Two satellites A and B have masses m and 2m respectively. A is in a circular orbit of radius R and B

Q: Why doesn't the mass of the satellite affect its orbital velocity?

Gravity provides the necessary centripetal force for the orbit. If you equate the gravitational force ($GMm/r^2$) to the centripetal force ($mv^2/r$), the mass of the satellite ($m$) cancels out on both sides. Thus, all objects at the same orbital radius must travel at the exact same speed, regardless of their mass.

Q: Does a higher orbit mean a satellite has more or less kinetic energy?

Less. The kinetic energy formula is $K = \frac{GMm}{2r}$. Because the radius $r$ is in the denominator, a larger radius means a smaller kinetic energy. Satellites further away move slower.

Q: But satellite B is heavier, shouldn't it have more energy?

It would have more energy if it were in the *same* orbit as satellite A. However, it is twice as far away. Being twice as far away exactly halves its kinetic energy per unit mass. The doubled mass and the halved energy-per-mass perfectly cancel each other out.

Q: What about the potential energy of these satellites?

The potential energy is given by $U = -\frac{GMm}{r}$. It is always negative and is exactly twice the magnitude of the kinetic energy. For satellite A, $U_A = -\frac{GMm}{R}$. For satellite B, $U_B = -\frac{GM(2m)}{2R} = -\frac{GMm}{R}$. So, their potential energies are also equal!

Q: Is the total mechanical energy of a satellite positive or negative?

The total energy is $E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$. It is always negative. A negative total energy indicates that the satellite is in a 'bound state'—it is trapped by Earth's gravity and cannot escape to infinity without added energy.

Question

Two satellites $A$ and $B$ have masses $m$ and $2m$ respectively. $A$ is in a circular orbit of radius $R$ and $B$ is in a circular orbit of radius $2R$ around the earth. The ratio of their kinetic energies, $T_A / T_B$ is

Options:

(A) $\frac{1}{2}$

(B) $2$

(C) $\frac{1}{\sqrt{2}}$

(D) $1$

Sigma Percentile · Accepted Answer

### The Cosmic Balancing Act: Mass vs. Orbit

Imagine you are an astrophysicist observing two satellites, $A$ and $B$, gracefully orbiting the Earth. Satellite $A$ is a standard model with mass $m$, cruising in a tight circular orbit of radius $R$. Satellite $B$, on the other hand, is a heavy-duty model with twice the mass ($2m$), but it orbits much further out, at a radius of $2R$.

Our mission is to compare their kinetic energies. At first glance, you might think the heavier satellite has more energy, or perhaps the one closer to Earth moves so much faster that it wins out. Let's let the physics decide.

### The Master Equation for Kinetic Energy

To find the kinetic energy, we first need to understand how fast these satellites are moving. The orbital velocity $v_0$ of a satellite in a circular orbit is dictated purely by the mass of the central body (Earth, $M$) and the radius of the orbit ($r$):

$$v_0 = \sqrt{\frac{GM}{r}}$$

Notice that the mass of the satellite itself doesn't affect its speed! Now, let's plug this into the standard kinetic energy formula, $T = \frac{1}{2}mv^2$:

$$T = \frac{1}{2} m \left( \sqrt{\frac{GM}{r}} \right)^2$$

$$T = \frac{GMm}{2r}$$

This beautiful, compact equation is our master key. It tells us that the kinetic energy of a satellite is directly proportional to its mass and inversely proportional to its orbital radius.

### Evaluating Satellite A

Let's apply our master equation to Satellite $A$. We simply substitute its specific mass ($m$) and radius ($R$):

$$T_A = \frac{GMm}{2R}$$

This is our baseline.

### Evaluating Satellite B

Now, let's look at Satellite $B$. We must be careful to substitute its specific parameters: a mass of $2m$ and a radius of $2R$.

$$T_B = \frac{GM(2m)}{2(2R)}$$

Look closely at what happens here. The factor of $2$ in the numerator (from the doubled mass) perfectly cancels out the factor of $2$ in the denominator (from the doubled radius).

$$T_B = \frac{GMm}{2R}$$

### The Revelation

Fascinatingly, we arrive at the exact same expression!

$$T_A = T_B$$

Despite Satellite $B$ being twice as massive, its orbit is twice as large, which means it moves slower. These two factors perfectly counterbalance each other, resulting in identical kinetic energies. Therefore, the ratio of their kinetic energies is simply:

$$\frac{T_A}{T_B} = 1$$

**The correct option is (d).**

Animated Solution for Physics - Gravitation: Two satellites $A$ and $B$ have masses $m$ and $2 m$ respectively. $A$ is in a circular orbit of radius $R$ and $B$ is in a circular orbit of radius $2 R$ around the earth. The ratio of their kinetic energies, $T_{A} / T_{B}$ is

Select Answer:

Visualized Solution

Visualizing the Orbits

Orbital Velocity Formula

Kinetic Energy of a Satellite

Kinetic Energy of Satellite A

Kinetic Energy of Satellite B

Simplifying $T_{B}$

Calculating the Ratio

The Sigma Insight: Orbital Motion of a Satellite

The Cosmic Balancing Act

The Master Equation for Kinetic Energy

Evaluating Satellite A

Evaluating Satellite B

The Revelation

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