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JEE Advanced 2023 Paper 1 · Physics · Gravitation · easy

A satellite moves in a circular orbit of radius r about a planet of mass M. A short thrust along its direction of motion increases its speed by a…

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A satellite moves in a circular orbit of radius rr about a planet of mass MM. A short thrust along its direction of motion increases its speed by a factor of 72\dfrac{\sqrt7}{2} without changing its direction. The new orbit is an ellipse with the point of thrust as one of its apsides. The ratio of the new period of revolution to the original period is
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Answer: 8

The circular speed satisfies vc2=GM/rv_c^2=GM/r. After the thrust the speed squared is v2=74vc2=7GM4rv^2=\tfrac74v_c^2=\dfrac{7GM}{4r}. The vis-viva equation v2=GM(2r−1a)v^2=GM\left(\dfrac2r-\dfrac1a\right) then gives 1a=2r−74r=14r\dfrac1a=\dfrac2r-\dfrac{7}{4r}=\dfrac{1}{4r}, so the semi-major axis of the new orbit is a=4ra=4r (bound, as v2<2vc2v^2<2v_c^2). Kepler's third law says T∝a3/2T\propto a^{3/2}, so TnewTold=(4rr)3/2=8\dfrac{T_{new}}{T_{old}}=\left(\dfrac{4r}{r}\right)^{3/2}=8. Check: the apogee is at 2a−r=7r2a-r=7r, giving a perigee at the thrust point (as the velocity is perpendicular to the radius and exceeds circular speed). Numerically a3/2=41.5=8a^{3/2}=4^{1.5}=8.
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