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)

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

At the earth's surface, mg = k $\times$x

At a height h, ${\mathrm{mg}}^{\text{'}} = \mathrm{k} \times {\mathrm{x}}^{\text{'}}$

         $\frac{{\mathrm{g}}^{\text{'}}}{\mathrm{g}} = \frac{{\mathrm{x}}^{\text{'}}}{\mathrm{x}} = \frac{{{\mathrm{R}}_{\mathrm{e}}}^2}{{({\mathrm{R}}_{\mathrm{e}}+\mathrm{h})}^2} = \frac{{\left(6400\right)}^2}{{(6400+1600)}^2}$

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

         ${\mathrm{x}}^{\text{'}} = \frac{16}{25}\times \mathrm{x} = \frac{16}{25}\times 1 \mathrm{cm} = 0.64 \mathrm{cm}$