2 answers

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

Question:

what is the distance from the moon to the point between earth and moon where the gravitational pulls of earth and moon are equal? mass of earth is 5.97x10^24 kg, mass of moon is 7.35x10^22 kg, distance between earth and moon is 3.84x10^8 m, and G=6.67x10^-11 Nm^2/kg^2.
How do I do this problem?


Answers

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.

.

gmoon=GMmoon/R12

gearth=GMearth/R22

gearth=gmoon so: GMmoon/R12=GMearth/R22 → G cancel out:

image from custom entry tool

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

All you need to do is solve for x

.

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