Question

When a falling meteoroid is at a distance above the Earth's surface of 3.00 times the Earth's radius, what is its acceleration due to the Earth's gravitation?

Answer 1

Answer

Acceleration at height h is given

$g_h=\frac{g_s}{(1+\frac{h}{R_e}) ^2}$

$=\frac{g_s}{(1+\frac{3R_e}{R_e}) ^2}=\frac{g_s}{16}$