A particle hanging from a spring stretches it by 1 cm at earth's surface. How much will the same particle stretch the spring at a place 800 km above the earth's surface/ Radius of the earth=6400 km.

Explanation

Here, `x = 2 cm, h = 4800 km, R = 6400 km, x' = ?`. Let `k` is the force constant of the spring.
In equilbrium, weight of the suspended body = stretching force
`:. Mg = kx` and `mg' = kx'` or `(g')/(g) = (x')/(x)`
But `(g')/(g) = (R^(2))/((R +h)) = ((6400)^(2))/((6400 + 4800)^(2)) = (16)/(49)`
`:. (x')/(x) = (16)/(49)` or `x' = (16)/(49) xx x = (16)/(49) xx 2 = 0.65 cm`