Q. The acceleration due to gravity about the earth’s surface would be half of its value on the surface of the earth at an altitude of ($R = 4000\ mile$)
A) $1200\ mile$
B) $2000\ mile$
C) $1600\ mile$
D) $4000\ mile$
Explanation
(c) $\frac{g^{'}}{g} = \left( \frac{R}{R + h} \right)^{2} \Rightarrow \frac{1}{2} = \left( \frac{R}{R + h} \right)^{2} \Rightarrow \frac{1}{2} = \left( \frac{4000}{4000 + h} \right)^{2}$
By solving we get $h = 1656.85\ mile\ \approx 1600\ mile$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.