Question.
Why is the weight of an object on the Moon $\frac{1}{6}$ th of its weight on the earth ?
Why is the weight of an object on the Moon $\frac{1}{6}$ th of its weight on the earth ?
Solution:
Acceleration due to gravity $\left(\mathrm{g}_{\mathrm{m}}\right)$ on Moon is $\frac{1}{6}$ th of the acceleration due to gravity $\left(\mathrm{g}_{\mathrm{e}}\right)$ on earth. i.e.,
$\mathrm{g}_{\mathrm{m}}=\frac{1}{6} \mathrm{~g}_{\mathrm{e}}$
or $m g_{\mathrm{m}}=\frac{1}{6} \mathrm{mg}_{\mathrm{e}} \quad$ where, $\mathrm{m}=$ mass of object
or $\quad W_{\mathrm{m}}=\frac{1}{6} W_{\mathrm{e}} \quad[$ weight $=\mathrm{mg}]$
Since, $g_{m}$ is $\frac{1}{6}$ th of $g_{e}$, thus, weight of an object on the Moon is also $\frac{1}{6}$ th of the weight of
the object on earth.
Acceleration due to gravity $\left(\mathrm{g}_{\mathrm{m}}\right)$ on Moon is $\frac{1}{6}$ th of the acceleration due to gravity $\left(\mathrm{g}_{\mathrm{e}}\right)$ on earth. i.e.,
$\mathrm{g}_{\mathrm{m}}=\frac{1}{6} \mathrm{~g}_{\mathrm{e}}$
or $m g_{\mathrm{m}}=\frac{1}{6} \mathrm{mg}_{\mathrm{e}} \quad$ where, $\mathrm{m}=$ mass of object
or $\quad W_{\mathrm{m}}=\frac{1}{6} W_{\mathrm{e}} \quad[$ weight $=\mathrm{mg}]$
Since, $g_{m}$ is $\frac{1}{6}$ th of $g_{e}$, thus, weight of an object on the Moon is also $\frac{1}{6}$ th of the weight of
the object on earth.