A simple pendulum has a periodic time T_1 whe | vedclass

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(C) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.On the surface of the Earth,$T_1 = 2 \pi \sqrt{\frac{l}{g}}$,where

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$2$

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(C) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.On the surface of the Earth,$T_1 = 2 \pi \sqrt{\frac{l}{g}}$,where $g$ is the acceleration due to gravity at the surface.At a height $h = R$ above the Earth's surface,the acceleration due to gravity $g_h$ is given by $g_h = \frac{GM}{(R+h)^2} = \frac{GM}{(R+R)^2} = \frac{GM}{4R^2} = \frac{g}{4}$.The time period at height $h$ is $T_2 = 2 \pi \sqrt{\frac{l}{g_h}}$.Taking the ratio,$\frac{T_2}{T_1} = \sqrt{\frac{g}{g_h}} = \sqrt{\frac{g}{g/4}} = \sqrt{4} = 2$.

$A$ simple pendulum has a periodic time $T_1$ when it is on the surface of the Earth of radius $R$. Its periodic time is $T_2$ when it is taken to a height $R$ above the Earth's surface. The value of $\frac{T_2}{T_1}$ is

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