Q 11-07-039JEE MainEasy
The weight of a body on the Earth's surface is $90$ N. Find its weight, in newtons, at a height equal to twice the Earth's radius above the surface.
Numerical value type. Enter your answer.
Answer: 10
$$W = W_0\left(\frac{R}{R + 2R}\right)^2 = \frac{90}{9} = 10\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics