Q. The height at which the acceleration due to gravity decreases by 36% of its value on the surface of the earth. (The radius of the earth is $R$)
A) $\frac{R}{6}$
B) $\frac{R}{4}$
C) $\frac{R}{2}$
D) $\frac{2}{3}R$
Explanation
(b) The value of acceleration due to gravity at a height $h$ reduces to
$= \ 100 - 36 = 64\% = \frac{64}{100}g$
$\therefore\frac{64}{100}g = \frac{gR^{2}}{(R + h)^{2}}$
or $\frac{8}{10} = \frac{R}{R + h}\ \ \ or\ \ h = \frac{R}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.