The value of acceleration due to gravity at a height of $4 R_E$ from surface of earth is $R_E$ is radius of…
The value of acceleration due to gravity at a height of $4 R_E$ from surface of earth is $R_E$ is radius of earth and acceleration due to gravity on the surface of the earth = $10 \mathrm{~ms}^{-2}$ )
$0.2 \mathrm{~ms}^{-2}$
$0.3 \mathrm{~ms}^{-2}$
$0.4 \mathrm{~ms}^{-2}$
$3 \mathrm{~ms}^{-2}$
Solution
Acceleration due to gravity ' $\mathrm{g}$ ' varies with height as
$
g^{\prime}=g\left(\frac{R}{R+h}\right)^2
$
Given $\mathrm{R}+\mathrm{h}=5 \mathrm{R}$ and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$
$
\therefore g^{\prime}=10\left(\frac{\mathrm{R}}{5 \mathrm{R}}\right)^2=\frac{10}{25}=0.4 \mathrm{~m} / \mathrm{s}^2
$