The maximum and minimum distances of a comet from the Sun are 1 . 6 × 10 12 m and 8 . 0 × 10 10 m respectively. If the speed of the comet at the nearest point is 6 × 10 4 m s - 1 , the speed at the farthest point is

By angular momentum conservation : m v 1 r 1 = m v 2 r 2 v 1 = 48 × 10 14 1 . 6 × 10 12 = 3000 m s - 1 = 3 × 10 3 m s - 1 .

Physics - Rotation Question with Solution | TestHub

PhysicsRotationAngular momentumEasy2 minPYQ_2021

PhysicsEasysingle choice

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

Options:

Answer:

Solution:

By angular momentum conservation :

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

$v_{1} = \frac{48 \times 10^{14}}{1.6 \times 10^{12}} = 3000 m s^{- 1}$

$= 3 \times 10^{3} m s^{- 1}$ .

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Angular momentum

⏱ 2mℹ️ Source: PYQ_2021

Doubts & Discussion

Loading discussions...