A particle is moving with a uniform speed in a circular orbit of radius R in a central force inverse

Q: Why do we equate the central force to the centripetal force?

Centripetal force is not a new, separate force of nature; it is simply the *name* we give to whatever net inward force is keeping an object in circular motion. In this case, the central force $F = k/R^n$ is the only force acting inward, so it *is* the centripetal force.

Q: Why did we use $m\omega^2 R$ instead of $mv^2/R$?

Both formulas are perfectly valid! However, the question specifically asks for the relationship involving the time period $T$. Since $\omega = 2\pi/T$, using the $\omega$ version of the formula gets us to $T$ much faster than using $v = 2\pi R/T$.

Q: What happens to the time period if $n$ is negative?

If $n$ is negative, the force actually *increases* with distance. For example, if $n = -1$, the force is $F \propto R$, which is Hooke's Law for a spring! Plugging $n = -1$ into our formula gives $T \propto R^0$, meaning the time period is independent of the radius—a classic property of simple harmonic motion.

Q: Is the constant $k$ important for the final answer?

No, because the question only asks for the *proportionality* between $T$ and $R$. The constant $k$, along with $4\pi^2$ and $m$, gets absorbed into the proportionality sign. They affect the exact value of $T$, but not how $T$ scales with $R$.

Q: Can this formula be applied to elliptical orbits?

Strictly speaking, no. This derivation assumed a perfect circular orbit where the radius $R$ and speed $v$ are constant. For elliptical orbits, the math becomes more complex, though for gravity ($n=2$), Kepler's Third Law still holds if you replace the radius $R$ with the semi-major axis $a$.

Question

A particle is moving with a uniform speed in a circular orbit of radius $R$ in a central force inversely proportional to the nth power of $R$. If the period of rotation of the particle is $T$, then :

Options:

(A) $T \propto R^{3/2}$ for any $n$

(B) $T \propto R^{\frac{n}{2}+1}$

(C) $T \propto R^{(n+1)/2}$

(D) $T \propto R^{n/2}$

Sigma Percentile · Accepted Answer

### The Beauty of Central Forces

Imagine a universe where forces only know one direction: straight towards or away from a central point. These are **central forces**, and they are the invisible architects of our cosmos. From the gravitational pull that keeps the Earth tethered to the Sun, to the electrostatic attraction binding electrons to a nucleus, central forces dictate the dance of matter.

In this problem, we are introduced to a generalized central force. Instead of a specific law like gravity, we are given a force $F$ that is inversely proportional to the $n$-th power of the distance $R$. Mathematically, this is expressed as:

$$F \propto \frac{1}{R^n}$$

By introducing a constant of proportionality $k$, we can write this as an exact equation:

$$F = \frac{k}{R^n}$$

### The Mechanics of Uniform Circular Motion

When a particle of mass $m$ moves in a perfect circle with a uniform speed $v$, its velocity vector is constantly changing direction. This continuous change in direction requires a continuous inward pull—a **centripetal force**.

Without this force, the particle would fly off in a straight line, obeying Newton's First Law of inertia. The magnitude of this required centripetal force $F_c$ is given by the classic formula:

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

However, the problem asks us about the **time period** $T$, which is the time taken to complete one full revolution. It is often more elegant to express the centripetal force in terms of the angular velocity $\omega$. Since $v = \omega R$, we can rewrite the force as:

$$F_c = m\omega^2 R$$

We also know that the angular velocity is the total angle of a circle ($2\pi$ radians) divided by the time period $T$:

$$\omega = \frac{2\pi}{T}$$

Substituting this into our force equation gives us the master equation for centripetal force in terms of the time period:

$$F_c = m\left(\frac{2\pi}{T}\right)^2 R$$

### The Master Equation: Bridging the Gap

Now comes the crucial logical bridge. The abstract central force provided in the problem is the *exact same force* that acts as the centripetal force keeping the particle in its circular orbit. Therefore, we must equate the two expressions:

$$\frac{k}{R^n} = m\left(\frac{2\pi}{T}\right)^2 R$$

This single equation contains all the physics of the problem. The rest is pure algebraic manipulation. Let's carefully expand the squared term on the right side:

$$\frac{k}{R^n} = \frac{4\pi^2 m}{T^2} R$$

### Algebraic Manipulation: Isolating the Time Period

Our goal is to find how the time period $T$ depends on the radius $R$. To do this, we need to isolate $T^2$ on one side of the equation. Let's cross-multiply to bring $T^2$ to the numerator on the left, and move the $R^n$ to the right:

$$T^2 = \frac{4\pi^2 m}{k} \cdot R \cdot R^n$$

Using the laws of exponents, when we multiply terms with the same base, we add their powers. Thus, $R^1 \cdot R^n = R^{n+1}$:

$$T^2 = \frac{4\pi^2 m}{k} R^{n+1}$$

### The Grand Conclusion

We are almost there. The term $\frac{4\pi^2 m}{k}$ is entirely composed of constants. Therefore, we can strip away the equality and return to a proportionality:

$$T^2 \propto R^{n+1}$$

To find the proportionality for $T$ itself, we take the square root of both sides. Remember that taking the square root is equivalent to raising a term to the power of $\frac{1}{2}$:

$$T \propto \left(R^{n+1}\right)^{\frac{1}{2}}$$

$$T \propto R^{\frac{n+1}{2}}$$

This perfectly matches option (c).

### Beyond the Problem: Kepler's Third Law

The true power of physics lies in its generality. Let's test our derived formula on a familiar scenario: **Gravity**.

Newton's Law of Universal Gravitation states that the gravitational force is an inverse-square law. This means that for gravity, $n = 2$. Let's plug $n = 2$ into our newly minted formula:

$$T \propto R^{\frac{2+1}{2}}$$

$$T \propto R^{\frac{3}{2}}$$

Squaring both sides yields:

$$T^2 \propto R^3$$

This is exactly **Kepler's Third Law of Planetary Motion**! By solving an abstract problem with a generalized power $n$, we have inadvertently proven one of the most famous laws in astronomy. This is the elegance of mathematical physics—a single generalized equation can hold the secrets of the entire solar system.

Animated Solution for Physics - Laws of Motion: A particle is moving with a uniform speed in a circular orbit of radius $R$ in a central force inversely proportional to the nth power of $R$ . If the period of rotation of the particle is $T$ , then :

Select Answer:

Visualized Solution

$Visualizing the Orbit$

$Centripetal Force$

$Given Central Force$

$Equating the Forces$

$Solving for Time Period T$

$Final Proportionality$

$The Way Forward: Kepler’s Law$

The Sigma Insight: Dynamics of Circular Motion

The Beauty of Central Forces

The Mechanics of Uniform Circular Motion

The Master Equation

Algebraic Manipulation

The Grand Conclusion

Beyond the Problem

Similar Questions

A particle is moving with uniform speed along the circumference of a circle of radius $R$ under the action of a central fictitious force $F$ which is inversely proportional to $R^{3}$ . Its time period of revolution will be given by

A particle of mass $m$ is suspended from a ceiling through a string of length $L$ . The particle moves in a horizontal circle of radius $r$ such that $r = \frac{L}{2}$ . The speed of particle will be

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that

Consider a binary star system of star A and star B with masses $m_{A}$ and $m_{B}$ revolving in a circular orbit of radii $r_{A}$ and $r_{B}$ , respectively. If $T_{A}$ and $T_{B}$ are the time period of star A and star B respectively, then

A tube of length $L$ is filled completely with an in-compressible liquid of mass $M$ and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $ω$ . The force exerted by the liquid at the other end is