What would happen to the moon if we reduce its velocity slightly?

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    $\begingroup$ This might help. $\endgroup$
    – user243267
    Dec 19, 2019 at 15:31
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    $\begingroup$ There is an amount by which you could reduce the Moon's velocity that would cause it to collide with the Earth. It would need to be a good fraction of its velocity but it exists nonetheless. This is not an answer to your question, but just a sidenote. $\endgroup$
    – Charlie
    Dec 19, 2019 at 16:44
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    $\begingroup$ Three things come to mind regarding your question. First, can you put a bound on "slightly"? Second, you are aware that the Moon's orbit is getting higher, right? The tides provide a slight increase in tangential velocity which moves the moon to a higher orbit. Third, are you trying to find a way to crash the Moon into the Earth? Because that would be bad; I live on the Earth so please do not. $\endgroup$ Dec 20, 2019 at 7:22
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    $\begingroup$ This seems like another question that is easily answered by playing Kerbal Space Program. $\endgroup$
    – Rich
    Dec 20, 2019 at 15:29
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    $\begingroup$ This also might help. $\endgroup$
    – lvella
    Dec 20, 2019 at 16:30

2 Answers 2


No, the moon will not eventually fall to earth. Reducing the tangential velocity by a small amount will affect the orbital trajectory of the moon. Since the path followed by the moon is already elliptical ($e=0.00549$), the actual affect depends on where the tangential velocity is reduced. If reduced at the apogee, the orbit will become more elliptical but if it is reduced at the perigee, the orbit will become more circular.

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    $\begingroup$ Also see en.wikipedia.org/wiki/Hohmann_transfer_orbit $\endgroup$
    – PM 2Ring
    Dec 19, 2019 at 16:42
  • $\begingroup$ It would fall to earth if you dropped the tangential velocity to zero. That is, if the trajectory of the moon got close enough to the Earth that the two bodies would interfere. (Yes, I know the OP said "slightly"). $\endgroup$ Feb 13, 2020 at 2:18

No, not for any amount that could reasonably be called "slight".

The thinking error I see here is that I believe what you are imagining happening is the Moon to start "spiraling out of orbit", like a plane going down without its engines, if it is disturbed somehow. And that this is something one might often pick up from movies or other such casual, pop-culture media or references thereto, or simply again by very Earth-bound based intuition of how things work such as that moving vehicles like planes need constant control or they will crash, but that is not actually correct in astrodynamical situations.

The more correct intuition is this. In astrodynamics, no matter what sort of "pull" you feel when thinking about this, a small body orbiting a larger one in vacuum, under the sole influence of gravitational forces alone, assuming at least that larger one is very close to spherical or one is far enough away that it is "pointlike" enough, will never fall out of orbit by itself. "Nudging" it won't work either, because orbits are stable in the sense that there is an extremely wide range of parameters of speed and distance over which the two bodies will just keep on orbiting. Hence, the "nudge" will simply shift the shape of the orbit slightly.

Generally speaking, a nudge of the orbiter "from behind" will cause it to travel a bit further out, because it's now going a bit faster and hence can more "successfully" fight the gravity of its parent body before being pulled back. Conversely, a nudge the other way, slowing it down, will cause the orbit to shrink somewhat and it will fall closer to the parent body for the opposite reason. A nudge from an oblique angle will have an effect somewhere in between.

Also, note that when I say "shrink", I don't necessarily mean the orbit simply uniformly gets smaller. (Closed) orbits are ellipses, and such nudging will actually tend to stretch or squeeze the ellipse.

Of course, one might, then ask, what this "orbital decay" one may have heard of is. Well, orbital decay is what happens when you add some kind of friction or drag process into the system that causes it to steadily lose energy. In that case, the orbit will shrink, so in effect the orbiter "spirals" inward, and it will eventually crash into its parent body. In effect, what is happening is the object is being continually "nudged" by the drag in the direction exactly opposite its orbital motion, until it runs out of speed and finally crashes.

