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Question: Periodic time of a satellite revolving above Earth’s surface at a height equal to R, where R the rad...

Periodic time of a satellite revolving above Earth’s surface at a height equal to R, where R the radius of Earth, is [gis acceleration due to gravity at Earth’s surface]

A

2π2Rg2\pi\sqrt{\frac{2R}{g}}

B

42πRg4\sqrt{2}\pi\sqrt{\frac{R}{g}}

C

2πRg2\pi\sqrt{\frac{R}{g}}

D

8πRg8\pi\sqrt{\frac{R}{g}}

Answer

42πRg4\sqrt{2}\pi\sqrt{\frac{R}{g}}

Explanation

Solution

T=2π(R+h)3GM=2π(R+R)3gR2=2π8Rg=42πRg\mathbf{T = 2\pi}\sqrt{\frac{\mathbf{(R + h}\mathbf{)}^{\mathbf{3}}}{\mathbf{GM}}} = 2\pi\sqrt{\frac{(R + R)^{3}}{gR^{2}}} = 2\pi\sqrt{\frac{8R}{g}} = 4\sqrt{2}\pi\sqrt{\frac{R}{g}} [Ash=Rh = R (given)]