Question

A body of weight $72\text{N}$ moves from the surface of earth to a height equal to half the radius of earth. Gravitational force will be

• A. $36\text{N}$
• B. $32\text{N}$
• C. $144\text{N}$
• D. $50\text{N}$

Answer 1

The correct option is B $32\text{N}$

Gravitational force on body at the surface of earth,

$F=m\frac{GM}{R^2}=mg=72\text{N}$

Gravitational force at height, $h=\frac{R}{2}$ is given by

$F^{\prime }=m{g}^{\prime }=m\frac{GM}{(R+h{)}^2}=\frac{mGM}{{\left(\frac{3R}{2}\right)}^2}$

$\Rightarrow F^{\prime }=\frac{mGM}{R^2}\times \frac{4}{9}=\frac{4}{9}\times 72=32\text{N}$

Alternate:
$g^{\prime }=\frac{g}{{(1+\frac{h}{R})}^2}$

Here $h=\frac{R}{2}$

$\Rightarrow g^{\prime }=\frac{g}{{(1+\frac{R}{2R})}^2}=\frac{4}{9}g$

Force $F=m{g}^{\prime }=\frac{72}{g}\times \frac{4}{9}g=32\text{N}$

Hence, option $\left(b\right)$ is correct.

Why this question? Caution - The result $g^{\prime }=g(1-\frac{2h}{Re})$ can be used only when $h<<R$. Key concept: The acceleration due to gravity due to a planet or point mass = $\frac{GM}{r^2}$.