A body hanging from a spring stretches it by `2 cm` at the earth's surface. How much will the same body stretch the spring at a place `800 cm` above the earth's surface? Radius of the earth is `6400 km`.

Explanation

Let `k` be the force constant of the spring. `M` be the mass of the body hanging from a spring and `x` be the stretching in the spring. In equilibrium,
weight of the suspended body = stretching force
At the surface of earth, `mg = kx`
At a height `h` from the surface of earth
`mg' = kx'`
`:. (g')/(g) = (x')/(x) = (R^(2))/((R + h)^(2)) = ((6400)^(2))/((6400 + 800)^(2))`
`= ((6400)^(2))/((7200)^(2)) = 0.79`
`x' = 0.79 xx x 0.79 xx 2 = 1.58 cm`