Imagine the following scenarios:

A. We have a spaceship at x distance from a star. It faces directly away from the star, and fires its engines such that it remains at exactly the same distance from the star until it runs out of fuel. At this point, it falls into the star.

B. We have a spaceship at x distance from a star. It faces directly away from the star, and fires its engines such that it depletes the fuel extremely quickly - and as a result - gains enough speed to reach escape velocity.

So let's compare the two scenarios:

In scenario A, the final state is that the star gains the momentum from the fuel fired, plus the momentum of the ship once it hits the star.

In scenario B, the final state is that the star gains the momentum from the fuel fired - however, the ship now has a large amount of momentum travelling away from the star.

We can take things to the extreme and say in scenario A, the ship stays in a fixed position 1km away from the star firing its engines for 10 years. In scenario B, the ship depletes its fuel in 10 seconds. Here, it's clear the differences in momentum between the two ships would be significant - far outweighing the momentum transferred to the star by the ship on collision.

So where does this additional energy go?

And what if we make this more complicated and repeat the same experiment.. but now, we have two identical ships; one on each side of the star. Thus, the final momentum of the star is unchanged - it only gains mass from the fuel and the ships.

In scenario A:

The ship uses all its fuel to stay 1km away from the star. In the process, it transfers momentum to the star from the expelled propellant. Once the fuel is exhausted, the momentum of the ship gained from acellerating towards the star is transferred to the star.

Thus, the star's final momentum is -fuel -ship.

In scenario B:

The ship immediately uses all its fuel to escape the star. In the process, it transfers momentum to the star from the expelled propellant. It then escapes with a large amount of momentum.

Thus, the star's final momentum is -fuel, and the ship's final momentum is +fuel.

Both ships used the same amount of energy. My assertion is that the momentum of ShipA when it collides with the star is less than the momentum of ShipB as it escapes. If both input the same amount of energy into the system, how can the final momentums not be equal?

  • $\begingroup$ Where is this "additional energy"? I can understand your first seven paragraphs, but don't see any additional energy. $\endgroup$
    – Allure
    Commented Jul 19, 2019 at 5:36
  • $\begingroup$ @Allure The additional energy I'm referring to is the discrepancy between the energy of the escaping ship, and the energy transferred to the star from the in-falling ship. If the ships start very close to the star, the in-falling ship won't have much momentum at the point it collides with the star. At least, not compared to the escaping ship $\endgroup$
    – Rob
    Commented Jul 19, 2019 at 5:45
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    $\begingroup$ I still don't see any additional energy. Are you using "momentum" and "energy" interchangeably? They're not the same concept. $\endgroup$
    – Allure
    Commented Jul 19, 2019 at 5:52
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    $\begingroup$ Rob, momentum and energy are not the same thing. $\endgroup$ Commented Jul 19, 2019 at 6:09
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    $\begingroup$ @MichaelWalsby This scenario is a purely newtonian/kinematic problem. I see no reason to invoke EM here. $\endgroup$ Commented Jul 19, 2019 at 7:43

3 Answers 3


I am confused (and I think you may be confused) about whether you are asking about energy or momentum. However the momentum aspects are simple. Let's assume:

  • all the fuel ejected by the spacecraft hits the star;
  • the spacecraft and star are initially momentarily at rest in some inertial frame, and all velocities are measured with respect to this frame;
  • Newtonian gravitation;
  • all collisions inelastic (fuel & spacecraft stick to the star on collision);
  • everything happens along a line, so I will use scalars when I mean vectors.

Let the masses be:

  • spacecraft, without fuel, $m_s$;
  • fuel $m_f$;
  • star $M$;

Then the total momentum in the initial state is trivially $0$.

Final state 1: spacecraft and all its fuel hits the star. Final velocity of the combined star, spacecraft & fuel is $V = 0$ by conservation of momentum.

Final state 2: spacecraft escapes to infinity, asymptotic final velocity of spacecraft $v$, of star + fuel $V$. By conservation of momentum:

$$v m_s + V(M + m_f) = 0$$

& hence

$$V = -v\frac{m_s}{M + m_f}$$

& it's as simple as that.

Note that in the second case I am computing the asymptotic velocities: the velocities after the spacecraft has escaped to infinity. However, in fact, once all the fuel has hit the star the expression is good at any time after that, although the two velocities change over time of course.


If I understand your question right, we don't need a star or spaceship, we just need a gravitational field. Let's imagine we're on the surface of the Earth and we have a VTOL (vertical take off and landing) aircraft. It could hover or it could take off. The question then asks how it's possible that the two "final" states could be so different: one ends with the VTOL aircraft crashing to the ground, the other ends with it far away, and both have used up all their fuel.

The answer is: why can't they? In the scenario you describe, where the fuel's energy is going to is to ejecting the exhaust. Giving these exhaust particles speed requires force, and the reaction force is what allows the VTOL aircraft to hover. If you kick the exhaust particles harder, or if you kick more of them, then you need more force, and so generate more reaction force. This lets the VTOL aircraft take off.

The total energy received by the exhaust particles in both cases equals the total energy contained by the fuel. It's just that in one case the exhaust receives all the energy at once and in the other it's received slower.

You might be able to get a more intuitive picture of this by imagining a battery-operated fan. If you set the speed to slow, you'll feel only a slight breeze, but the battery will last longer. That's natural: the slight breeze takes less energy to produce, so the battery (which contains a fixed amount of energy) will take longer to deplete. Similarly, if you set the speed up, the reverse happens. Analogously, in the case of the VTOL aircraft, the slight breeze suffices to let it hover, but not to let it take off.

So there's no additional energy and no problem.

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    $\begingroup$ This is just wrong. Suppose the VTOL uses up all its fuel, and either lands on top of a cliff or it hovers and then descends to the ground. The energy in the exhaust in the first case has to contain slightly less energy than in the second case, because some of the energy went into lifting the craft to the top of the cliff. $\endgroup$ Commented Oct 29, 2019 at 20:46

I think I misunderstood your question - you're asking about a rocket with variable thrust, and comparing two cases:

  • Escaping the primary by using all the fuel quickly.
  • Hovering over the primary until all its fuel is gone.

You're wondering about the apparent difference in energy in the two final states.

I think the Tsiolkovsky Rocket equation might still apply - but one of the terms (rocket mass) is not what you think it is...

Where is the difference in energy? I don't think there is a difference, it's just hard to see where the energy has gone in one case:

  • When escaping, we end up with a rocket at infinity so it is pretty clear that all the energy is in the rocket (actually, some of it is in the exhaust gases as well).
  • When hovering, it looks like the energy has disappeared since we end up where we started - hovering at some fixed point above the primary (and about to fall!). However, we actually transferred the thrust to the primary via the gravitational field during the burn. We effectively towed the primary along for a bit. So the primary now has accelerated. Its velocity is tiny, but it is not zero and there is the missing kinetic energy.

Have you seen Wandering Earth?

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    $\begingroup$ Suppose I start trying to climb up the down escalator. Your analysis will say that whether I succeed depends only on how much energy I use. But if I climb at the same rate that the escalator is going down, I will never get to the top no matter how much energy I use, while if I run up, I will get to the top using a finite amount of energy. $\endgroup$ Commented Oct 31, 2019 at 10:22

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