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

Question

Aicharya · Accepted Answer

According to question $\frac{G M m}{(R+h)^2}=\frac{1}{16} \frac{G M m}{R^2}$ $\begin{array}{l}
\frac{1}{(R+h)^2} =\frac{1}{16 R^2} \\
\text {or } \frac{R}{R+h} =\frac{1}{4} \\
\text {or } \frac{R+h}{R} =4 \\
h =3 R
\end{array}$ ~

The height at which the weight of a body becomes $1 / 16$ th, its weight on the surface of earth (radius $R$…

Solution