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A satellite revolves in elliptical orbit around a planet of mass `M`. Its time period is `T` and `M` is at the centre of the path. The length of the major axis of the path is: (neglect the gravitational effect of other object in space)A. `2[(GMT^(2))/(4pi^(2))]^(1//3)`B. `[(GMT^(2))/(4pi^(2))]^(1//3)`C. `1/2[(GMT^(2))/(4pi^(2))]^(1//3)`D. none of these |
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Answer» Correct Answer - A Since the planet is at the centre, the focus and centre of the elliptical path coincides and the elliptical path becomes circular and the major axis is nothing but the diameter. For a circular path: `(mv^(2))/r=sqrt((GM)/(r^(2)))m` Also `T=(2pir)/v=(2pir^(3//2))/(sqrt(GM))implies T^(2)=(4pi^(2)r^(3))/(GM)` `implies ((GMT^(2))/(4pi^(2)))^(1//3)`=Radius `implies` Diameter (major axis) `=2((GMT^(2))/(4pi^(2)))^(1//3)` |
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