Time period of a satellite revolving above Earth’s surface at a height equal to R, where R the radius of Earth, is

4 π \sqrt{\frac{R}{g}}

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4 π \sqrt{\frac{2 R}{g}}

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8 π \sqrt{\frac{R}{g}}

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8 π \sqrt{\frac{2 R}{g}}

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Solution

The correct option is C $8 π \sqrt{\frac{R}{g}}$
$T = 2 π \sqrt{\frac{(R + h)^{3}}{G M}} = 2 π \sqrt{\frac{(R + R)^{3}}{g R^{2}}} = 2 π \sqrt{\frac{8 R}{g}} = 4 π \sqrt{\frac{2 R}{g}} [A s h = R (g i v e n)]$

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Time period of a satellite revolving above Earth’s surface at a height equal to R, where R the radius of Earth, is

The correct option is C 8π√RgT=2π√(R+h)3GM=2π√(R+R)3gR2=2π√8Rg=4π√2Rg [As h=R (given)]

The correct option is C $8 π \sqrt{\frac{R}{g}}$
$T = 2 π \sqrt{\frac{(R + h)^{3}}{G M}} = 2 π \sqrt{\frac{(R + R)^{3}}{g R^{2}}} = 2 π \sqrt{\frac{8 R}{g}} = 4 π \sqrt{\frac{2 R}{g}} [A s h = R (g i v e n)]$