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My 5 year old asked a question today that I wasn't sure how to answer, "What would happen if people took so many pieces of the Moon that it was gone?"

I'm curious specifically about what a gradual joining of the Earth and Moon would mean.

In order to remove the effect of rockets from the equation let's say that a near perfect method of transporting Moon rocks to Earth has been devised: such that 100% of energy is expended in the physical action of moving Moon matter to Earth without it burning up in the atmosphere.

Theoretically it seems like a space elevator that transports matter without it disintegrating in the atmosphere would serve as an approximation of the process I'm trying to understand. A Moon-based catapult or railgun that launches parachuted rocks to Earth (so they don't burn up in the atmosphere but instead land gently) also seems to fit what I'm going for.

Clearly Moon rocks have energy from:

  1. Gravitational potential relative to the Moon
  2. Gravitational potential relative to the Earth
  3. Rotational angular momentum around the Moon's axis
  4. Rotational angular momentum around the Earth

Once Moon rocks are moved to Earth they would only have 2 & 4.

My first instinct is that the transfer of momentum from the Moon to the Earth would cause the Moon to move further away and would slow the Earth's rotation.

What would really happen?

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A partial, but simple answer:

Currently, the Earth-Moon system has a center of gravity (not sure exactly where) and a total angular momentum (which is the sum of: the orbital angular momentum; the Earth's angular momentum due to rotation; and the Moon's angular momentum due to rotation), and note these three the are not all necessarily pointing in exactly the same direction.

Now, lasso the moon and reel it in to Earth. It doesn't matter if you do it in one piece or many, fast or slow, but, the center of gravity and total angular momentum cannot change.

I won't do the math, but it's not immediately clear if the final angular velocity of the resultant planet will be higher or lower, so, same goes for the length of day.

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A partial answer (only partial because I'm making some grand approximations about how the matter is transferred) -

Let's assume that the Earth is stationary as the moon orbits it as opposed to the two orbiting the center of mass (this approximation gets better as the mass of the moon decreases). Suppose the moon's mass goes down by 1/2 and Earth's mass goes up by the same amount. If we assume conservation of angular momentum (i.e. the method used to transfer the matter did not apply a torque on the system) then the moon must now go faster. By $L = 1/2 I \omega^2$ and the fact that $I$ is $m_{moon} r^2$, as $m_{moon}$ decreases, $\omega$ must increases to keep $L$ constant.

Since the moon must go faster, by Kepler's 3rd law, its distance to the Earth must also increase. So your intuition is correct.

As for whether the Earth's rotation will speed up or slow down, it depends on the angular momentum of the moon rocks when they reach the Earth. This will depend on the method used to transport the rocks. If humans are carrying the rocks, then the angular momentum is likely to be low, since the spacecraft will slow down (= applying torque and angular momentum conservation does not hold) before it lands. If we're haphazardly shooting rocks off the moon at the Earth, then they might impact the Earth at an angle with high speed, in which case it's conceivable it speeds the Earth's rotation up instead.

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  • $\begingroup$ I think we can assume that once the moon rocks reach Earth, they will stop orbiting the Earth. The given problem is equivalent to attaching a rope to the Moon and winching the whole thing down to Earth's surface in one piece. $\endgroup$ – Solomon Slow Mar 6 '18 at 23:17
  • $\begingroup$ @jameslarge yes, they stop orbiting the Earth. I don't understand though, did my answer imply the rocks were still orbiting the Earth? $\endgroup$ – Allure Mar 6 '18 at 23:29
  • $\begingroup$ @jameslarge, so you analyze the situation similar to my answer? $\endgroup$ – Tom B. Mar 6 '18 at 23:56
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To translate your question into a more physically addressable form: We will wave a wand or using a teleportation booth to transport moon matter to the surface of the earth, in such a way that the gravitational potential energy, orbital kinetic energy, and momentum of the transported matter disappears and the matter lands on the earth's surface moving precisely the same way as you would be moving due to the earth's rotation. What happens to the moon's orbit and the earth's rotation?

Answer: the earth would continue to rotate at an unchanged rate. The earth would become more massive, and would have a stronger gravitational field, so the moon would move in closer to the earth and would move faster in its orbit as a result. If the mass were transported all at once instead of gradually, the moon's orbit would become more elliptical, with the highest point roughly at the location where the moon was at the moment when the mass was transported. The center of mass of the earth-moon system would shift toward the earth because of the mass transfer from earth to moon.

If a non-magic method were used such as a rail gun (or a really amazing catapult), the story would be a bit different. The moon's orbital velocity is a little over 1 km/sec. The Navy has tested a rail gun with a muzzle velocity of 2.5 km/s, so it's physically possible. Launching a projectile at 1.022 km/sec in a retrograde direction from the moon would let the projectile (which would at that point have zero orbital momentum) drop straight toward the center of the earth.

A good way to think about these things is to consider the extremes. Let's say we don't worry about energy conservation but insist on conservation of momentum. Now suddenly reduce to moon to the mass of a pea by using the rail gun to move almost all the original mass to the earth. The pea-moon would have to be moving at very nearly the speed of light to have that much angular momentum (with respect to the earth). We would never see it again. The earth, on the other hand, would have increased its mass by about 1.2 percent (the mass of the moon), without any angular momentum being added – which means it would have to be rotating a bit slower. That supports your intuition.

HOWEVER we really can't ignore any part of the conservation of energy and momentum; and it's important to consider carefully the mechanisms by which the system changes. For example, a slight excess or shortage of launch speed would cause the projectile to have a net negative or positive orbital angular momentum which would be transferred to the earth on impact (by hitting the "leading" or "trailing" edge of the earth at the equator as seen from the moon). So, it could either speed or slow the earth's rotation.

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  • $\begingroup$ Actually, as opposed the magic wand method, I'm specifically interested in the physical consequences of taking orbiting matter and joining it with Earth matter, from the question a perfect method of transporting Moon rocks to Earth has been devised: such that the majority of energy used goes towards the physical displacement of Moon matter. It seems that this would require at least some sort of change in the angular momentum of Earth. I'll try to clarify this in my question better. $\endgroup$ – Cory Klein Mar 6 '18 at 17:45
  • $\begingroup$ Maybe it would be better to postulate something that is physically possible. For example, use a catapult to toss rocks from the moon so that they are "stopped" and without orbital motion. They would drop straight down to the earth, hit the ocean, and come to rest on the bottom. A pretty clear answer can be given in that case. $\endgroup$ – S. McGrew Mar 6 '18 at 18:47
  • $\begingroup$ I did update my question with more specificity surrounding this issue. I don't think there is any variant of this question that is actually "physically possible", but the same goes for questions involving "frictionless pulleys" or "no air resistance", which can be beneficial in isolating a physical specific effect of a scenario. BTW - your downvote didn't come from me. $\endgroup$ – Cory Klein Mar 6 '18 at 18:50
  • $\begingroup$ It's a bit risky to try to answer questions that some folks might consider silly. No worries. I've edited the answer to clarify. $\endgroup$ – S. McGrew Mar 6 '18 at 21:28

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