Show that the weight of an object on the moon is one-sixth of its weight on the earth. [Given: mass of earth $= 5.98 \times 10^{24} \ kg$,mass of moon $= 7.36 \times 10^{22} \ kg$,radius of earth $= 6.37 \times 10^{6} \ m$,radius of moon $= 1.74 \times 10^{6} \ m$]

(N/A) Let the mass of an object be $m$. Let its weight on the moon be $W_{m}$ and its weight on the earth be $W_{E}$.
Using the universal law of gravitation,the weight of an object on the moon is given by $W_{m} = \frac{GM_{m}m}{R_{m}^{2}}$,where $M_{m}$ is the mass of the moon and $R_{m}$ is its radius.
Similarly,the weight of the object on the earth is $W_{E} = \frac{GM_{E}m}{R_{E}^{2}}$,where $M_{E}$ is the mass of the earth and $R_{E}$ is its radius.
Taking the ratio of the two weights:
$\frac{W_{m}}{W_{E}} = \frac{GM_{m}m}{R_{m}^{2}} \times \frac{R_{E}^{2}}{GM_{E}m} = \frac{M_{m}}{M_{E}} \times \left( \frac{R_{E}}{R_{m}} \right)^{2}$
Substituting the given values:
$\frac{W_{m}}{W_{E}} = \frac{7.36 \times 10^{22}}{5.98 \times 10^{24}} \times \left( \frac{6.37 \times 10^{6}}{1.74 \times 10^{6}} \right)^{2}$
$\frac{W_{m}}{W_{E}} \approx 0.0123 \times (3.66)^{2} \approx 0.0123 \times 13.4 = 0.165 \approx \frac{1}{6}$
Thus,the weight of an object on the moon is one-sixth of its weight on the earth.