Question

The height at which the weight of a body becomes 1/16th its weight on the surface of earth (Radius R ) is:

Answer 1

$\frac{GMm}{{(R+h)}^2}=\frac{1}{16}\frac{GMm}{R^2}\frac{1}{{(R+h)}^2}=\frac{1}{16R^2}or,{\left(\frac{R}{R+h}\right)}^2=\frac{1}{16}\Rightarrow \frac{R}{R+h}=\frac{1}{4}.\therefore R+h=4R\Rightarrow h=3R$