Question

A point P lies on the axis of a fixed ring of mass M and radius R at a distance from centre O. The particle starts from P and reach at O under gravitational attraction between ring and particle. The speed of particle at distance R from centre is

Answer 1

Given:

A point P lies on the axis of a fixed ring of mass M and radius R at a distance 2R from centre O.

The particle, of mass m, starts from P and reach at O under gravitational attraction between ring and particle

To Find:

The speed of particle at distance R from center.

Solution:

Gravitational Potential at point P due to the ring ,

• Pp = -GM/√5R
• Potential Energy = mPp

Gravitational Potential at a distance R from the center is,

• V = -GM/√2R
• Potential Energy = mP

We know, By

Work Energy theorem,

• Change in potential Energy = Kinetic energy .

• -GMm/√5R - -GMm/√2R = 0.5mV²
• 2$\frac{GM}{R }$( 1/√2 - 1/√5) = V²
• V = $\sqrt{2\frac{GM}{R }(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{5} } ) }$m/s

The speed of particle at distance R from centre is V = $\sqrt{2\frac{GM}{R }(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{5} } ) }$m/s.

Answer 2

Answer:

hope it helps you please mark as brainliest and thank you

Explanation:

option c is the answer