A planet of mass M, has two natural satellites with masses m1 and m2. The radii of their circular or

Q: Why does the mass of the satellite cancel out when calculating orbital velocity?

The gravitational force acting on the satellite is proportional to its mass ($F_g \propto m$), and the required centripetal force to keep it in a circle is also proportional to its mass ($F_c \propto m$). Because both forces scale linearly with the satellite's mass, the mass cancels out. This means all objects at a given orbital radius must travel at the exact same speed to maintain a circular orbit, regardless of how heavy they are.

Q: I got confused with the fractional exponent in $(1/4)^{3/2}$. How do I calculate this easily?

A fractional exponent of $3/2$ can be broken down into two simpler operations: a square root (the $/2$ part) and a cube (the $3$ part). So, first take the square root of $1/4$, which is $1/2$. Then, raise that result to the power of $3$, which gives $(1/2)^3 = 1/8$. Breaking it down this way prevents calculation errors.

Q: Why is the angular momentum ratio exactly 1 even though the orbits and masses are different?

This is a beautiful coincidence of the specific values given in this problem. Since $L \propto m\sqrt{R}$, the ratio is $\frac{L_1}{L_2} = \left(\frac{m_1}{m_2}\right) \sqrt{\frac{R_1}{R_2}}$. Substituting the given values, we get $2 \times \sqrt{1/4} = 2 \times 1/2 = 1$. The larger mass of the inner satellite perfectly compensates for its smaller orbital radius.

Q: Does Kepler's Third Law apply to all orbits, or only circular ones?

Kepler's Third Law ($T^2 \propto a^3$) applies to all elliptical orbits as well, where $a$ is the semi-major axis of the ellipse. For circular orbits, the semi-major axis is simply the radius $R$, so it simplifies to $T^2 \propto R^3$.

Q: What would happen to the mechanical energy of the satellites if we didn't ignore the gravitational force between them?

If we did not ignore the gravitational force between the satellites, they would exert forces on each other, causing perturbations in their orbits. Their individual mechanical energies would no longer be constant, and their orbits would deviate from perfect circles, potentially leading to chaotic motion or orbital decay over long periods.

Question

A planet of mass $M$, has two natural satellites with masses $m_1$ and $m_2$. The radii of their circular orbits are $R_1$ and $R_2$, respectively. Ignore the gravitational force between the satellites. Define $v_1, L_1, K_1$ and $T_1$ to be respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite 1; and $v_2, L_2, K_2$ and $T_2$ to be the corresponding quantities of satellite 2. Given, $\frac{m_1}{m_2} = 2$ and $\frac{R_1}{R_2} = \frac{1}{4}$, match the ratios in column-I to the numbers in column-II.

List-I:

P. A. $\frac{v_1}{v_2}$

Q. B. $\frac{L_1}{L_2}$

R. C. $\frac{K_1}{K_2}$

S. D. $\frac{T_1}{T_2}$

List-II:

1. p. $\frac{1}{8}$

2. q. $1$

3. r. $2$

4. s. $8$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the center of a solar system, watching two satellites orbit a massive central planet of mass $M$.

This problem, sourced from the prestigious **JEE Advanced 2018**, is a beautiful test of your fundamental understanding of circular orbital mechanics.

Instead of testing a single formula, it asks you to compare four key physical quantities—**orbital speed**, **angular momentum**, **kinetic energy**, and **orbital time period**—for two satellites moving at different distances.

Let's write down our given parameters clearly:

$$\frac{m_1}{m_2} = 2 \quad 	ext{and} \quad \frac{R_1}{R_2} = \frac{1}{4}$$

This means satellite 1 is twice as heavy as satellite 2, but it orbits much closer to the planet—at only one-fourth of the distance of satellite 2.

---

### The Master Equation: Orbital Velocity

What keeps a satellite in a stable circular orbit? It is the delicate balance between the gravitational pull of the planet and the inertia of the satellite trying to fly off in a straight line.

Mathematically, the gravitational force provides the necessary centripetal force:

$$\frac{G M m}{R^2} = \frac{m v^2}{R}$$

Notice how the mass of the satellite $m$ cancels out from both sides of the equation. This is a profound physical realization: **the speed required to stay in a stable orbit depends only on the mass of the central body and the radius of the orbit, not on the mass of the satellite itself.**

Solving for the orbital velocity $v$, we get:

$$v = \sqrt{\frac{G M}{R}}$$

This tells us that orbital velocity is inversely proportional to the square root of the orbital radius:

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

Now, let's find the ratio of their speeds:

$$\frac{v_1}{v_2} = \sqrt{\frac{R_2}{R_1}}$$

Since we are given $\frac{R_1}{R_2} = \frac{1}{4}$, the reciprocal ratio is $\frac{R_2}{R_1} = 4$. Substituting this in, we get:

$$\frac{v_1}{v_2} = \sqrt{4} = 2$$

Thus, the inner satellite travels at exactly **twice** the speed of the outer satellite. This matches **A $ightarrow$ r**.

