The lengths of seconds pendulums on the surfa | vedclass

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(A) The time period of a seconds pendulum is $T = 2 	ext{ s}$.The formula for the length of a pendulum is $T = 2\pi \sqrt{l/g}$, which implies $l = g

Vedclass · Accepted Answer

$\frac{h\sqrt{l_h}}{\sqrt{l_s} - \sqrt{l_h}}$

Vedclass · Answer

(A) The time period of a seconds pendulum is $T = 2 	ext{ s}$.The formula for the length of a pendulum is $T = 2\pi \sqrt{l/g}$, which implies $l = g(T/2\pi)^2$.On the surface of the earth, $l_s = g(T/2\pi)^2$.At an altitude '$h$',$l_h = g_h(T/2\pi)^2$, where $g_h$ is the acceleration due to gravity at height '$h$'.Taking the ratio, $l_h/l_s = g_h/g = (R/(R+h))^2$, where '$R$' is the radius of the earth.Taking the square root on both sides, $\sqrt{l_h/l_s} = R/(R+h)$.Rearranging the equation: $(R+h)\sqrt{l_h} = R\sqrt{l_s}$.$R\sqrt{l_h} + h\sqrt{l_h} = R\sqrt{l_s}$.$h\sqrt{l_h} = R(\sqrt{l_s} - \sqrt{l_h})$.Therefore, $R = \frac{h\sqrt{l_h}}{\sqrt{l_s} - \sqrt{l_h}}$.

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

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