Here gs= gravitational acceleration at surface $\begin{aligned}
& \text {Value of } g=g_5\left(1+\frac{h}{R}\right)^{-2} \
& =g_5(1+2)^{-2}=\frac{g_5}{9}
\end{aligned}$
Force =mg=90×9g5=100 N
JEE Main 2024 — Physics Mechanics
A 90 kg body placed at 2R distance from surface of earth experiences gravitational pull of : ( R= Radius of earth, g=10 m s−2 )
Held on 4 Apr 2024 · Verified 6 Jul 2026.
100 N
300 N
225 N
120 N
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