A satellite is moving in a low nearly circular orbit around the… | JEE Main 2020 (3 Sep, Shift 1)

Q22JEE Main 2020 · 3 SEP · SHIFT 1Difficult

Satellite Energy & Weightlessness

A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth’s radius Re. By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it become $\sqrt{\frac{3}{2}}$ times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is R. Value of R is:

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Answer: (b)

Initial orbital speed $v_0 = \sqrt{\frac{GM}{R_e}}$ ; new speed $v = \sqrt{\frac{3}{2}}\,v_0$ at $R_e$ (perigee).
Angular momentum: $v R_e = v' R \Rightarrow v' = \frac{v R_e}{R}$ . Let $x = \frac{R}{R_e}$ .
Energy: $\frac{1}{2}v^2 - \frac{GM}{R_e} = \frac{1}{2}v'^2 - \frac{GM}{R} \Rightarrow \frac{3}{4} - 1 = \frac{3}{4x^2} - \frac{1}{x}$ .
$x^2 - 4x + 3 = 0 \Rightarrow x = 3$ (the root $x = 1$ is the perigee).
So the answer is (b) $R = 3R_e$ .

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Satellite Energy & Weightlessness question from JEE Main 2020 (3 Sep, Shift 1)

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