Running a thought experiment, I considered someone's weight when they were standing on the Earth with the Moon directly overhead.

My initial thought was that the person weighs less because the Moon's gravity is working "against" the Earth's.

Conversely, someone standing on the Earth with the Moon above the opposite side would conceivably weigh more, since the Moon's and Earth's gravity are working "together".

But, I think this is an incomplete analysis, and here's why:

When considering both the Earth's and Moon's gravity, one must consider the two bodies as an integrated whole, with a mass equal to the sum of each individual body's mass. Now, here's the kicker: when performing gravity calculations, the center of this (unified) mass must be used to establish the distance in the formula. Also known as the barycenter.

Thus, in $ F = G \dfrac{m_{1} m_{2}}{r^{2}}$, $m_{1}$ is the combined mass of the Earth and Moon, $m_{2}$ is the mass of the person standing on the Earth (for whom we're measuring the force of gravity), and $r$ is the distance between this person (on the surface of the Earth) and the barycenter of the Earth-Moon system.

So, someone standing on the Earth directly underneath the Moon is closer to the barycenter than someone standing on the opposite side of the Earth, and my intuitive thinking is that whatever pull the two bodies have working against each other is offset by the reduced $r$. Conversely for someone standing on the opposite side of the Earth.


Is the concept I've described above accurate?

Taking this a step further (no pun intended): Consider someone on the near side vs. far side of the Moon. I can't conceptualize the above phenomenon being the same since the barycenter is (still) inside Earth, thus a really long distance away from someone on either the near side or far side of our Moon.


Would someone standing on the far side of the Moon weigh substantially (measurably) different than someone on the near side?


No, I don't think that question (and its answers) answered mine.

I even linked it below. But it doesn't mention anything about integrated masses of the two bodies or the concept of center-of-mass-as-barycenter.

I assert that my questions (especially the 2nd one, which is the primary motivation for me asking this to begin with) is not even remotely addressed by the heavier day-or-night question.

Finally, I explicitly did not mention "day" nor "night"; I'm aware of the fact that the Moon can be directly above someone standing on the Earth in broad daylight.

For the record, I have read the below Physics SE questions:


marked as duplicate by John Rennie, Yashas, Kyle Kanos, sammy gerbil, Jon Custer Jun 19 '17 at 13:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Since the moon is in freefall in the Earth's gravitational field, there'd be no difference on either side except tidal effects (pretty small).


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