## In This post you will get answer of Q.

A body hanging from a spring stretches it by $1, cm$ at the earths surface. How much will the same body stretch the spring at a place $16400 ,km$ above the earths surface? (Radius of the earth $= 6400, km$)

Solution:

## In equilibrium, weight of the suspended body $=$ stretching force.

$therefore$ At the earths surface, $mg = k times x$

At a height $h$, $mg’ = k times x’$

$frac{g’}{g} = frac{x’}{x} = frac{R^{2}_{e}}{left(R_{e}+hright)^{2}}$

$ = frac{left(6400right)^{2}}{left(6400+1600right)^{2}} = left(frac{6400}{8000}right)^{2} = frac{16}{25}$

$x’ = frac{16}{25} times x = frac{16}{25 } times1,cm$

$= 0.64,cm$