The lengths of seconds pendulums on the surface of the earth and at an altitude ' h ' from the surface of the earth are l s and l h respectively. The radius of the earth is

Question

The lengths of seconds pendulums on the surface of the earth and at an altitude ' h ' from the surface of the earth are l s ​ and l h ​ respectively. The radius of the earth is

TestHub · Accepted Answer

The time period of a simple pendulum is given by $T = 2 π \frac{l}{g}$ .

For a seconds pendulum, the time period is constant ( $T = 2$ s) at both locations.

Therefore, $T_{s} = T_{h} \Rightarrow 2 π \frac{l _{s}}{g _{s}} = 2 π \frac{l _{h}}{g _{h}}$ .

This gives $\frac{l _{s}}{l _{h}} = \frac{g _{s}}{g _{h}}$ .

The acceleration due to gravity on the surface of the earth is $g_{s} = \frac{GM}{R ^{2}}$ and at an altitude $h$ is $g_{h} = \frac{GM}{( R + h ) ^{2}}$ .

Substituting these values, we get:

$\frac{l _{s}}{l _{h}} = \frac{\frac{GM}{R ^{2}}}{\frac{GM}{( R + h ) ^{2}}} = \frac{( R + h ) ^{2}}{R ^{2}} = (1 + \frac{h}{R})^{2}$

Taking the square root on both sides:

$\frac{l _{s}}{l _{h}} = 1 + \frac{h}{R}$

$\frac{h}{R} = \frac{l _{s}}{l _{h}} - 1 = \frac{l _{s} - l _{h}}{l _{h}}$

$R = \frac{h l _{h}}{l _{s} - l _{h}}$

Answer: $\frac{h l _{h}}{l _{s} - l _{h}}$

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