Q. At what height in km over the earth’s pole the free fall acceleration decreases by one percent? (Assume the radius of the earth to be 6400 km)
A) 32
B) 64
C) 80
D) 1.253
Explanation
(a) At a height $h$, (Taking $h < \ < R$) from the surface of earth
$g_{h} = g\left( 1 - \frac{2h}{R} \right)$ or $\frac{g_{h}}{g} = 1 - \frac{2h}{R} = \frac{90}{100}$
or $\frac{2h}{R} = 1 - \frac{99}{100} = \frac{1}{100}$
or $g = \frac{R}{100} = \frac{6400}{200} = 32$ km
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.