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All the places that I've seen an answer to this question simply state that it's because there is a stronger gravitational pull on the near side of the moon, facing the earth. But I'm wondering how this gravitational pull actually manages to shift the centre of gravity of the moon, and how does it tie into the idea that the centre of gravity is where a body's weight is evenly distributed?

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marked as duplicate by anna v, John Rennie, Jon Custer, ZeroTheHero, Kyle Kanos Jan 8 at 11:10

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.

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    $\begingroup$ Can you link to some of the other explanations you have encountered? $\endgroup$ – rob Jan 7 at 2:58
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    $\begingroup$ Are you sure you're not thinking of the Moon's center of figure (its geometric center), which is offset from its center of mass by a couple of kilometers? $\endgroup$ – David Hammen Jan 7 at 3:10
  • $\begingroup$ the explanation is given in the answer here physics.stackexchange.com/questions/151402/… , so i voted it as duplicate $\endgroup$ – anna v Jan 7 at 8:05
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The near side of the moon experiences more gravity because it is closer to Earth. The force of gravity from the Earth acting on the moon has an inverse relationship to the distance between the two bodies. So the part of the moon that is closer to Earth experiences greater gravity than the far side of the moon.

If there were no other bodies of mass acting on the moon, then the centre of gravity would be the centre of mass. But that's not the case here.

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A simple way to shoe the difference is to consider two particles of equal mass $m$ located along the same radius vector at distances $r_1$ and $r_2$ from the centre of the Earth.

The position of the centre of mass of those two object $r_{\rm com}$ is found using the defining equation for centre of mass $$2m \,r_{\rm com} = m\, r_1 + m\, r_1 \Rightarrow r_{\rm com} = \frac 12 \left(r_1 + r_1\right) $$ ie midway between the two masses.

If the gravitational field strength $g$ is constant then the centre of gravity is also in the same position $$2m \,r_{\rm cog} g = m\, g\, r_1 + m\, g\, r_1 \Rightarrow r_{\rm cog} = \frac 12 \left(r_1 + r_1\right) $$

However for a non-uniform gravitational field things change $$\frac{GM_{\rm Earth}\,2m}{r^2_{\rm cog}} = \frac{GM_{\rm Earth}\,m}{r^2_1} +\frac{GM_{\rm Earth}\,m}{r^2_2}\Rightarrow r_{\rm cog} = r_1\,r_2 \sqrt{\frac{2}{r_1^2+r_2^2}}$$

The important thing to note here is that the first term on the right hand side of the first equation has a larger effect than the second term because $r_1<r_2$ as the gravitational field strength in the region of $r_1$ is greater than that in the region of $r_2$ so that the position of the centre of gravity will be less than $\frac 12 \left(r_1 + r_1\right) $

As a numerical example if $r_1 = 3$ and $r_2=4$ then $r_{\rm com} = 3.5$ and $r_{\rm cog} \approx 3.4$.

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