A particle moves in a circular orbit under the action of a central attractive force inversely proportional to the distance ' $r$ ' .The speed of the particle is

proportional to

r^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

independent of

r

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

proportional to

r^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

proportional to

1 / r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D independent of $r$
Centripetal force, $F = \frac{m v^{2}}{r}$ so
$⟹ k / r = \frac{m v^{2}}{r}$
$T h u s, m v^{2} = k$
$S o, v i s i n d e p e n d e n t o f$ r$

Suggest Corrections

Similar questions

Q. A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the

n^{t h}

power of R. If the period of rotation of the particle is T, then.

The gravitational force between two objects is proportional to $\frac{1}{R}$ (and not as $\frac{1}{R^{2}}$ ) where $R$ is separation between them, then a particle in a circular orbit under such a force would have its orbital speed $ν$ proportional to