Question

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

• A. $1.28 \,cm$
• B. $0.64 \,cm$
• C. $3.6 \,cm$
• D. $0.12 \,cm$

Answer 1

Answer: (B)

In equilibrium, weight of the suspended body = stretching force.
At the earth’s surface, $mg = k \times x$
At a height $h, mg' = k \times x'$
$\frac{g'}{g} = \frac{x'}{x} = \frac{R^{2}_{e}}{\left(R_{e} + h\right)^{2}}$
$= \frac{\left(6400\right)^{2}}{\left(6400 + 1600\right)^{2}}$
$= \left(\frac{6400}{8000}\right)^{2} = \frac{16}{25} $
$x' = \frac{16}{25}\times x = \frac{16}{25 }\times 1 \,cm $
$ = 0.64 \,cm$