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A planet is in orbit around a larger planet, and I want to push it out of orbit.

The planet has an average radius of 250 miles, and a mass of about 1.47x10^22 kg. The planet it is orbiting has a radius of about 3000 miles, and is about 1.5 times as dense. Assume they have a proportionate gravitational pull. I'm plucking the figures out of the air partly, but they give a general idea of size.

What force would I realistically need to break the planet free from it's orbit? Assume I have access to several tug-spaceships, what power would they need to be able to shift this?

You can include hard science to show what you've based your answer on, but I'm looking for some ideas of size of the ships and the power output required so I can make sure they are realistic.

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    $\begingroup$ How far away are the two planets from one another? $\endgroup$ Oct 24, 2016 at 21:22
  • $\begingroup$ Am I correct in saying the density of that smaller planet has a density of 53 tonnes per cubic metre? That seems a bit much. $\endgroup$
    – user6511
    Oct 24, 2016 at 21:27
  • $\begingroup$ Assume a non-rotating Earth moves at constant velocity around a star your ship must wrestle the enormous total kinetic energy to kick out the planet, and constantly applying additional energy to ensure the planet is banned forever. When the star finally goes supernova the planet goes bye bye! $\endgroup$
    – user6760
    Oct 25, 2016 at 0:38
  • $\begingroup$ worldbuilding.stackexchange.com/questions/59574/… $\endgroup$
    – AutoBaker
    Oct 25, 2016 at 19:27
  • $\begingroup$ I've worded the question differently as the answers i'm getting indicate it's likely to be impossible!! See the link above this comment. $\endgroup$
    – AutoBaker
    Oct 25, 2016 at 19:28

4 Answers 4


It cannot be done realistically.

The most efficient energy source is matter-antimatter reaction allowing us to get 100% possible efficiency, matter will be destroyed and create massive amounts of gamma rays.

Matter m1 gives energy: E = m*c^2
Matter m2 must be accelerated to velocity v from start velocity v0:
E = (1/2)m2(v- v0)^2

Solution: sqrt(m1/m2) = 1/sqrt(2) * (v-v0)/c.

So you need the amount of matter which is equal to the square root ratio of the desired velocity difference and light speed. While it sounds high, let's look at the numbers: You need escape velocity which will be always in the km/s range. Let's say you need 5 km/s difference.

So with your estimated 1.47x10^22 kg mass the necessary ratio of antimatter is 2e-10. This means 2x10^12 kg of antimatter, so even if your antimatter had the density of uranium, it would still be a sphere of approximately 300 m. Only the necessary juice.

The other thing is that you cannot safely radiate away this amount of energy. One billet of antimatter is able to destroy a capital city. Any planet in the path of your radiated gamma rays will be roasted. So no, celestial objects are by far too heavy, even for sophisticated civilisations.

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    $\begingroup$ Yet celestial objects do change orbits over time, and planets can even be ejected from their native solar systems. All this answer says is that a brute force approach doesn't work well. $\endgroup$
    – jamesqf
    Oct 25, 2016 at 4:42
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    $\begingroup$ Yes, celestial objects. All those interactions are happening with objects of comparable (planet) or much greater size (sun). The barycenter of all involved objects moves still with constant speed, without collision only impulse and angular momentum are changed. Because the objects are of comparable size, their exchange of impulse and a.momentum are noticeable. It is a sign of extreme efficiency that you only need less than a billionth of the mass to achieve the same effect, but a billionth of a planet is still unbelievable large. So as long you have no death star available.... $\endgroup$ Oct 25, 2016 at 15:24

Basic orbital mechanics: an orbit has a specific energy, to change the orbit, one has to either increase (lift) or decrease (lower) that energy and thus change the orbit in the same manner.

Planets are heavy, moons (as you suggested), are too very heavy. I can't give numbers here, but Scott Manley can, and he does here. Even as KSP has densities times 10 and radii divided by 10 in comparison to the real world, it can possibly help in estimating what you need: a HUGE load of fuel.


You would need to increase the orbital speed of that moon to exceed the escape velocity at that distance from the main planet. It greatly depends on the size of that orbit, so your numbers aren't sufficient.

