, 200-400 km) would require a lot less fuel than actually getting into stable orbit [obligatory xkcd: what-if.xkcd.com/58 ] But that's rather pointless: your whole payload would fall back to the ground. Anyway, even if you don't actually orbit the Earth on your way to the Moon (though all of the Apollos have, just to go through checklists and stuff), you must have the speed to do this at some point, because the speed to get to the Moon is higher. $\endgroup$
    – IMil
    Commented Nov 11, 2020 at 23:10

The short answer is: Tsiolkovsky rocket equation. You need some velocity to achieve some position (an orbit or a body) in space. Farther a position - more velocity. More velocity - more propellant mass, and this relation is not linear and not in favor of velocity.

$$\Delta v=v_e \ln(m_0/m_f)$$


$\Delta v$ - theoretical maximum increment of velocity,
$m_0$ - the initial mass, including tanks, engines, avionics, propellants and (of course) payload,
$m_f$ - the final mass, it can be payload only, depending of the rocket purpose and construction (payload can be very broad term, including a stage to fly to the Moon, Mars an so on, with it's own payload),
$v_e$ - the exhaust velocity of the selected type of the propellant for the selected type of the engine,
$\ln()$ - the natural logarithm.

If you add a stage, you add an initial mass, and yes, you can take the SAME payload to the Moon, but for the price of much heavier rocket. And it will be another rocket than ones for LEO. Or rocket may be the same but with less payload.

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    $\begingroup$ "for the price of much heavier rocket": which may not be able to lift itself off the pad. Realistically it's the other option, the mass for the added stage comes out of the payload. $\endgroup$ Commented Nov 9, 2020 at 13:33
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    $\begingroup$ If one is new to rocketry, it's worth noting that this effect is such a brutal dominating force that it is known as "the tyranny of the rocket equation." $\endgroup$
    – Cort Ammon
    Commented Nov 9, 2020 at 15:37
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    $\begingroup$ Indeed, the key is to recognize that "if rocket has sufficient thrust to take a heavier object into space", it does not have sufficient thrust to carry that object and additional stages into space. $\endgroup$ Commented Nov 9, 2020 at 18:09
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    $\begingroup$ Adding a stage may reduce the size of engine needed for what used to be the first stage. Until a rocket approaches orbital speeds, reducing the amount of thrust produced by an engine will increase the amount of time spent fighting gravity, thus increasing the total delta-V requirement, thus limiting how small an engine can be. Once a rocket is in any orbit outside the atmosphere, however, shrinking the engine would increase the amount of time needed to reach a higher orbit, but not the total delta-T required to get there. $\endgroup$
    – supercat
    Commented Nov 10, 2020 at 18:14

It perhaps become clearer when stating what rockets do. They change velocity. In space terms, that's delta-v.

A rocket stage can only change your velocity some limited amount. Different targets in space require different amounts of velocity change (Low orbit: 8km/s, low Moon orbit: 12km/s)

If your rocket stage can not give all the velocity change you need, a trick is needed. The trick is to replace some of the payload of the first rocket with another rocket, which can fire after the first one is spent.
The good news: You now have both the velocity change of the first rocket and the second rocket.
The bad news: Much of the payload is now the second rocket.

So in your example, when reaching low orbit, all the payload can be payload if you only intended to go that far. But if you need to go further to the Moon, extra velocity change is needed, so some of that payload must be an extra rocket.


Why is it easier to go halfway up a mountain than all the way to the top?

Assume that you are already in low Earth orbit. You have a payload X that you want to put in a higher orbit, say GEO. Then you can compute the deltaV needed to move that payload between orbits, and the fuel needed to create that deltaV.

But the fuel needed is not magically placed in LEO. You need to launch it (and probably the rocket stage that will use it) from the ground to LEO. That takes more fuel, which means you need either a bigger rocket, or several launches of smaller rockets.

  • $\begingroup$ You are logical, but I am looking forward to a small increase in velocity, just to escape the LEO, and not bothered about time required to reach the destination, hopefully it keeps moving towards the moon, with smaller thrusts for course correction etc. this is what I meant by "some additional thrust".. $\endgroup$
    – Niranjan
    Commented Nov 11, 2020 at 18:11
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    $\begingroup$ @Niranjan: Unfortunately, physics doesn't work like that. You need a specific amount of deltaV to transition from LEO to GEO (or translunar orbit, or whatever). It does not matter whether this deltaV is applied really slowly, as for instance with an ion engine, or quickly with a big rocket. If you apply more deltaV than that necessary amount, you will either overshoot, or will have to apply opposite deltaV to brake. $\endgroup$
    – jamesqf
    Commented Nov 13, 2020 at 2:18

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