Question

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.

• A. 1.2cm
• B. 0.79 cm.
• C. 0.29 cm.
• D. 0.99 cm.

Answer 1

The correct option is B

0.79 cm.

Suppose the mass of the particle is m and the spring constant of the spring is k. The acceleration due to gravity at earth's surface is g = $\frac{Gm}{r^2}$ with usual symbols. The extension in the spring is mg/k.

Hence , 1 cm = $\frac{GMm}{{kR}^2}$ ...(i)

At a heingh h = 800 km, the extension is given by

x = $\frac{GMm}{{k(R+h)}^2}$ ..(ii)

By (i and (ii) $\frac{x}{1cm}$) = $\frac{R^2}{{(R+h)}^2}$

=$\frac{{6400km}^2}{{\left(7200km\right)}^2}$ = 0.79

Hence , x = 0.79 cm.