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


standing right 'under' earth (exaclty between the moon and the earth) and
on the opposite side of the moon (So both, the moon and earth are exactly under you)
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 $$\begin{align}
\vec g &amp;= \vec a_M + \vec a_E - \vec a_{M,E} \\
&amp;= -\frac {\mu_M}{{r_M}^2}\left(\hat r_M-\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^2\left(\frac{\hat r_E - \frac{r_M}{R_E}\hat r_m}{||\hat r_E - \frac{r_M}{R_E}\hat r_m||^3}-\hat r_E\right)\right)
\end{align}\tag{1}$$
In the case where $\hat r_E = \hat r_M$ (the point on the Moon closest to the Earth) equation (1) simplifies to $$\begin{align}
\vec g
&amp;= -\frac {\mu_M}{{r_M}^2}\left(1-\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^2\left(\frac{1}{||1 - \frac{r_M}{R_E}||^2}-1\right)\right)\hat r_m \\
&amp;\approx -\frac {\mu_M}{{r_M}^2}\left(1-2\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^3\left(1+\frac32\frac{r_m}{R_E}+\cdots\right)\right)\hat r_m
\end{align}$$
In the case where $\hat r_E = -\hat r_M$ (the point on the Moon furthest from the Earth) equation (1) simplifies to $$\begin{align}
\vec g
&amp;= -\frac {\mu_M}{{r_M}^2}\left(1-\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^2\left(1-\frac{1}{||1 + \frac{r_M}{R_E}||^2}\right)\right)\hat r_m \\
&amp;\approx -\frac {\mu_M}{{r_M}^2}\left(1-2\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^3\left(1-\frac32\frac{r_m}{R_E}+\cdots\right)\right)\hat r_m
\end{align}$$
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 $2\times10^{-7}$.