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I've read that the total angular momentum of the Earth-Moon system is 3.41x10^41 g.cm^2/s (1). How much of that is due to the Moon? In other words, if the Moon ceased to exist and the Earth continued to rotate at its current rate, how much would the angular momentum decrease?

(1) https://www.sciencedirect.com/topics/earth-and-planetary-sciences/earth-moon-system

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According to the reference provided by Roger Wood (1), the orbital angular momentum of the Moon is 2.9x10^34 kg.m^2/s. The rotational angular momentum of the Earth is 7.1x10^33 kg.m^2/s. The rotational angular momentum of the Moon is negligible. So the total angular momentum of the Earth-Moon system is 3.61x10^34 kg.m^2/s. So the contribution of the Earth's rotation is approximately 20% with the orbit of the Moon providing the remainder. Converting units, the total angular momentum is 3.61x10^41 g.cm^2/s which is slightly different than the 3.41x10^41 g.cm^2/s from reference (2). So the answer to the original question is: The Moon contributes about 80%.

(1) people.ast.cam.ac.uk/~wyatt/poa_201415_originofmoon.pdf

(2) https://www.sciencedirect.com/topics/earth-and-planetary-sciences/earth-moon-system

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    $\begingroup$ We can't add angular momentum about different points together in a simple way. The quoted orbital angular momentum of the Moon is about the center of mass of the Earth-Moon system, but the quoted rotational angular momentum of the Earth is (likely) about its own axis, not the Earth-Moon center of mass. These points are very different, the EM CM is about 2/3 of the way to the Earth's surface. But "most of it" is probably not wrong. $\endgroup$ – uhoh Mar 3 at 12:56

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