JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Gravitation

A particle of mass m moves in a circular orbit under the central potential field,

V (r) = - \frac{C}{r}

, where C is a constant. The correct radius - velocity graph of the particle's motion is ;

(A)(see structure)
(B)(see structure)
(C)(see structure)
✓ Correct
(D)(see structure)

Answer: (C)

Solution

The central force is the negative gradient of the potential,

F_{r} = - \frac{d V}{d r} = - \frac{C}{r ^{2}}

, an attraction of magnitude

\frac{C}{r ^{2}}

. For a circular orbit it supplies the centripetal force:

\frac{m v ^{2}}{r} = \frac{C}{r ^{2}} \Rightarrow r = \frac{C}{m v ^{2}}

The orbital radius therefore falls off as

1/ v^{2}

, so the radius-velocity curve starts high at small speeds and decreases steadily, approaching the velocity axis without meeting it.

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option

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