The length of a seconds pendulum at a height h = 2R from earth surface will be:
(Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = $\pi$2 ms$-$2)
Solution
<p>$g = {{GM} \over {{{(R + h)}^2}}} = {{GM} \over {9{R^2}}} = {{{g_0}} \over 9}$</p>
<p>$$ \Rightarrow T = 2\pi \sqrt {{l \over g}} = 2\pi \sqrt {{l \over {{{{g_0}} \over 9}}}} $$</p>
<p>$\Rightarrow 2 = 2\pi \sqrt {{{9l} \over {{g_0}}}}$</p>
<p>$\Rightarrow l = {{{g_0}} \over {9{\pi ^2}}} = {1 \over 9}\,m$</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Kepler's Laws
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