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How much (if any) difference in gravity would you feel if you are on the surface of the moon and:

  1. standing right 'under' earth (exaclty between the moon and the earth) and
  2. on the opposite side of the moon (So both, the moon and earth are exactly under you)
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  • $\begingroup$ I suspect this question is motivated by the rather widely held concept that the Earth should make someone lighter as weighed by a scale at the point on the Moon closest to the Earth but heavier at the point on the Moon furthest from the Earth. This is incorrect. The Earth makes someone lighter at both points, and by almost the same amount. There are places where the Earth does someone on the Moon slightly heavier than nominal; these places are the set of points where the Earth is on the horizon. $\endgroup$ – David Hammen Dec 19 '19 at 23:55
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Very, very little

It's important to keep in mind that the Moon is constantly accelerating toward the Earth. The component of gravitation that one feels on the surface of the Moon is the difference between the gravitational acceleration toward the Earth at the point of interest and the Moon's gravitational acceleration toward thbe Earth as a whole.

At the sub-Earth point (the point on the Moon closest to the Earth), the gravitational acceleration toward the Earth is slight greater than is the gravitational acceleration of the Moon as a whole toward the Earth. The Earth very slightly reduces the gravitational acceleration toward the center of the Earth at the Moon's sub-Earth point.

At the antipode to the sub-Earth point (the point on the Moon furthest from the Earth; I'll call this the anti-sub-Earth point), the gravitational acceleration toward the Earth is slight less than is the gravitational acceleration of the Moon as a whole toward the Earth. The Earth once again very slightly reduces the gravitational acceleration toward the center of the Earth at the Moon's anti-sub-Earth point. The effect is nearly (but not exactly) identical to that at the sub-Earth point.

There are places on the Moon where the gravitational acceleration toward the Earth does slight increase the observed gravitational acceleration, which is the set of points where the Earth is more or less on the horizon. This effect here is about half that (in magnitude) of the effects at the sub-Earth and anti-sub-Earth points.

A qualitative discussion

Ignoring that the Moon is rather lumpy in a gravitational sense, the gravitational acceleration on the surface of the Moon due to the Moon itself is $$\vec a_M = -\frac{\mu_M}{{r_M}^2}\hat r_M$$ where $\mu_M=G M_M$ is the Moon's gravitational parameter, $r_M$ is the Moon's radius, $\hat r_M$ is the vector from the center of the Moon to some point on the surface of the Moon, and $\boldsymbol a_M$ is the gravitational acceleration due to the Moon at that point.

At that point, the gravitational acceleration toward the Earth is $$\vec a_E = \frac{\mu_E}{||R_E \hat r_E - r_M \hat r_M||^3}(R_E \hat r_E - r_M \hat r_M)$$ where $\mu_E=G M_E$ is the Earth's gravitational parameter, $R_E$ is the distance between the centers of mass of the Moon and the Earth, $\hat r_E$ is the vector from the center of the Moon to the center of the Earth, and $\boldsymbol a_E$ is the Newtonian gravitational acceleration toward the Earth at that point.

The Moon itself accelerates Earthward due to Earth gravity by $$\vec a_{M,E} = \frac{\mu_E}{||R_E \hat r_E||^2}\hat r_E$$ The specific weight (force measured by a spring scale divided by mass) at the point of interest is $$\begin{align} \vec g &= \vec a_M + \vec a_E - \vec a_{M,E} \\ &= -\frac {\mu_M}{{r_M}^2}\left(\hat r_M-\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^2\left(\frac{\hat r_E - \frac{r_M}{R_E}\hat r_m}{||\hat r_E - \frac{r_M}{R_E}\hat r_m||^3}-\hat r_E\right)\right) \end{align}\tag{1}$$ In the case where $\hat r_E = \hat r_M$ (the point on the Moon closest to the Earth) equation (1) simplifies to $$\begin{align} \vec g &= -\frac {\mu_M}{{r_M}^2}\left(1-\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^2\left(\frac{1}{||1 - \frac{r_M}{R_E}||^2}-1\right)\right)\hat r_m \\ &\approx -\frac {\mu_M}{{r_M}^2}\left(1-2\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^3\left(1+\frac32\frac{r_m}{R_E}+\cdots\right)\right)\hat r_m \end{align}$$ In the case where $\hat r_E = -\hat r_M$ (the point on the Moon furthest from the Earth) equation (1) simplifies to $$\begin{align} \vec g &= -\frac {\mu_M}{{r_M}^2}\left(1-\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^2\left(1-\frac{1}{||1 + \frac{r_M}{R_E}||^2}\right)\right)\hat r_m \\ &\approx -\frac {\mu_M}{{r_M}^2}\left(1-2\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^3\left(1-\frac32\frac{r_m}{R_E}+\cdots\right)\right)\hat r_m \end{align}$$ The dominant perturbing factor, $2\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^3$ is the same for both points and is rather small, about $1.5\times10^{-5}$. People at these two points would weigh a tiny bit smaller than they would if the Earth was not present. The difference between the two points is tinier still, a factor of about $2\times10^{-7}$.

