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The orbital velocity of a satellite in a circular orbit just above the earth's surface is v_(0). The orbital velocity for a satellite orbiting in a circular orbit at an altitude of half of earth's radius is: |
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Answer» `sqrt((3)/(2))v_(0)` and orbital speed at a height h above the surfaceof earth is `v=sqrt((GM)/(R+h))=sqrt((gR^(2))/(R+(R )/(2)))=sqrt((2)/(3)gR)` `v= sqrt((2)/(3))*v_(0)` So CORRECT choice is (b). |
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