However, drag requires some kind of medium against which to rub, and the vacuum of space is just that: vacuum. This phenomenon is of most concern for something like the International Space Station, which is close enough to Earth that there is actually still some very tenuous atmosphere present around it and for which the slight drag it provides adds up over time. If there were enough friction in space to bring down something as remote, big, and heavy as the Moon (by comparison, at a distance of about 377 Mm from Earth's surface versus the ISS at 0.4 Mm(*), and with a mass, and hence inertia and resistance to being slowed down, of roughly $7 \times 10^{19}$ Mg, against a paltry 420 Mg for the station), it would have been brought down already long ago - and likely our Universe would have to be quite different. Moreover, even with friction present, "nudging" won't suddenly cause some sort of dramatic increase in the orbital decay: again, this goes back to what I said before about "stability". Indeed, regarding the ISS - and also satellites, too - "nudging" is actually used on purpose to maintain the orbit and counteract the effects of drag-induced orbital decay!

(*) Regarding the full extent of Earth's "atmosphere", you might be interested to know that the very thinnest part, the exosphere, also called the "geocorona", extends to at least 10 Mm from the surface, if not even as far as 200 Mm - roughly if not a bit more than halfway between the Earth and Moon. It can even be imaged, see:


where it is illuminated by sunlight from behind, and using that Earth is ~13 Mm in diameter you can see the recordable part extends to around 40 Mm or so (~3 Earths). The 200 Mm boundary is where it gives way to the solar wind.

  • $\begingroup$ But moon is stealing energy from earth to get further away from it, surely there could exist a configuration where the opposite is the case? $\endgroup$ Dec 20, 2019 at 9:58
  • $\begingroup$ @TomášZato-ReinstateMonica No, there can't be. At least not for long as long as earth rotates the way it does. The energy that the moon "steals" from earth is its rotational energy, because the earth is rotating much faster than the moon is orbiting. When the earth and moon were formed in a fully apocalyptic cosmic collision (en.wikipedia.org/wiki/Theia_(planet) ), the earth was rotating much faster than today, and the moon was orbiting very much closer than today. The moon would only spiral down to impact if the earth were rotating in the opposite direction, and pretty fast at that. $\endgroup$ Dec 20, 2019 at 11:17
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    $\begingroup$ @cmaster-reinstatemonica: The Moon would also (slowly) spiral closer to the Earth if it started with an orbital period of less than one Earth day. But it was never that close to begin with. The geostationary orbit is an unstable equilibrium, with tidal forces pushing things orbiting inside it inwards and things outside it outwards. Notably, Mars's moon Phobos orbits below the stationary orbit around Mars, and is thus slowly spiraling inwards. If undisturbed, it's estimated to crash into Mars in about 40 million years. $\endgroup$ Dec 20, 2019 at 12:28
  • $\begingroup$ @IlmariKaronen I guess it depends on the mass of the moon whether the geostationary orbit is an unstable or stable equilibrium. If the moon is heavy enough, it would absorb so much angular momentum from earth as it goes higher that the earth's angular velocity drops faster than the orbiting moon's angular velocity. In this case, geostationary orbit is an attractor. On the other hand, if the moon is not heavy enough, earth's angular velocity would change slower than the moon's, and we have an unstable equilibrium as you said. $\endgroup$ Dec 20, 2019 at 13:16
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    $\begingroup$ @cmaster-reinstatemonica: I haven't done the math, but that makes intuitive sense. As the Earth's rotation slows down (or speeds up) due to tidal forces, the geostationary orbit also moves outwards (or inwards). If the Moon was massive enough (and close enough to stationary orbit to start with), the geostationary orbit would move faster than the Moon's orbit and catch up to it, and the end-state would be a mutual tidal lock like Pluto and Charon are in. $\endgroup$ Dec 20, 2019 at 18:02

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