A planet of small mass `m` moves around the sun of mass `M` along an elliptrical orbit such that its minimum and maximum distance from sun are `r` and `R` respectively. Its period of revolution will be:

Options

A. `2pisqrt(((r+R)^2)/(6GM))`
B. `2pisqrt(((r+R)^2)/(3GM))`
C. `pisqrt(((r+R)^2)/(2GM))`
D. `2pisqrt(((r+R)^2)/(GM))`

Answer

To find the period of revolution of a planet moving in an elliptical orbit around the Sun, we can use Kepler's Third Law of planetary motion. The law states that the square of the period of revolution (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit.

### Step-by-Step Solution:

1. **Identify the semi-major axis (a)**:
The semi-major axis of an elliptical orbit is the average of the minimum distance (periapsis, r) and the maximum distance (apoapsis, R) from the Sun.
\[ a = \frac{r + R}{2} \]

2. **Apply Kepler's Third Law**:
According to Kepler's Third Law, the period \( T \) of revolution is given by:
\[ T^2 = \frac{4\pi^2}{G M} a^3 \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Sun.

3. **Substitute the value of a**:
Substitute the expression for \( a \) into the equation:
\[ T^2 = \frac{4\pi^2}{G M} \left(\frac{r + R}{2}\right)^3 \]

4. **Simplify the equation**:
\[ T^2 = \frac{4\pi^2}{G M} \cdot \frac{(r + R)^3}{8} \]
\[ T^2 = \frac{\pi^2 (r + R)^3}{2 G M} \]

5. **Take the square root to find T**:
\[ T = \sqrt{\frac{\pi^2 (r + R)^3}{2 G M}} = \frac{\pi (r + R)^{3/2}}{\sqrt{2 G M}} \]

### Final Answer:
The period of revolution \( T \) of the planet is:
\[ T = \frac{\pi (r + R)^{3/2}}{\sqrt{2 G M}} \]