Two satellites of masses M and 16 M are orbiting a planet in a circular orbitl of radius R. Their time periods of revolution will be in the ratio of

Options

A. `1 : 1`
B. `1 : 4`
C. `4 : 1`
D. `1 : 16`

Answer

To solve the problem of finding the ratio of the time periods of two satellites of masses M and 16M orbiting a planet in a circular orbit of radius R, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between gravitational force and centripetal force**:
The gravitational force acting on a satellite is what keeps it in orbit, and this force can be equated to the centripetal force required for circular motion.

2. **Write down the gravitational force equation**:
The gravitational force \( F_g \) acting on a satellite of mass \( m_s \) is given by:
\[ F_g = \frac{G \cdot M_p \cdot m_s}{R^2} \]
where \( G \) is the universal gravitational constant, \( M_p \) is the mass of the planet, and \( R \) is the radius of the orbit.

3. **Write down the centripetal force equation**:
The centripetal force \( F_c \) required to keep the satellite in circular motion is given by:
\[ F_c = \frac{m_s \cdot v^2}{R} \]
where \( v \) is the orbital velocity of the satellite.

4. **Set the gravitational force equal to the centripetal force**:
Since the gravitational force provides the necessary centripetal force, we can set these two equations equal:
\[ \frac{G \cdot M_p \cdot m_s}{R^2} = \frac{m_s \cdot v^2}{R} \]

5. **Cancel the mass of the satellite \( m_s \)**:
Since \( m_s \) appears on both sides of the equation, we can cancel it out (assuming \( m_s \neq 0 \)):
\[ \frac{G \cdot M_p}{R^2} = \frac{v^2}{R} \]

6. **Rearrange to find the velocity**:
Multiply both sides by \( R \):
\[ v^2 = \frac{G \cdot M_p}{R} \]

7. **Relate velocity to the time period**:
The orbital velocity \( v \) can also be expressed in terms of the time period \( T \):
\[ v = \frac{2\pi R}{T} \]
Substituting this into the equation gives:
\[ \left(\frac{2\pi R}{T}\right)^2 = \frac{G \cdot M_p}{R} \]

8. **Solve for the time period \( T \)**:
Squaring the left side:
\[ \frac{4\pi^2 R^2}{T^2} = \frac{G \cdot M_p}{R} \]
Rearranging gives:
\[ T^2 = \frac{4\pi^2 R^3}{G \cdot M_p} \]
Taking the square root:
\[ T = 2\pi \sqrt{\frac{R^3}{G \cdot M_p}} \]

9. **Analyze the result**:
Notice that the time period \( T \) depends on the radius \( R \) and the mass of the planet \( M_p \), but it does not depend on the mass of the satellite \( m_s \).

10. **Conclusion**:
Since both satellites (one of mass \( M \) and the other of mass \( 16M \)) are orbiting at the same radius \( R \), their time periods will be the same. Therefore, the ratio of their time periods is:
\[ \text{Ratio} = \frac{T_1}{T_2} = 1:1 \]