Question

The mass of the moon is $\left( \dfrac{1}{8} \right)$ of the earth but the gravitational pull is $\left( \dfrac{1}{6} \right)$ of the earth it is due to the fact that:
A. moon is the satellite of the earth
B. the radius of the earth is $\left( \dfrac{8}{6} \right)$ of the moon
C. the radius of the earth is $\left( \sqrt{\dfrac{8}{6}} \right)$ of the moon
D. the radius of the moon is $\left( \dfrac{6}{8} \right)$ of the earth

Answer 1

Hint: We will use the relationships of gravitational force with the mass of the bodies and the distances between them as given by Newton and use them to figure out what the ration in the radius of the two bodies given here must for the conditions given in the question to be valid. The gravitational force is proportional to the masses of the bodies inversely proportional to the inverse of the square of the distance between them.

Formula used:
Relationships of gravitational force with the mass of the bodies and the distances between them
$\begin{align}
  & F\propto m \\
 & F\propto \dfrac{1}{{{r}^{2}}} \\
\end{align}$

Complete step by step answer:
We will use the universal law of gravitation given by Newton according to which the force of gravity between two bodies will be proportional to their masses and inversely proportional to the distances between them. Let us assume that the same body is at the surface of both moon and earth. The gravitational pull on the body will be proportional to the masses of the moon and earth and inversely proportional to the square of their radii as that will be the distance considered between the centre of mass of the body and earth or moon. So, the ratio of the forces will be
     \[\dfrac{{{g}_{m}}}{{{g}_{e}}}=\dfrac{1}{6}=\dfrac{\dfrac{{{M}_{m}}}{{{R}_{m}}^{2}}}{{{\dfrac{{{M}_{e}}}{{{R}_{e}}}}^{2}}}=\dfrac{{{M}_{m}}}{{{M}_{e}}}\times \dfrac{\ {{R}_{e}}^{2}}{{{R}_{m}}^{2}}=\dfrac{1}{8}\times \dfrac{\ {{R}_{e}}^{2}}{{{R}_{m}}^{2}}\]
And using this we will get the relation between their radii as
$\begin{align}
 & \dfrac{{{\ {R}}_{e}}^{2}}{{{R}_{m}}^{2}}=\dfrac{1}{6}\times 8=\dfrac{8}{6} \\
 & \therefore \dfrac{{{\ {R}}_{e}}}{{{R}_{m}}}=\sqrt{\dfrac{8}{6}} \\
\end{align}$
Hence, the correct option is C, i.e. the radius of the earth is $\left( \sqrt{\dfrac{8}{6}} \right)$ of the moon.

Note:
The gravitational pull in this question will be considered at just the surface of the two celestial bodies i.e. earth and moon given here as there is no information provided about where the gravitational pull must be considered and we will take it to be their surface.