The period of revolution of a planet around the sun in a circular orbit is same as that of period of similar planet revolving around a star of twice the raduis of first orbit and if `M` is the mass of the sun then the mass of star is

Options

A. `2M`
B. `4M`
C. `8M`
D. `16 M`

Explanation

To solve the problem, we will use Kepler's Third Law of planetary motion, which relates the period of revolution of a planet around a star to the radius of its orbit and the mass of the star.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( T_1 \) be the period of the planet revolving around the Sun.
- Let \( T_2 \) be the period of the planet revolving around the star.
- Let \( r \) be the radius of the orbit of the planet around the Sun.
- Let \( M \) be the mass of the Sun.
- Let \( r' = 2r \) be the radius of the orbit of the planet around the star.
- Let \( M_s \) be the mass of the star.

2. **Apply Kepler's Third Law**:
According to Kepler's Third Law:
\[ T^2 \propto \frac{r^3}{M} \]
For the planet around the Sun:
\[ T_1^2 = \frac{4\pi^2 r^3}{G M} \]
For the planet around the star:
\[ T_2^2 = \frac{4\pi^2 (r')^3}{G M_s} \]

3. **Substitute \( r' \)**:
Since \( r' = 2r \):
\[ T_2^2 = \frac{4\pi^2 (2r)^3}{G M_s} = \frac{4\pi^2 (8r^3)}{G M_s} = \frac{32\pi^2 r^3}{G M_s} \]

4. **Set the Periods Equal**:
Given that \( T_1 = T_2 \):
\[ \frac{4\pi^2 r^3}{G M} = \frac{32\pi^2 r^3}{G M_s} \]

5. **Cancel Common Terms**:
Cancel \( 4\pi^2 r^3 \) from both sides:
\[ \frac{1}{M} = \frac{8}{M_s} \]

6. **Solve for \( M_s \)**:
Rearranging gives:
\[ M_s = 8M \]

### Final Answer:
The mass of the star \( M_s \) is \( 8M \).