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JEE Main 2021 (16 Mar, Shift 1) · Kepler's Laws & Planetary Motion

Kepler's Laws & Planetary Motion question from JEE Main 2021 (16 Mar, Shift 1)

Q18JEE Main 2021 · 16 MAR · SHIFT 1Easy

Kepler's Laws & Planetary Motion

The maximum and minimum distances of a comet from the Sun are 1.6 $\times$ 1012 m and 8.0 $\times$ 1010 m respectively. If the speed of the comet at the nearest point is 6 $\times$ 104 ms ${}^{-1}$ , the speed at the farthest point is:

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Answer: (a)

Conservation of angular momentum: $v_{\min} r_{\max} = v_{\max} r_{\min}$ .
$v_{\min} = \dfrac{6\times10^{4} \times 8.0\times10^{10}}{1.6\times10^{12}} = \dfrac{6\times10^{4}}{20}$
$v_{\min} = 3.0\times10^{3}\text{ m/s}$ , so the answer is (a).

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