Question

A simple pendulum has a time period \(T_{1}\) when on the earth's surface, and \(T_{2}\) when taken to a height \(R\) above the earth's surface, where \(R\) is the radius of the earth. The value of \(T_{2} / T_{1}\) is .....

Answer 1

The time period of the pendulum is given by
$$
T=2 \pi \sqrt{\frac{1}{g}}
$$
The variation of \(g\) with the height above the earth is given by
$$
g=\frac{G M}{\left(R_{e}+h\right)^{2}}
$$
When \(h=0 \Rightarrow g_{1}=\frac{G M_{v}}{R_{v}^{2}}\) and \(h=R_{r}\)
\(\Rightarrow g_{2}=\frac{G M_{r}}{4 R_{?}^{2}} \Rightarrow \frac{T_{2}}{T_{1}}\)
\(=\sqrt{\frac{g_{1}}{g_{2}}}=\sqrt{\frac{\frac{G_{i} M_{v}}{R_{v}^{2}}}{\frac{G_{i} M_{v}}{4 R_{e}^{2}}}}=2\)

Answer 2

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