Question

Assuming that the mass of Earth is $100$ times larger than that of Moon and radius of Earth is about $4$ times as that of Moon, show that the weight of an object on Moon is $\frac{1}{6}\mathrm{th}$ of that on the earth.

Answer 1

Step 1: Given data:

Mass of earth (${\mathrm{M}}_{\mathrm{e}}$) $=100\times$ Mass of moon $\left({\mathrm{M}}_{\mathrm{m}}\right)$ $=100{\mathrm{M}}_{\mathrm{m}}$

Radius of earth (${\mathrm{R}}_{\mathrm{e}}$) $=4\times$ Radius of moon $\left({\mathrm{R}}_{\mathrm{m}}\right)$ $=4{\mathrm{R}}_{\mathrm{m}}$

Acceleration due to gravity on earth, ${\mathrm{g}}_{\mathrm{e}}=\frac{{\mathrm{GM}}_{\mathrm{e}}}{{{\mathrm{R}}_{\mathrm{e}}}^2}$

Acceleration due to gravity on moon, ${\mathrm{g}}_{\mathrm{m}}=\frac{{\mathrm{GM}}_{\mathrm{m}}}{{{\mathrm{R}}_{\mathrm{m}}}^2}$

Step 2: Finding the relation between ${\mathrm{g}}_{\mathrm{e}}$ and ${\mathrm{g}}_{\mathrm{m}}$:

As it is given ${\mathrm{g}}_{\mathrm{e}}=\frac{{\mathrm{GM}}_{\mathrm{e}}}{{{\mathrm{R}}_{\mathrm{e}}}^2}$

In the above equation put the values of mass and radius of the earth in terms of the mass and radius of the moon respectively-

$$\begin{gathered} {\mathrm{g}}_{\mathrm{e}}=\frac{\mathrm{G}\left(100{\mathrm{M}}_{\mathrm{m}}\right)}{(4{{\mathrm{R}}_{\mathrm{m}})}^2} \\ =\frac{100}{16}\left(\frac{{\mathrm{GM}}_{\mathrm{m}}}{{{\mathrm{R}}_{\mathrm{m}}}^2}\right) \\ =\frac{100}{16}{\mathrm{g}}_{\mathrm{m}} \\ {\mathrm{g}}_{\mathrm{m}}=\frac{16}{100}{\mathrm{g}}_{\mathrm{e}}=0.16{\mathrm{g}}_{\mathrm{e}} \end{gathered}$$

Step 3: Finding the weight of an object on earth and moon:

Let the weight of an object on earth and moon be ${\mathrm{W}}_{\mathrm{e}}$ and ${\mathrm{W}}_{\mathrm{m}}$ respectively.

The formula to calculate weight is $\mathrm{W}=\mathrm{mg}$.

${\mathrm{W}}_{\mathrm{m}}=\mathrm{m}\times {\mathrm{g}}_m={\mathrm{mg}}_{\mathrm{m}}$ where $m$ is the mass of the body which is constant everywhere.

Putting the value of ${\mathrm{g}}_{\mathrm{m}}$ in terms of ${\mathrm{g}}_{\mathrm{e}}$ in the above equation, we get

$$\begin{gathered} {\mathrm{W}}_{\mathrm{m}}=\mathrm{m}(0.16{\mathrm{g}}_{\mathrm{e}}) \\ =0.16{\mathrm{mg}}_{\mathrm{e}} \\ =0.16{\mathrm{W}}_{\mathrm{e}} \\ {\mathrm{W}}_{\mathrm{m}}\approx \frac{1}{6}{\mathrm{W}}_{\mathrm{e}} \end{gathered}$$

Hence, the weight of an object on Moon is $\frac{1}{6}\mathrm{th}$ of that on the earth.