Question

The acceleration due to gravity on the moon is $\dfrac{1}{6}th$ the acceleration due to gravity on the earth. If the ratio of densities on the earth and the moon is $\dfrac{5}{3}$ then radius of the moon in terms of radius of the earth will be

Answer 1

Hint: First only consider the acceleration due to gravity of both the moon and the earth nearer to their respective surfaces and not any other values then compare the value of their acceleration due to gravity by taking their ratio. Further rearrange the terms such that the final expression is in terms of radius of the moon and the radius of the earth.

Complete step by step answer:
The acceleration due to gravity due to a body of mass M and radius R near its surface is given by,
$g=\dfrac{GM}{{{R}^{2}}}m{{s}^{-2}}$. Let us denote the mass of the earth as M and the radius of the earth as R. Similarly, let the mass of the moon be m and the radius of the moon be r. Hence acceleration due to gravity of earth is ${{g}_{E}}=\dfrac{GM}{{{R}^{2}}}m{{s}^{-2}}$ and acceleration due to gravity of moon is ${{g}_{M}}=\dfrac{Gm}{{{r}^{2}}}m{{s}^{-2}}$.
Taking the ratio of ${{\text{g}}_{\text{M}}}\text{ with respect to }{{\text{g}}_{\text{E}}}$ we get,
$\dfrac{{{g}_{M}}}{{{g}_{E}}}=\dfrac{\dfrac{Gm}{{{r}^{2}}}}{\dfrac{GM}{{{R}^{2}}}}$ after canceling g and rearranging the terms we get,
$\dfrac{{{g}_{M}}}{{{g}_{E}}}=\dfrac{m}{{{r}^{2}}}\times \dfrac{{{R}^{2}}}{M}$ it is given that this ratio is equal to the $\dfrac{1}{6}$. After substituting in the equation we get,
$\dfrac{1}{6}=\dfrac{m}{{{r}^{2}}}\times \dfrac{{{R}^{2}}}{M}$ Now let us write this equation in such a manner that we can write the equation in terms of density.
Multiplying and dividing the above equation on the right hand side by R, r, $\dfrac{4}{3}\pi $we get,
$\dfrac{1}{6}=\dfrac{mr}{\dfrac{4}{3}\pi {{r}^{3}}}\times \dfrac{\dfrac{4}{3}\pi {{R}^{3}}}{MR}$
$\dfrac{1}{6}=\dfrac{mr}{\dfrac{4}{3}\pi {{r}^{3}}}\times \dfrac{1}{\dfrac{M}{\dfrac{4}{3}\pi {{R}^{3}}}R}$
The term $\dfrac{m}{\dfrac{4}{3}\pi {{r}^{3}}}\text{is the density of moon }{{\text{D}}_{\text{M}}}$ and $\dfrac{M}{\dfrac{4}{3}\pi {{R}^{3}}}\text{is the density of earth }{{\text{D}}_{E}}$ Hence substituting in the above equation we get,
$\dfrac{1}{6}=\dfrac{{{\text{D}}_{\text{M}}}}{{{\text{D}}_{E}}}\times \dfrac{r}{R}....(1)$ It is given in the question that $\dfrac{{{\text{D}}_{\text{M}}}}{{{\text{D}}_{E}}}=\dfrac{3}{5}$, hence substituting in equation 1 we get,
$\dfrac{1}{6}=\dfrac{3}{5}\times \dfrac{r}{R}$ hence radius of the moon with respect to earth is i.e. $r=\dfrac{5}{18}R$.

Note:
We have considered the acceleration due to gravity near the surface of the moon as well as the earth. But as the height increases the value of g is different in case of moon as well as in case of earth. But still the radius of the moon obtained with respect to the earth will remain the same.