AS particle hagin form a spring stretches it by 1 cm at earth's surface. How much will hesame partile stretch the spring at a place 800 km above the earth's surface/ Radius of the earth=6400 km.

Explanation

Suppose the mass of the particle is m and the spring constant f the spring is k. The ccelertion due to gravity at earth's surface si g=(GM)/R^2` with usual symbols The extension in the spring is mg/k.
Hence 1cm=(GMm)/(kR^2)`……….i
At a heigh h=800 km the extension is given by
`x=(GMm)/(k(R+h)^2`.........ii
by i and ii
x/(1cm)=R^2/((R+H)^2)
`((6400km)^2)/((7200km)^2)=0.79`
`Hence x=0.79cm