Question

At what altitude $\left(h\right)$ above the earth's surface would the acceleration due to gravity be one fourth of its value at the earth's surface ?

• A. $h=R$
• B. $h=4R$
• C. $h=2R$
• D. $h=16R$

Answer 1

The correct option is A $h=R$

Considering $g$ be the acceleration due to gravity at the surface, same at the altitude $h$ is
$g_h=\frac{g}{4}$

At altitude $\left(h\right)$ above the earth's surface, acceleration due to gravity will be

$g_h=\frac{g}{{(1+\frac{h}{R})}^2}$

Here $g_h=\frac{g}{4}$

$\frac{g}{4}=\frac{g}{{(1+\frac{h}{R})}^2}$

${(1+\frac{h}{R})}^2=\frac{1}{4}$

$1+\frac{h}{R}=\pm \frac{1}{2}$

Taking positive value,

$\Rightarrow h=R$

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

Caution - The result $g^{\prime }=g(1-\frac{2h}{Re})$ can be used only when $h<<R$.