A particle of mass $$m$$ moves in a circular orbit under the central potential field, $$U(r) = \frac{C}{r}$$, where $$C$$ is a positive constant. The correct radius - velocity graph of the particle's motion is :

We need to determine the correct relationship between the radius $$r$$ and the orbit velocity $$v$$ of a particle moving in a stable circular orbit under a given central potential field.

1. Relate Potential Energy to Force

From the problem, the central potential field is given by:

$$U(r) = \frac{C}{r} = C \cdot r^{-1}$$

Where $$C$$ is a positive constant.

The conservative central force $$\vec{F}$$ acting on the particle is related to the gradient of this potential energy by $$F = -\frac{dU}{dr}$$:

$$F = -\frac{d}{dr}\left(C \cdot r^{-1}\right) = -C \cdot (-1 \cdot r^{-2}) = \frac{C}{r^2}$$

This tells us that the particle experiences an attractive central force directed towards the origin whose magnitude is inversely proportional to the square of the radius.

2. Apply the Circular Orbit Condition

For a particle of mass $$m$$ maintaining a stable circular orbit of radius $$r$$ with a constant tangential velocity $$v$$, this conservative central force provides the necessary centripetal force:

$$F_{\text{centripetal}} = \frac{m v^2}{r}$$

Equating the two forces gives:

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

3. Determine the Radius-Velocity ($$r-v$$) Relationship

Simplify the equation by multiplying both sides by $$r$$:

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

Isolate the radius $$r$$:

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

Since both $$C$$ and $$m$$ are positive constants, we can express this relationship as a proportionality:

$$r \propto \frac{1}{v^2}$$

4. Identify the Correct Graph

Let's analyze how the radius behaves as a function of velocity based on $$r = \frac{\text{constant}}{v^2}$$:

• As the velocity $$v$$ approaches zero ($$v \to 0$$), the orbit radius shoots up towards infinity ($$r \to \infty$$).
• As the velocity $$v$$ becomes very large ($$v \to \infty$$), the orbit radius decays asymptotically towards zero ($$r \to 0$$).

This inverse-square relationship yields a decreasing, non-linear curve that stays within the first quadrant, asymptotically approaching both the vertical $$r$$-axis and horizontal $$v$$-axis. This behavior matches the curve shown in the first graph option.

Conclusion

The correct radius-velocity graph of the particle's motion is represented by Option A.