Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a p

Q: Why does the mass of the planet $m$ not affect the final proportionality?

In any orbital motion where gravity is the only force acting, the gravitational force is directly proportional to the mass of the orbiting body ($F_g \\propto m$), and the required centripetal force is also proportional to the mass ($F_c = \\frac{mv^2}{R}$). When we equate them, the mass $m$ cancels out from both sides, meaning the orbital speed and time period are completely independent of the planet's mass.

Q: Can we use Kepler's Third Law ($T^2 \\propto R^3$) directly here?

No! Kepler's Third Law ($T^2 \\propto R^3$) is strictly derived from Newton's Inverse-Square Law of Gravitation ($F_g \\propto R^{-2}$). Since this problem specifies a non-standard force law ($F_g \\propto R^{-5/2}$), Kepler's standard law does not apply, and we must derive the relationship from first principles.

Q: What happens to the stability of the orbit under a force law like $F_g \\propto R^{-5/2}$?

According to Bertrand's Theorem, stable, closed circular orbits only exist for inverse-square forces ($F \\propto R^{-2}$) and radial harmonic oscillator forces ($F \\propto R$). Under a force law like $F_g \\propto R^{-5/2}$, a perfect circular orbit is mathematically possible but highly unstable; any tiny perturbation would cause the planet to either spiral into the star or escape to infinity.

Q: How do I handle negative exponents when simplifying $T \\propto \\frac{R}{R^{-3/4}}$?

Remember the exponent rule: $\\frac{x^a}{x^b} = x^{a-b}$. Here, the exponent in the numerator is $1$ (since $R = R^1$) and the exponent in the denominator is $-3/4$. Thus, we get $R^{1 - (-3/4)} = R^{1 + 3/4} = R^{7/4}$.

Q: Is there a quick way to remember the generalized formula for exams?

Yes! For any central force of the form $F \\propto R^{-n}$, the relationship is always $T^2 \\propto R^{n+1}$. This is a highly useful shortcut for JEE.

Question

Imagine a light planet revolving around a very massive star in a circular orbit of radius $R$ with a period of revolution $T$. If the gravitational force of attraction between the planet and the star is proportional to $R^{-5/2}$, then

Options:

(A) $T^2$ is proportional to $R^2$

(B) $T^2$ is proportional to $R^{7/2}$

(C) $T^2$ is proportional to $R^{3/2}$

(D) $T^2$ is proportional to $R^{3.75}$

Sigma Percentile · Accepted Answer

### Exploring Non-Standard Gravitational Fields

Imagine stepping into an alternate universe where the laws of physics are subtly rewritten.

In our universe, gravity is governed by Newton's famous **Inverse-Square Law**, where the force of attraction between two bodies falls off as the square of the distance between them, i.e., $F_g \propto R^{-2}$.

This inverse-square nature is incredibly special; it is the reason why planetary orbits are stable, closed ellipses (as proven by Kepler and Newton).

But what if gravity behaved differently? What if the force of attraction was proportional to $R^{-5/2}$?

In this problem, we explore the dynamics of a light planet orbiting a massive star under this hypothetical force law and determine how the orbital time period $T$ scales with the orbital radius $R$.

---

### The Physics of Circular Orbits

For any object of mass $m$ to maintain a stable circular orbit of radius $R$ with a constant speed $v$, a net inward force must act on it.

This force is called the **centripetal force**, and its magnitude is given by:

$$F_c = \frac{m v^2}{R}$$

In our orbital system, this centripetal force is provided entirely by the gravitational attraction between the planet and the star.

Therefore, we can set the centripetal force proportional to the gravitational force:

$$F_c = F_g$$

Given that the gravitational force in this universe scales as $R^{-5/2}$, we write:

$$\frac{m v^2}{R} \propto R^{-5/2}$$

Notice that the mass of the planet $m$ is a constant, so we can absorb it into our proportionality relation.

---

### Solving for Orbital Velocity

To understand how the time period scales, we first need to find how the orbital speed $v$ depends on the radius $R$.

