A satellite revolves in a circular orbit at a height of 200km from the surface of Earth. If the period of revolution of satellite is 90 minutes, G=6.66*10^-11 and mean radius is 6*10^6 m , calculate the average density of Earth

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Khimraj · Answer

Time period of revolution of satellite T = 2 π(r 3 /G*m E ) 1/2 So T 2 = 4π 2 (r 3 /G*m E ) (G*m E )/r 3 = 4π 2 /T 2 So (4π/3)G* ρ E = 4 π 2 /T 2 Then ρ E = 3 π/(T 2 *G) = 3π/[(90*60) 2 * 6.66*10 -11 ] SO ρ E = 4853 Kg/m 3 Hope it clears. If still you have doubt you can ask me. If you like answer then please approve it.

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A satellite revolves in a circular orbit at a height of 200km from the surface of Earth. If the period of revolution of satellite is 90 minutes, G=6.6610^-11 and mean radius is 610^6 m , calculate the average density of Earth

A satellite revolves in a circular orbit at a height of 200km from the surface of Earth. If the period of revolution of satellite is 90 minutes, G=6.6610^-11 and mean radius is 610^6 m , calculate the average density of Earth

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A satellite revolves in a circular orbit at a height of 200km from the surface of Earth. If the period of revolution of satellite is 90 minutes, G=6.66*10^-11 and mean radius is 6*10^6 m , calculate the average density of Earth

A satellite revolves in a circular orbit at a height of 200km from the surface of Earth. If the period of revolution of satellite is 90 minutes, G=6.66*10^-11 and mean radius is 6*10^6 m , calculate the average density of Earth

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A satellite revolves in a circular orbit at a height of 200km from the surface of Earth. If the period of revolution of satellite is 90 minutes, G=6.6610^-11 and mean radius is 610^6 m , calculate the average density of Earth

A satellite revolves in a circular orbit at a height of 200km from the surface of Earth. If the period of revolution of satellite is 90 minutes, G=6.6610^-11 and mean radius is 610^6 m , calculate the average density of Earth