The maximum and minimum distances of a comet from the Sun are 1.6 × 1 0 12 m and 8.0 × 1 0 10 m respectively. If the speed of the comet at the nearest point is 6 × 1 0 4 ms − 1 , the speed at the farthest point is :

By angular momentum conservation: m 1 v 1 r 1 = m 2 v 2 r 2 v 1 = 1.6 × 1 0 12 48 × 1 0 14 = 3000 m / s v 1 = 3 × 1 0 3 m / s

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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 :

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Answer:

Solution:

By angular momentum conservation:

$m_{1} v_{1} r_{1} = m_{2} v_{2} r_{2}$

$v_{1} = \frac{48 \times 1 0 ^{14}}{1.6 \times 1 0 ^{12}} = 3000 m / s$

$v_{1} = 3 \times 1 0^{3} m / s$

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Angular momentum

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