When looking on Earth & Moon as examples, it seems that you'd need ~300 m/s delta-v to do that; and assuming chemical rockets you would need to use fuel with something like 8% of the Moon's mass to break it free, which cannot be achieved even by e.g. strip-mining the whole Earth's crust and converting it to rocket fuel. With ion thrusters you might need only 1% of Moon's mass as propellant, but you'd need enough power to expel that 1% of Moon's mass with these ion thrusters which isn't really feasible, things like earth-sized solar panels or using up all our uranium supplies are not sufficient to do it quickly.

Of course, the numbers may vary depending on tech and circumstances, but the rough ballpark to accelerate your mass is something like 10^27 J. So, it all depends on the timeframe. Converting the whole moon surface to effectively a solar-powered spaceship engine could do it over many thousands of years. A Kardashev type-2 civilization that is advanced enough to harness all the power of a star (not the tiny fraction that is covered by earth) could do it with some seconds of that output; but building something that can do that is a much harder task than just breaking some moon away.

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    $\begingroup$ Planetary engine was also my first idea :) $\endgroup$
    – Sonic
    Oct 25, 2016 at 13:22
  • $\begingroup$ How exactly do you come with 300 m/s delta-v ? It is sqrt(2*G*M/distance) for Earth-Moon which gives us 1440 m/s escape velocity. If we assume chemical reaction with hydrogen/oxygen, we get 140 MJ per kilogram which is, as you said, 1% of moon mass. And do not forget, you need not only energy (melting the planet does not move it), you need impulse to propel it. $\endgroup$ Oct 25, 2016 at 15:54
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    $\begingroup$ @ThorstenS the average orbital velocity of the moon is 1000+ m/s so you need to add 300-400m/s delta-v, depending on how exactly you'd make the burns; and getting that amount of delta-v in a single stage with the impulse of chemical rockets requires ~8% of the mass (of that single stage, no matter if it's a capsule or a moon) as fuel/propellant; doing that with impulse of ion thrusters uses up ~1% of the mass as propellant, assuming that you get the energy from some other source (solar/nuclear/magic/whatever). It may be 50% more or less, I was aiming for an order of magnitude estimate. $\endgroup$
    – Peteris
    Oct 25, 2016 at 16:17
  • $\begingroup$ @Peteris Ah, ok, now I understand it. I was irritated about 300 m/s, because 1440 m/s escape velocity minus 1022 m/s orbital velocity is still more than 400 m/s. No further objections from me. $\endgroup$ Oct 25, 2016 at 16:26
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    $\begingroup$ It's worth noting that "applying" 300m/s of delta-v to something with the structural integrity of Earth's moon in a short interval is going to be incredibly destructive to the moon, regardless of the Kardashev level of the civilization. $\endgroup$
    – jdunlop
    Dec 7, 2020 at 22:43

Simple: Speed up the moon so that it gets out of the planet's orbit. The moon has a density of 54629 kg/m^3, and we can find that the planet it orbits has a mass of 3.8110^25 kg. Those are some dense worlds, like what's up with them? Did they form around neutron stars or something? Anyways, let's assume that the moon has an orbital period of 30 Earth days, and with zero eccentricity, they moon would be orbiting at 756095 km. That puts the moon's escape velocity at 2.59 km/s. The orbital velocity is 1.83 km/s, so we're more than halfway there. We will need to add 4.2510^27 J of kinetic energy to the system to get the required velocity.

So yeah, it is possible, but it will require a lot of energy. And that's assuming all the energy put in goes into pushing the moon out.

  • $\begingroup$ You have calculated how much energy is needed to push the satellite out; which is nice, but of course it was known from the very beginning that some amount of energy will do the trick. It have not shown that a spaceship can deliver that energy; and this is what the question was asking. $\endgroup$
    – AlexP
    Dec 8, 2020 at 2:32
  • $\begingroup$ You'd need a pretty big spaceship. Or one that works with zero-point energy or some other sort of unproven concept. $\endgroup$ Dec 8, 2020 at 5:09

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