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Question

A satellite having time period same as that of the earth's rotation about its own axis is orbiting the earth at a height $$8$$R above the surface of earth, where, R is radius of earth. What will be the time period of another satellite at a height $$3.5$$R from the surface of earth?


Solution

Time period of satellite $$T^2= (\dfrac{4\pi^2}{GM})R^3$$  Where  $$R$$ is radius of orbit.
$$\Rightarrow T^2 \propto R^3$$
Given Time period of satellite at height $$8R$$ is $$T_1=24$$ hours, 
Let $$T_2$$ be time period of satellite at height.
$$\Rightarrow \dfrac{T_2^2}{T_1^2}=\dfrac{(3.5R+R)^3}{(8R+R)^3}$$
$$\Rightarrow T_2= T_1 (\dfrac{1}{2})^{\frac{3}{2}}$$   $$\Rightarrow T_2= \dfrac{24}{2\sqrt 2}=6\sqrt 2$$ hours

Physics

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