A simple pendulum has a time period T₁when on the earths surface and T₂ when taken to a height R above the earths surface,where R is the radius of the earth The value of T₂/T₁ is---------

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Solution

$T i m e p e r i o d o f t h e s i m p l e p e n d u l u m n i s g i v e n b y : T = 2 π \sqrt{\frac{l}{g}} w h e r e l i s t h e l e n g t h o f t h e p e n d u l u m n a n d g i s t y h e a c c e l e r a t i o n d u e t o g r a v i t y o n t h a t p o i n t . T_{1} = 2 π \sqrt{\frac{l}{g_{1}}} T_{2} = 2 π \sqrt{\frac{l}{g_{2}}} g_{1} = a c c e l e r a t i o n d u e t o g r a v i t y o n e a r t h s u r f a c e = \frac{G M}{R^{2}} g_{2} = a c c e l e r a t i o n d u e t o g r a v i t y a t h e i g h t R = \frac{G M}{{(R + R)}^{2}} = \frac{G M}{4 R^{2}} \frac{T_{1}}{T_{2}} = \sqrt{\frac{g_{2}}{g_{1}}} = \sqrt{\frac{\frac{G M}{4 R^{2}}}{\frac{G M}{R^{2}}}} = \frac{1}{2}$ ,

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A simple pendulum has a time period T1when on the earths surface and T2 when taken to a height R above the earths surface,where R is the radius of the earth The value of T2/T1 is---------

A simple pendulum has a time period T₁when on the earths surface and T₂ when taken to a height R above the earths surface,where R is the radius of the earth The value of T₂/T₁ is---------