If a planet of mass `m` is revolving around the sun in a circular orbit of radius `r` with time period, `T` then mass of the sun is

Options

A. `4pi^(2)r^(3)//GT`
B. `4pi^(2)r^(3)//GT^(2)`
C. `4pi^(2)r//GT`
D. `4pi^(2)r^(3)//G^(2)T^(2)`

Explanation

To find the mass of the sun (denoted as \( M \)) when a planet of mass \( m \) is revolving around it in a circular orbit of radius \( r \) with a time period \( T \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces Involved**:
The gravitational force provides the necessary centripetal force for the planet to maintain its circular orbit. Therefore, we can set up the equation:
\[ \frac{m v^2}{r} = \frac{G M m}{r^2} \]
Here, \( G \) is the gravitational constant, \( v \) is the orbital speed of the planet, and \( r \) is the radius of the orbit.

2. **Cancel Out the Mass of the Planet**:
Since the mass of the planet \( m \) appears on both sides of the equation, we can cancel it out:
\[ \frac{v^2}{r} = \frac{G M}{r^2} \]
This simplifies to:
\[ v^2 = \frac{G M}{r} \]

3. **Relate Orbital Speed to Time Period**:
The orbital speed \( v \) can also be expressed in terms of the time period \( T \):
\[ v = \frac{2 \pi r}{T} \]
Substituting this expression for \( v \) into the equation from step 2 gives:
\[ \left(\frac{2 \pi r}{T}\right)^2 = \frac{G M}{r} \]

4. **Square the Equation**:
Squaring both sides results in:
\[ \frac{4 \pi^2 r^2}{T^2} = \frac{G M}{r} \]

5. **Rearranging to Solve for Mass of the Sun**:
Multiply both sides by \( r \) to isolate \( M \):
\[ 4 \pi^2 r^3 = G M T^2 \]
Now, divide both sides by \( G T^2 \):
\[ M = \frac{4 \pi^2 r^3}{G T^2} \]

### Final Answer:
The mass of the sun is given by:
\[ M = \frac{4 \pi^2 r^3}{G T^2} \]