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For a geostationary satellite orbiting around the earth identify the necessary condition.

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it period of revolution must be $2 π \sqrt{\frac{R}{g}}$ , where $R$ is the radius of earth

it must lie in the equatorial plane of earth

its period of revolution must be $24 h r s$

its height from the surface of earth must be $36000 k m$

answer is A, B, D.

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Detailed Solution

Direction of rotation of geostationary satellite and direction of rotation of earth around its own axis must be the same .

Line of action of earth's gravitational pull on geostationary satellite must lie iie in the orbital plane of the satellite

Height of geostationary satellite above the surface of earth h = ${(\frac{g R^{2} T^{2}}{4 π^{2}})}^{\frac{1}{3}} - R = 36000 k m$

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