Let's isolate $v^2$ by multiplying both sides of our proportionality by $R$:

$$v^2 \propto R \cdot R^{-5/2}$$

Using the laws of exponents, we add the powers of $R$:

$$v^2 \propto R^{1 - 5/2}$$

$$v^2 \propto R^{-3/2}$$

To find the velocity $v$, we take the square root of both sides:

$$v \propto \left(R^{-3/2}\right)^{1/2}$$

$$v \propto R^{-3/4}$$

This tells us that as the radius of the orbit increases, the orbital speed of the planet decreases, but at a different rate than in our universe (where $v \propto R^{-1/2}$).

---

### Connecting Velocity to Time Period

Now, let's relate the orbital speed $v$ to the time period $T$ of one complete revolution.

The time period is simply the total distance traveled in one orbit (the circumference of the circle) divided by the speed:

$$T = \frac{2\pi R}{v}$$

Since $2\pi$ is a constant, we can write:

$$T \propto \frac{R}{v}$$

Now, we substitute our expression for $v$ ($v \propto R^{-3/4}$) into this relation:

$$T \propto \frac{R}{R^{-3/4}}$$

Using exponent rules to simplify this fraction:

$$T \propto R^{1 - (-3/4)}$$

$$T \propto R^{1 + 3/4}$$

$$T \propto R^{7/4}$$

---

### Finding the Final Proportionality for $T^2$

To match the options provided in the question, we need to find the relationship for $T^2$.

Let's square both sides of our proportionality:

$$T^2 \propto \left(R^{7/4}\right)^2$$

$$T^2 \propto R^{7/2}$$

Thus, the square of the time period is proportional to $R^{7/2}$. This perfectly matches **Option (b)**.

---

### Generalizing the Result: A Powerful Shortcut

What if you encounter a similar question in JEE with a different exponent, say $F_g \propto R^{-n}$?

Let's derive a general shortcut so you can solve these questions in seconds!

For any central force $F \propto R^{-n}$:

1. Centripetal force balance: 
   $$\frac{v^2}{R} \propto R^{-n} \implies v^2 \propto R^{1-n} \implies v \propto R^{\frac{1-n}{2}}$$

2. Time period relation:
   $$T \propto \frac{R}{v} \implies T \propto \frac{R}{R^{\frac{1-n}{2}}} \implies T \propto R^{1 - \frac{1-n}{2}} \implies T \propto R^{\frac{n+1}{2}}$$

3. Squaring both sides:
   $$T^2 \propto R^{n+1}$$

Let's test our shortcut with our current problem where $n = 5/2$:

$$T^2 \propto R^{5/2 + 1} = R^{7/2}$$

It works flawlessly!

What about our real universe where $n = 2$ (Newton's Inverse-Square Law)?

$$T^2 \propto R^{2+1} = R^3$$

This is none other than **Kepler's Third Law of Planetary Motion** ($T^2 \propto R^3$)!

By understanding the underlying physics, we have not only solved this specific problem but also mastered a generalized framework for any central force law.

Animated Solution for Physics - Gravitation: Imagine a light planet revolving around a very massive star in a circular orbit of radius $R$ with a period of revolution $T$ . If the gravitational force of attraction between the planet and the star is proportional to $R^{- 5/2}$ , then

Select Answer:

Visualized Solution

Visualizing the Orbital Setup

The Centripetal Force Requirement

Equating Gravitational and Centripetal Forces

Finding Velocity Proportionality

Relating Time Period to Velocity

Substituting Velocity into Time Period

Simplifying the Exponent of $R$

Squaring to Find $T^{2}$

The Way Forward: Generalizing for $F \propto R^{- n}$

The Sigma Insight: Kepler's Laws of Planetary Motion

Exploring Non-Standard Gravitational Fields

The Physics of Circular Orbits

Solving for Orbital Velocity

Connecting Velocity to Time Period

Finding the Final Proportionality for $T^{2}$

Generalizing the Result

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