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    $\begingroup$ This answer is pretty vague. How about adding some quantitative discussion? $\endgroup$ – Ben Crowell Dec 14 '19 at 1:41
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    $\begingroup$ @BenCrowell - Added. $\endgroup$ – David Hammen Dec 20 '19 at 1:29
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Very little

Earth's surface gravity is $1g$, felt at a radius of about $4,000$ miles from the planet's center. The distance to the moon is roughly $240,000$ miles, which is ~$60$ times the earth's radius. Because gravity decreases with the square of distance, the earth's pull on the moon is only $\frac{1}{3600}g$.

The moon's surface gravity is ~$\frac{1}{6}g$, and the earth's gravity will make you lighter/heavier by ~$\frac{1}{3600}g$ depending on if you're standing on the near, where the Earth is pulling away, so the gravity is $(\frac{1}{6}-\frac{1}{3600})g$ or far side of the moon, where it is $(\frac{1}{6}+\frac{1}{3600})g$. Your apparent weight will only vary by a few tenths of a percent between the near and far side of the moon.

I've ignored the radius of the moon itself here (~$1000$ miles), as it is relatively small compared to the earth-moon distance. But there is an even smaller tug from earth on the far side than the near side, further reducing the change in apparent weight.

EDIT: This answer is wrong, but will remain here as testament to incorrect reasoning. @David Hammen has it correct - the earth's pull actually decreases apparent weight on both the near and far side of the moon, much like how the moon creates a tidal bulge on both sides of the earth. The near side of the moon gets pulled more than the moon as a whole, while the moon as a whole gets pulled more than the far side. The overall scale of the effects is correct though, with the earth only exerting a few ten-thousandths of a g at that distance.

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  • $\begingroup$ I edited your answer to include an additional clarification and added latex ... feel free to roll back if you do not like it. $\endgroup$ – Tosic Dec 13 '19 at 15:49
  • $\begingroup$ This answer is incorrect. $\endgroup$ – David Hammen Dec 13 '19 at 16:16
  • $\begingroup$ Yes it doesn't seem right ... I think this would be okay if the Moon was stationary relative to the Earth (or moving at constant velocity) ... but it rotates around the Earth and one needs to take centrifugal force into account, is that right? $\endgroup$ – Tosic Dec 13 '19 at 17:21
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    $\begingroup$ This is wrong. The moon is in free fall, which eliminates the effect to first order. $\endgroup$ – Ben Crowell Dec 14 '19 at 1:40
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To a good first approximation, there is no difference. There's an easy way to see this with a thought experiment: Imagine the Moon was a sphere of liquid. Its geoid (selenoid?) is the surface where the water has no tendency to flow elsewhere due to gravity. If gravity was higher in one place, water would flow that direction raising the surface elevation and decreasing the pull of gravity until it was the same everywhere again.

Dirt and rock flows also, though much more slowly. The Moon's surface adjusts through isostasy so that it follows the geoid surface over large areas. (Over smaller areas, the strength of the rock is enough to hold things in place for a very long time, hence craters and mountains.) Large areas of highland is mostly due to the rock underneath being less dense and hence bouyant

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  • $\begingroup$ This argument works only because the moon is tidally locked (it is also the mechanism that keeps it locked). This should probably be noted in the answer. $\endgroup$ – mmeent Dec 13 '19 at 23:36
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    $\begingroup$ Note that the perceived gravitational force even on the surface of a body in perfect isostatic equilibrium is not necessarily constant. All we can say is that, by definition, the net force vector is everywhere normal to the surface (or else stuff on the surface would flow/roll along the tangential component). In particular, the perceived gravity on the surface of a rotating body is lower on the equator than at the poles, a fact that has been notably used in science fiction but can actually be measured even on Earth. Tidal forces have a similar effect. $\endgroup$ – Ilmari Karonen Dec 14 '19 at 0:08
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    $\begingroup$ This doesn't seem right to me. The fact that the moon's surface is an equipotential is a statement about the gravitational potential $\phi$. That doesn't tell us anything about constancy of the gravitational field $\nabla\phi$. $\endgroup$ – Ben Crowell Dec 14 '19 at 1:40
  • $\begingroup$ This answer is wrong, and for many reasons. (1) The Moon is not in isostatic equilibrium. It has a frozen equatorial bulge and a frozen tidal bulge that reflect the Moon's rotation rate and closeness to the Earth four billion of years ago rather than now, along with marked disparities at the Moon's basins. (2) The Moon doesn't change shape anywhere near fast enough to counter these tidal forces. (3) This is a question about tidal forces, and this doesn't answer that question. $\endgroup$ – David Hammen Dec 20 '19 at 2:04

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