A satellite is going round a planet in circular orbit of radius ‘r’. Its kinetic energy is made double instantaneously, the direction of motion at that instant being unaltered. Then

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The satellite will leave the gravitational field of the planet.

The satellite will describe elliptical motion with minimum and maximum distances from the planet being r and $r \sqrt{2}$

The angular momentum of the satellite about the planet is increased by a factor $\sqrt{2}$

The satellite will describe circular orbit of radius $r \sqrt{2}$

answer is C, D.

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Detailed Solution

$T . E_{i} = - \frac{G M m}{2 r} = - \frac{G M m}{r} + \frac{1}{2} m {(\sqrt{\frac{G M}{r}})}^{2} T . E_{f} = - \frac{G M m}{r} + 2 \times \frac{1}{2} m \frac{G M}{r} = 0$

Hence parabolic escape
Hence (D) is correct option
Also $L_{1} = m v r, L_{f} = m (\sqrt{2} v) r = \sqrt{2} L_{i}$
Hence (C) is also correct option

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