A $90 \mathrm{~kg}$ body placed at $2 \mathrm{R}$ distance from surface of earth experiences gravitational…
A $90 \mathrm{~kg}$ body placed at $2 \mathrm{R}$ distance from surface of earth experiences gravitational pull of : $\text { ( } \mathrm{R}=\text { Radius of earth, } \mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2} \text { ) }$
$100 \mathrm{~N}$
$300 \mathrm{~N}$
$225 \mathrm{~N}$
$120 \mathrm{~N}$
Solution
Here $g_s=$ gravitational acceleration at surface
$\begin{aligned}
& \text {Value of } g=g_5\left(1+\frac{h}{R}\right)^{-2} \\
& =g_5(1+2)^{-2}=\frac{g_5}{9}
\end{aligned}$ $\text {Force }=\mathrm{mg}=90 \times \frac{\mathrm{g}_5}{9}=100 \mathrm{~N}$