Q. The value of $g$ on the earth’s surface is $980\ cms^{- 2}$. Its value at a height of 64 km from the earth’s surface is
A) $960.40\ cms^{- 2}$
B) $984.90\ cms^{- 2}$
C) $982.45\ cms^{- 2}$
D) $977.55\ cms^{- 2}$
Explanation
(a) $g^{'} = \frac{gR^{2}}{(R + h)^{2}}$
$= 980 \times \left( \frac{6400}{6400 + 64} \right)^{2} = 960cms^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.