Question

If the height at which the weight of a body becomes ${\left(\frac{1}{16}\right)}^{th}$ of its weight on the surface of the earth $\left(\text{Radius}R\right)$ is $pR$, then find $p$.

Answer 1

Formula used: $W^{\prime }=W{\left(\frac{R}{R+h}\right)}^2$
Given,
$W^{\prime }=\frac{W}{16}=\frac{mg}{16}$

As we know,
$W^{\prime }=m{g}^{\prime }=\frac{mg}{{(1+\frac{h}{R})}^2}$

$\Rightarrow \frac{1}{16}=\frac{1}{{(1+\frac{h}{R})}^2}$

$\Rightarrow 1+\frac{h}{R}=4$
$\Rightarrow \frac{h}{R}=3$
$\Rightarrow h=3R$

Final Answer: $3$