Two satellites of earth `S_(1)` and `S_(2)` are moving in the same orbit. The mass of `S_(1)` is four times the mass of `S_(2)`. Which one of the following statements is true?

Options

A. The potential energies of earth and satellite in the two cases are equal .
B. `S_(1)` and `S_(2)` are moving with the same speed.
C. The kinetic energies of the two satellites are equal .
D. The time period of `S_(1)` is four times that `S_(2)`.

Explanation

To solve the problem, we need to analyze the situation of two satellites, \( S_1 \) and \( S_2 \), moving in the same orbit around the Earth. The mass of satellite \( S_1 \) is four times the mass of satellite \( S_2 \). Let's denote the mass of \( S_2 \) as \( m \) and the mass of \( S_1 \) as \( 4m \).

### Step 1: Understand the gravitational force acting on the satellites
The gravitational force \( F \) acting on a satellite in orbit is given by the formula:
\[ F = \frac{G M m}{r^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( r \) is the distance from the center of the Earth to the satellite.

### Step 2: Apply the centripetal force condition
For a satellite in circular motion, the gravitational force provides the necessary centripetal force:
\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \]
where \( v \) is the orbital velocity of the satellite.

### Step 3: Derive the expression for orbital velocity
From the equation above, we can simplify to find the orbital velocity \( v \):
\[ v^2 = \frac{G M}{r} \]
Thus, the orbital velocity \( v \) is given by:
\[ v = \sqrt{\frac{G M}{r}} \]
This shows that the orbital velocity depends only on the mass of the Earth \( M \) and the radius \( r \), and not on the mass of the satellite.

### Step 4: Analyze the kinetic energy of the satellites
The kinetic energy \( KE \) of a satellite is given by:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the expression for \( v^2 \):
\[ KE = \frac{1}{2} m \left(\frac{G M}{r}\right) \]
For satellite \( S_1 \) (mass \( 4m \)):
\[ KE_{S_1} = \frac{1}{2} (4m) \left(\frac{G M}{r}\right) = 2m \left(\frac{G M}{r}\right) \]
For satellite \( S_2 \) (mass \( m \)):
\[ KE_{S_2} = \frac{1}{2} m \left(\frac{G M}{r}\right) = \frac{1}{2} m \left(\frac{G M}{r}\right) \]
Thus, the kinetic energy of \( S_1 \) is four times that of \( S_2 \).

### Step 5: Analyze the gravitational potential energy
The gravitational potential energy \( U \) of a satellite is given by:
\[ U = -\frac{G M m}{r} \]
For satellite \( S_1 \):
\[ U_{S_1} = -\frac{G M (4m)}{r} = -\frac{4G M m}{r} \]
For satellite \( S_2 \):
\[ U_{S_2} = -\frac{G M m}{r} \]
Thus, the gravitational potential energy of \( S_1 \) is four times that of \( S_2 \).

### Step 6: Analyze the time period of the satellites
The time period \( T \) of a satellite in orbit is given by:
\[ T = 2\pi \sqrt{\frac{r^3}{G M}} \]
This expression shows that the time period depends only on the radius \( r \) and the mass of the Earth \( M \), and is independent of the mass of the satellite.

### Conclusion
Based on the analysis:
- The potential energy of \( S_1 \) is four times that of \( S_2 \).
- The kinetic energy of \( S_1 \) is four times that of \( S_2 \).
- The time period of both satellites is the same.

Thus, the correct statement is that the time period of both satellites is the same.

### Final Answer
The true statement is that both satellites have the same time period.

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