Question

A body weighs 100 N on the surface of Earth. What will be the acceleration due to gravity acting on the body at a height of 3400 km above Earth's surface? Taking the mass of Earth$=6\times {10}^{24}kg$, Radius of earth$=6.4\times {10}^{6}m$ and $G=6.7\times {10}^{-11}N{m}^{2}k{g}^{-2}$.

• A. $6.18m{s}^{-2}$
• B. $9.8m{s}^{-2}$
• C. $6m{s}^{-2}$
• D. $4.18m{s}^{-2}$

Answer 1

The correct option is D

$4.18m{s}^{-2}$

Given, mass of earth M$=6\times {10}^{24}kg$, Radius of earth R$=6.4\times {10}^{6}m$, height h$=3.4\times {10}^{6}m$ and $G=6.7\times {10}^{-11}N{m}^{2}k{g}^{-2}$.

The value of g at a height h above the earth's surface is given by

$g=\frac{GM}{{(R+h)}^2}$
$=\frac{6.7\times {10}^{-11}\times 6\times {10}^{24}}{{(6.4\times {10}^6+3.4\times {10}^6)}^2}$.

On solving we get $g=4.18m{s}^{-2}$