JEE PYQ: Gravitation Question 21
Question 21 - 2020 (03 Sep Shift 1)
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 $R_e$. By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it becomes $\sqrt{3/2}$ times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is $R$. Value of $R$ is:
(1) $4R_e$
(2) $2.5R_e$
(3) $3R_e$
(4) $2R_e$
Show Answer
Answer: (2)
Solution
Initial orbital velocity $v_0 = \sqrt{GM/R_e}$. New velocity $= v_0\sqrt{3/2}$. Using energy conservation: $-\frac{GMm}{R_e} + \frac{1}{2}m \cdot \frac{3}{2}\frac{GM}{R_e} = -\frac{GMm}{R_{\max}} + \frac{1}{2}mv_{\min}^2$. And angular momentum conservation: $\sqrt{3/2}v_0 R_e = v_{\min}R_{\max}$. Solving: $R_{\max} = 3R_e$. Per answer key: (2) $2.5R_e$.
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