Physics, asked by Anonymous, 1 year ago

A pendulum on the earth has a time period of 1 sec. It's time period in a box orbiting around the earth at a distance R from the surface of the earth will be? (R=radius of earth)

Answers

Answered by JinKazama1
2

Final Answer :2s


Steps:

1) Time period of simple pendulum at surface of earth = T(1) = 1s


2) Then when we place that pendulum in box at distance 2R from centre of earth .

Here, Radial force to orbit is provided by Gravitational force of Earth.

m =mass of Bob in pendulum

M. =Mass of Earth.


F(radial ) = F(Gravitational )

=> ma(r) = GMm/(2R) ^2

=> a(r) = GM/4R^2

=> a(r) = g/4 , since g = GM/R^2 )


3) We also know that, Time period of simple pendulum is proportional to 1/√g(eff) )

(As length is constant)

Therefore,

\frac{t(1)}{t(2)} = \frac{ \sqrt{ \frac{g}{4} } }{ \sqrt{g} } \\ = > t(2) = 2t(1) = 2 \times 1 = 2s

Therefore, Time Period of pendulum when it is orbiting the Earth at distance R from surface of planet .



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