A body stretches a spring by a particular length at the earth's surface at the equator. At what height above the south pole will it stretch the same spring by the same length? Assume the earth to be spherical.

#### Solution

At the equator, g' = g − ω^{2}R ...(i)

Let h be the height above the South Pole where the body stretch the spring by the same length.

The acceleration due to gravity at this point is \[g' = g\left( 1 - \frac{2h}{R} \right)\]

Weight of the body at the equator = weight of the body at height h above the South Pole

\[\therefore g - \omega^2 r = g\left( 1 - \left( \frac{2h}{R} \right) \right)\]

\[ \Rightarrow 1 - \frac{\omega^2 R}{2g} = 1 - \frac{2h}{R}\]

\[ \Rightarrow h = \frac{\omega^2 R^2}{2g}\]

\[ = \frac{\left( 7 . 3 \times {10}^{- 5} \right)^2 \times \left( 6400 \times {10}^{- 3} \right)^2}{2 \times 9 . 81}\]

\[ = \frac{\left( 7 . 3 \right)^2 \times \left( 64 \right)^2}{19 . 62}\]

\[ = 11125 \text{ m }= 10 \text{ km} \left( \text { approx }. \right)\]