Question

Find the acceleration due to gravity at a height equal to half the radius of the earth, if g$g=9.8m{s}^{-2}$.

Answer 1

Step 1: Given data

Acceleration due to gravity, $g=\frac{GM}{R}$

Assuming Gravitational constant, $G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$

Assuming $g=9.8m{s}^{-2}$ (acceleration due to gravity)

$height=\frac{R}{2}$

where R is a radius of earth

Step 2: Finding acceleration at a height due to gravity

We know that, acceleration due to gravity at some height above the earth surface is given by the formula $g\text{'}=\frac{g}{{(1+{\displaystyle \frac{h}{R}})}^2}$

$g\text{'}=\frac{g}{{(1+{\displaystyle \frac{h}{R}})}^2}$

$=\frac{9.8}{{(1+{\displaystyle \frac{R}{2R}})}^2}$

$=\frac{9.8}{{(1+{\displaystyle \frac{1}{2}})}^2}$

$=\frac{9.8}{{\left({\displaystyle \frac{3}{2}}\right)}^2}$

$=\frac{9.8\times 4}{9}$

$=4.35m{s}^{-2}$

Therefore, the acceleration due to gravity at a height equal to half the radius of the earth is $4.35m{s}^{-2}.$