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Variation of G Due to the Rotation of Earth

A satellite r...

Question

A satellite revolves around the planet of mean density $10^{4} k g / m^{3}$ .if the radius of the orbit is only slightly greater than the radius of the planet, find the time revolution of the planet, Find the time revolution of the satellite. $G = 6.67 \times 10^{- 11}$ S.I. units.

1.044 h r

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2.044 h r

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5.23 h r

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120.4 h r

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Solution

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Similar questions

8000

m^{- 3}

G = 6.67 \times 10^{- 11}

m^{- 2} / k g^{- 2}

74 \times 10^{3} k m

16.7

27

ρ

_{1}

_{2}

T

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A satellite revolves around the planet of mean density 104kg/m3.if the radius of the orbit is only slightly greater than the radius of the planet, find the time revolution of the planet, Find the time revolution of the satellite. G=6.67×10−11 S.I. units.