The time period `T` of the moon of planet mars (mass `M_(m)`) is related to its orbital radius `R` as (`G`=gravitational constant)

Options

A. `T^(2)=(4pi^(2)R^(3))/(GM_(m))`
B. `T^(2)=(4pi^(2)GR^(3))/(M_(m))`
C. `T^(2)=(4piR^(3)G)/(M_(m))`
D. `T^(2)=4piM_(m)GR^(3)`

Explanation

To find the relationship between the time period \( T \) of the moon of planet Mars (with mass \( M_m \)) and its orbital radius \( R \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces Involved**:
The gravitational force acting on the moon due to Mars provides the necessary centripetal force for the moon's circular motion.

2. **Write the Gravitational Force Equation**:
The gravitational force \( F \) between Mars and its moon can be expressed as:
\[ F = \frac{G M_m M}{R^2} \]
where \( G \) is the gravitational constant, \( M_m \) is the mass of Mars, \( M \) is the mass of the moon, and \( R \) is the distance between the center of Mars and the moon (orbital radius).

3. **Write the Centripetal Force Equation**:
The centripetal force required to keep the moon in circular motion is given by:
\[ F_c = \frac{M v^2}{R} \]
where \( M \) is the mass of the moon and \( v \) is its orbital speed.

4. **Set the Gravitational Force Equal to Centripetal Force**:
For the moon to stay in orbit, the gravitational force must equal the centripetal force:
\[ \frac{G M_m M}{R^2} = \frac{M v^2}{R} \]

5. **Simplify the Equation**:
Cancel \( M \) from both sides (assuming \( M \neq 0 \)):
\[ \frac{G M_m}{R^2} = \frac{v^2}{R} \]
Multiplying both sides by \( R \):
\[ \frac{G M_m}{R} = v^2 \]

6. **Relate Orbital Speed to Time Period**:
The time period \( T \) of the moon's orbit is related to its speed and radius:
\[ T = \frac{2 \pi R}{v} \]
Squaring both sides gives:
\[ T^2 = \frac{4 \pi^2 R^2}{v^2} \]

7. **Substitute for \( v^2 \)**:
From the previous step, we have \( v^2 = \frac{G M_m}{R} \). Substitute this into the equation for \( T^2 \):
\[ T^2 = \frac{4 \pi^2 R^2}{\frac{G M_m}{R}} = \frac{4 \pi^2 R^3}{G M_m} \]

8. **Final Result**:
Thus, the relationship between the time period \( T \) and the orbital radius \( R \) is:
\[ T^2 = \frac{4 \pi^2 R^3}{G M_m} \]

### Conclusion:
The time period \( T \) of the moon of planet Mars is related to its orbital radius \( R \) as:
\[ T^2 = \frac{4 \pi^2 R^3}{G M_m} \]