The maximum and minimum distances of a comet from the Sun are $1.6 \times 10^{12} \mathrm{~m}$ and $8.0 \times 10^{10} \mathrm{~m}$ respectively. If the speed of the comet at the nearest point is $6 \times 10^{4} \mathrm{~ms}^{-1}$, the speed at the farthest point is:

(A) $3.0 \times 10^{3} \mathrm{~m} / \mathrm{s}$

(B) $1.5 \times 10^{3} \mathrm{~m} / \mathrm{s}$

(C) $4.5 \times 10^{3} \mathrm{~m} / \mathrm{s}$

(D) $6.0 \times 10^{3} \mathrm{~m} / \mathrm{s}$

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Gravitation

The maximum and minimum distances of a comet from the Sun are

1.6 \times 1 0^{12} m

and

8.0 \times 1 0^{10} m

respectively. If the speed of the comet at the nearest point is

6 \times 1 0^{4} ms^{- 1}

, the speed at the farthest point is:

(A) $3.0 \times 1 0^{3} m / s$ ✓ Correct
(B) $1.5 \times 1 0^{3} m / s$
(C) $4.5 \times 1 0^{3} m / s$
(D) $6.0 \times 1 0^{3} m / s$

Answer: (A)

Solution

The Sun's pull on the comet is always along the line joining them, so it exerts no torque about the Sun and the comet's angular momentum is conserved. At the nearest and farthest points the velocity is perpendicular to the radius, so

v_{1} r_{1} = v_{2} r_{2}

v_{2} = \frac{6 \times 1 0 ^{4} \times 8.0 \times 1 0 ^{10}}{1.6 \times 1 0 ^{12}} = 3.0 \times 1 0^{3} m s^{- 1}

At the farthest point the comet moves at

3.0 \times 1 0^{3}

m/s.

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Gravitation

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