The length of second pendulum at a height h = 2R from earth surface will be: 1k m then k=? (given R=Radius of earth and acceleration due to gravity at the surface of earth g=π2 m/s-2)

Time period of simplependulum is given as, T=2πLg1 g1=GM9R2=g9=π29 and 2=2πLπ2×9

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The length of second pendulum at a height h = 2R from earth surface will be: $\frac{1}{k} m$ then $k = ?$ (given R=Radius of earth and acceleration due to gravity at the surface of earth $g = π^{2} m / s^{- 2})$

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answer is 9.

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$Time period of simplependulum is given as, T = 2 π \sqrt{\frac{L}{g^{1}}}$

$g^{1} = \frac{GM}{9 R^{2}} = \frac{g}{9} = \frac{π^{2}}{9} and 2 = 2 π \sqrt{\frac{L}{π^{2}} \times 9}$

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