A particle of mass m moves in a circular orbit under the central potential field, U(r) = -C/r, where

Q: Why do we take the derivative of potential energy to find the force?

In physics, conservative forces are defined as the negative gradient (or spatial derivative) of potential energy. It mathematically describes how a particle is pushed towards regions of lower potential energy.

Q: What does the negative sign in $F = -\frac{C}{r^2}$ mean?

The negative sign indicates the direction of the force. In a radial coordinate system, a negative force points inwards towards the origin, meaning it is an attractive force.

Q: Why do we equate the central force to $\frac{mv^2}{r}$?

An object moving in a circle is constantly accelerating towards the center. This requires a net inward force called the centripetal force. Here, the central potential field is the only thing pulling the particle, so it must act as this centripetal force.

Q: How do we know the graph of $v \propto \frac{1}{\sqrt{r}}$ is a decreasing curve?

Because $r$ is in the denominator, as $r$ gets larger, the fraction $\frac{1}{\sqrt{r}}$ gets smaller. This means velocity decreases as radius increases, which corresponds to a downward-sloping curve.

Q: Is this similar to how planets orbit the sun?

Exactly! The gravitational potential energy is also of the form $U = -\frac{GMm}{r}$. Following the exact same steps, we find that the orbital velocity of planets is $v \propto \frac{1}{\sqrt{r}}$.

Question

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

Sigma Percentile · Accepted Answer

### The Dance of Central Potentials

Imagine a satellite orbiting a planet, or an electron whizzing around a nucleus. What keeps them in their elegant, never-ending circular dance? The answer lies in the invisible, pulling hand of a **central potential field**.

In this problem, we are given a particle of mass $m$ moving in a circular orbit under a central potential $U(r) = -\frac{C}{r}$. Our mission is to uncover the hidden relationship between the particle's velocity $v$ and its orbital radius $r$, and then translate that relationship into a visual graph.

### Unmasking the Force

The first step in our journey is to find the force acting on the particle. In physics, potential energy and conservative force are two sides of the same coin. The force is simply the negative spatial gradient (or derivative) of the potential energy.

Mathematically, this is expressed as:

$$F = -\frac{dU}{dr}$$

Let's plug in our given potential $U(r) = -\frac{C}{r}$:

$$F = -\frac{d}{dr}\left(-\frac{C}{r}ight)$$

Taking the derivative, we get:

$$F = -\frac{C}{r^2}$$

The negative sign here is crucial—it tells us that the force is **attractive**, pulling the particle radially inward towards the center. The magnitude of this force is $|F| = \frac{C}{r^2}$.

### The Centripetal Requirement

For any object to travel in a perfect circle, it requires a constant inward pull to continuously change its direction. This is the **centripetal force**, given by the famous formula:

$$F_c = \frac{mv^2}{r}$$

In our scenario, the attractive central force we just calculated is the *only* force acting on the particle. Therefore, it must be the one providing this necessary centripetal force.

Let's equate the magnitude of our central force to the centripetal force:

$$\frac{C}{r^2} = \frac{mv^2}{r}$$

### Decoding the Velocity-Radius Relationship

Now comes the elegant algebraic simplification. We can cancel one factor of $r$ from the denominators on both sides:

$$\frac{C}{r} = mv^2$$

Rearranging to solve for $v^2$, we get:

$$v^2 = \frac{C}{mr}$$

Since $C$ and $m$ are positive constants, we can clearly see the proportionality:

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

Taking the square root of both sides reveals the final relationship:

$$v \propto \frac{1}{\sqrt{r}}$$

### Visualizing the Math

What does the graph of $v \propto \frac{1}{\sqrt{r}}$ look like?

Let's analyze its behavior:
1. As $r$ becomes very small ($r 	o 0$), the velocity $v$ shoots up to infinity ($v 	o \infty$).
2. As $r$ becomes very large ($r 	o \infty$), the velocity $v$ approaches zero ($v 	o 0$).

This describes a curve that starts high up on the $v$-axis and gently slopes downward, asymptotically approaching the $r$-axis. It is a decreasing, concave-up curve, often referred to as a hyperbolic-type curve.

Looking at our options, the first graph perfectly captures this exact mathematical behavior. The physics of the central potential has beautifully dictated the geometry of the particle's motion!

Animated Solution for Physics - Work, Energy, and Power: 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

Select Answer:

Visualized Solution

$Visualizing the Orbit$

$F = - \frac{d U}{d r}$

$Calculating Force$

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

$Equating Forces$

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

$Graph Analysis$

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Dance of Central Potentials

Unmasking the Force

The Centripetal Requirement

Decoding the Velocity-Radius Relationship

Visualizing the Math

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