Question

Why is the weight of an object on the moon $\frac{1}{6}$ th its weight on the earth?

Answer 1

Step 1: Parameters at Moon

Mass of the moon $m_m=\frac{1}{100}{m}_e$

Radius of the moon $R_m=\frac{1}{4}{R}_e$

Step 2: Calculate the value of $g$at moon's surface

the value of $g$ at surface of earth,

$g_e=\frac{G{M}_e}{{R^2}_e}$

the value of $g$on surface of moon,

$g_m=\frac{G{M}_m}{{R^2}_m}$

$g_m=\frac{G\times {\displaystyle \frac{1}{100}}{M}_e}{{\displaystyle \frac{1}{16}}\times {R^2}_e}$

$g_m\simeq \frac{1}{6}\times \frac{G{M}_e}{{R^2}_e}$

$g_m=\frac{1}{6}{g}_e$

Hence, the weight of an object on the moon $\frac{1}{6}$ times its weight on the earth.