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I've seen a few times now that objects in orbit around the moon tend to crash into it in relatively short time frames. The orbits tend to become more elliptical until they intersect the surface, is the explanation that I recall. I understand the moon doesn't have a spherically symmetric mass distribution, so its gravitational field is also not spherically symmetric. But gravity is a conservative force. So I can see how a satellite would follow a wobbly orbit rather than an elliptical one, but I don't see how energy or angular momentum would be lost.

Is there a simple way to understand that?

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    $\begingroup$ The earth is also in the picture. You don't need to add or remove energy for an orbit to get elliptical to the point of perilune contacting the ground. $\endgroup$ – ikrase Jul 11 at 20:42
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The Moon is not a perfect sphere with homogenous density, there are mass concentrations (called the Mascons).

So the lunar orbit is not a perfect circle or ellipse. Low orbits change their shape without losing energy under the influence of the Mascons. If the orbit loses too much height at one point the object may crash into the lunar surface. The orbit is gaining height at another point so no energy is lost.

A circular orbit does not lose height while keeping its circular shape, it is transformed to an elongated ellipse.

"Lunar mascons make most low lunar orbits unstable," says Konopliv. As a satellite passes 50 or 60 miles overhead, the mascons pull it forward, back, left, right, or down, the exact direction and magnitude of the tugging depends on the satellite's trajectory. Absent any periodic boosts from onboard rockets to correct the orbit, most satellites released into low lunar orbits (under about 60 miles or 100 km) will eventually crash into the Moon. PFS-2 released by Apollo 16 was simply a dramatic worst-case example. But even its longer-lived predecessor PFS-1 (released by Apollo 15) literally bit the dust in January 1973 after less than a year and a half.

Source: https://science.nasa.gov/science-news/science-at-nasa/2006/06nov_loworbit

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    $\begingroup$ For those who aren't aware, "Mascon" is a portmanteau of Mass Concentration. $\endgroup$ – Erin Anne Jul 12 at 6:16
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Neither energy nor angular momentum is lost, although it can be transferred between the satellite and the moon. As the satellite's orbit gets more elliptical, the farthest point gets farther from the moon, and the nearest point gets closer. It changes from constant speed and constant radius to an orbit that varies from high speed/low radius to low speed/high radius. At some point, the low-radius part gets closer to the center than the surface, and a crash happens.

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    $\begingroup$ +1 for eccentricity, but I don't think "Neither energy nor angular momentum is lost." is really correct as a blanket statement. The inhomogeneities can apply all kinds of torques and angular momentum is constantly exchanged back and forth between the spacecraft and the Moon because of them. $\endgroup$ – uhoh Jul 12 at 3:24
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    $\begingroup$ What is true is that the orbit can change to intersect the surface without either energy or angular momentum being lost. $\endgroup$ – Steve Linton Jul 12 at 9:17
  • $\begingroup$ My intuition wants to tell me that if there's a mass concentration that causes a larger acceleration of the satellite toward it, that kinetic energy will still be there as the satellite pulls away. Do the inhomogeneities transfer angular momentum to the moon? $\endgroup$ – Greg Jul 12 at 17:24
  • $\begingroup$ I was vague both in what I wrote and in how I was thinking about it. I would assume the earth-moon-satellite system is very nearly a closed system, where we can neglect the sun, Jupiter, and certainly the other satellites. So then neither energy nor angular momentum is lost from that system, if I understand things correctly. $\endgroup$ – Mark Foskey Jul 12 at 18:42
  • $\begingroup$ ...edited to reflect my comment. $\endgroup$ – Mark Foskey Jul 12 at 18:45
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This is not a real answer, but it seems that in more than 3+1 dimensions we can't expect stable orbits. And the "inverse square" law for n dimensions is 1/r^(n-1). Which, for 3 dimensions plus time, give us 1/r^2.

What does that have to do with lunar orbits? Generally the 1/r^2 law is valid when the size of the source is small compared with the distance to the source. That is to say, the 1/r^2 law is always rigorously correct, but the interpretation of r might not be what you think it is when you're close to the source. That is true for gravity, for the exposure from a source of radioactive material (that's closer to my job), quite generally, and it's geometry, not specific to the properties of a radiating or gravitational source. If the moon has a non-spherical mass distribution, then it will have multipole moments, which means 1/r^4, 1/r^6, and so on. Which are not 1/r^2. That is, not stable. (I can't recall off-hand if odd moments are a thing.)

And, for reference, the diameter of the earth is about 8,000 miles, while an orbit could be around 100 miles. Or 200, or 400. At any rate, the distance of an orbit from the surface can be much smaller than the diameter of the body -- it's common for orbits to be not much larger than the size of the body. Which means you can't just assume a point source, you need to know something about the multipole moments, that is, the mass distribution.

So anything that deviates from 1/r^2 can't be considered stable. So lunar orbits are only stable if you are far enough away. The moon has a non-spherical mass distribution, so if you're close compared to the diameter of the moon, the orbit can't be stable.

I still haven't demonstrated to myself that angular momentum is transferred. It seems likely, since a non-spherical distribution gives the satellite something to grab on to, so to speak. But I haven't run simulations or anything to demonstrate it. So this is an incomplete answer. But it would seem you don't even need to show that in order to know that orbits aren't stable for 1/r^(n-1) where n>3.

Higher Dimensional Gravity

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    $\begingroup$ Is the orbit of the Moon around the Earth considered stable? $\endgroup$ – Organic Marble Aug 2 at 0:15
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    $\begingroup$ Organic Marble I would think so. The distance between the centers of mass are about 30 times the diameter of the Earth and about 120 times the diameter of the moon. The moon is slowly receding, but that's due to different physics. By comparison, Luna 10 had a 350 x 1,000 km orbit (compared to a lunar diameter of just over 2,000 km, so up to 3,000 km from the center of mass, 1.5 times) and crashed into the moon in the same year. $\endgroup$ – Greg Aug 2 at 1:25
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    $\begingroup$ Thanks for the info! $\endgroup$ – Organic Marble Aug 2 at 1:27

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