---

### Conservation and Ratios of Angular Momentum

Next, let's look at the angular momentum $L$ of the satellites about the center of the planet.

For a circular orbit, the position vector and velocity vector are always perpendicular. Therefore, the magnitude of angular momentum is simply:

$$L = m v R$$

Substituting our expression for orbital velocity $v = \sqrt{\frac{G M}{R}}$ into this formula, we get:

$$L = m \sqrt{\frac{G M}{R}} R = m \sqrt{G M R}$$

This reveals that the angular momentum of a satellite is directly proportional to its mass and the square root of its orbital radius:

$$L \propto m \sqrt{R}$$

Let's calculate the ratio of their angular momenta:

$$\frac{L_1}{L_2} = \left(\frac{m_1}{m_2}ight) \sqrt{\frac{R_1}{R_2}}$$

Substituting the given ratios $\frac{m_1}{m_2} = 2$ and $\frac{R_1}{R_2} = \frac{1}{4}$:

$$\frac{L_1}{L_2} = 2 	imes \sqrt{\frac{1}{4}} = 2 	imes \frac{1}{2} = 1$$

Remarkably, even though the two satellites have different masses and orbit at different distances, their angular momenta are **exactly equal**! This matches **B $ightarrow$ q**.

---

### Comparing Kinetic Energies

Now, let's analyze the kinetic energy $K$ of each satellite.

The kinetic energy is defined as:

$$K = \frac{1}{2} m v^2$$

Using the relation $v^2 = \frac{G M}{R}$, we can write the kinetic energy as:

$$K = \frac{G M m}{2 R}$$

This shows that kinetic energy is directly proportional to the mass of the satellite and inversely proportional to the orbital radius:

$$K \propto \frac{m}{R}$$

Let's find the ratio of their kinetic energies:

$$\frac{K_1}{K_2} = \left(\frac{m_1}{m_2}ight) \left(\frac{R_2}{R_1}ight)$$

Be careful here! Since radius is in the denominator, the ratio is inverted to $\frac{R_2}{R_1}$. Substituting our values:

$$\frac{K_1}{K_2} = 2 	imes 4 = 8$$

The inner satellite has **eight times** the kinetic energy of the outer satellite. This matches **C $ightarrow$ s**.

---

### Kepler's Third Law and Time Period

Finally, let's determine the time period of revolution $T$, which is the time taken to complete one full orbit.

The time period is the total distance of one orbit divided by the orbital speed:

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

Substituting $v = \sqrt{\frac{G M}{R}}$ gives:

$$T = 2 \pi \sqrt{\frac{R^3}{G M}}$$

Squaring both sides yields Kepler's famous Third Law:

$$T^2 \propto R^3 \implies T \propto R^{3/2}$$

Let's find the ratio of their time periods:

$$\frac{T_1}{T_2} = \left(\frac{R_1}{R_2}ight)^{3/2}$$

Substituting $\frac{R_1}{R_2} = \frac{1}{4}$:

$$\frac{T_1}{T_2} = \left(\frac{1}{4}ight)^{3/2} = \left(\sqrt{\frac{1}{4}}ight)^3 = \left(\frac{1}{2}ight)^3 = \frac{1}{8}$$

Thus, the inner satellite completes its orbit in only **one-eighth** of the time taken by the outer satellite. This matches **D $ightarrow$ p**.

---

### Final Synthesis

Let's collect all our matched pairs:
*   **A $ightarrow$ r** ($v_1/v_2 = 2$)
*   **B $ightarrow$ q** ($L_1/L_2 = 1$)
*   **C $ightarrow$ s** ($K_1/K_2 = 8$)
*   **D $ightarrow$ p** ($T_1/T_2 = 1/8$)

Comparing this with the given options, we find that it matches perfectly with **option (b)**.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Visualizing the Satellite System

The Physics of Orbital Velocity

Calculating the Velocity Ratio $\frac{v _{1}}{v _{2}}$

Understanding Angular Momentum

Calculating the Angular Momentum Ratio $\frac{L _{1}}{L _{2}}$

The Physics of Kinetic Energy

Calculating the Kinetic Energy Ratio $\frac{K _{1}}{K _{2}}$

Kepler's Third Law of Planetary Motion

Calculating the Time Period Ratio $\frac{T _{1}}{T _{2}}$

Matching the Columns and Selecting the Option

The Sigma Insight: Orbital Motion of a Satellite

Analyzing the Setup

The Master Equation

Conservation and Ratios of Angular Momentum

Comparing Kinetic Energies

Kepler's Third Law and Time Period

Final Synthesis

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