Question

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

Answer 1

Explanation: 2 Marks

The mass of moon is $\frac{1}{100}$ times and its radius $\frac{1}{4}$ times that of earth.
Let mass of earth be $m_e$ and mass of the moon be $m_m=\frac{m_e}{100}$
Let radius of earth be $r_e$ and radius of moon be $r_m=\frac{r_e}{4}$
Let $m$ be the mass of the object.

According to Universal law of gravitation, force between earth and object is
$F_1=\frac{G{m}_{e}m}{r_e^2}$
and force between moon and the object is
$F_2=\frac{G{m}_{m}m}{r_m^2}$ $=\frac{G\times m_e\times m\times 16}{100\times r_e^2}$ $\approx \frac{1}{6}\times \frac{G{m}_{e}m}{r_e^2}=\frac{1}{6}{F}_1$

Therefore, the gravitational attraction on the moon is about one sixth when compared to earth. Hence, the weight of an object on the moon is $\frac{1}{6}th$ its weight on the earth.