What is the distance from the moon to the point between earth and moon where the gravitational pulls of earth

Consider a mass \(\mathrm{m}\) is at the point \(\mathrm{P}\) where force due to moon and force due to earth are equal and opposite to each other. It is shown in the figure.

Consider that point \(\mathrm{P}\) is at the distance \(\mathrm{x}\) from earth.

Gravitational force on \(\mathrm{m}\) due to \(\mathrm{moon}=\vec{F}_{m}=\frac{\left(\mathrm{G} M_{m} \mathrm{~m}\right.}{(R-x)^{2}}\)

Gravitational force due to earth on mass \(\mathrm{m}=\vec{F}_{e}=\frac{G M_{e} m}{x^{2}}\)

Since, both forces are equal and opposite, then,

\(\left|\vec{F}_{m}\right|=\left|\vec{F}_{e}\right|\)

\(\frac{G M_{m} m}{(R-x)^{2}}=\frac{G M_{e} m}{x^{2}}\)

\(\left[\frac{R-x}{x}\right]^{2}=\frac{M_{m}}{M_{e}}\)

\(\frac{R}{x}-1=\sqrt{\frac{M_{m}}{M_{e}}}\)

\(\frac{R}{x}=\sqrt{\frac{M_{m}}{M_{e}}}+1\)

\(x=\frac{R}{\sqrt{\frac{M_{m}}{M_{e}}+1}}\)

now, lets plug all values,

$$ x=\frac{3.84 \times 10^{8} \mathrm{~m}}{\sqrt{\frac{7.35 \times 10^{22} \mathrm{~kg}}{5.97 \times 10^{24 \mathrm{~kg}}}+1}}=\frac{3.84 \times 10^{8} \mathrm{~m}}{1.13834}=3.373 \times 10^{8} \mathrm{~m} $$

So, we can see that this distance is very close to the centre of moon. Then, we can say that point where force due to moon and earth would be equal near the centre of moon roughly.

It happens because mass of moon is very very less comparision to earth. Mass of moon is about 100 times less than earth.

This is the location of the point from the center of the earth.

Note: here, we have assumed both masses as a point-like particles.

g_moon=GM_moon/R₁²

g_earth=GM_earth/R₂²

g_earth=g_moon so: GM_moon/R₁²=GM_earth/R₂^{2 →} G cancel out:

so R₁=x and R₂=d-x where d=3.84 x 10^8 m

All you need to do is solve for x