A planet of mass m is moving in an elliptical orbit about the sun (mass of sun = M). The maximum and minimum distances of the planet from the sun are `r_(1)` and `r_(2)` respectively. The period of revolution of the planet wil be proportional to :

Options

A. `r_(1)^(3//2)`
B. `r_(2)^(3//2)`
C. `(r_(1)-r_(2))^(3//2)`
D. `(r_(1)+r_(2))^(3//2)`

Answer

To solve the problem, we will use Kepler's Third Law of planetary motion, which 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**:
The semi-major axis (a) of an elliptical orbit is the average of the maximum distance (r1) and the minimum distance (r2) from the sun. Therefore, we can express it as:
\[ a = \frac{r_1 + r_2}{2} \]

2. **Apply Kepler's Third Law**:
According to Kepler's Third Law, the square of the period (T) is proportional to the cube of the semi-major axis (a):
\[ T^2 \propto a^3 \]

3. **Substitute the expression for a**:
Now, substituting the expression for the semi-major axis into Kepler's Third Law:
\[ T^2 \propto \left(\frac{r_1 + r_2}{2}\right)^3 \]

4. **Simplify the expression**:
Since \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\) is a constant, we can simplify the expression:
\[ T^2 \propto \frac{(r_1 + r_2)^3}{8} \]

5. **Remove the constant for proportionality**:
We can ignore the constant factor when discussing proportionality:
\[ T^2 \propto (r_1 + r_2)^3 \]

6. **Take the square root to find T**:
To find T, we take the square root of both sides:
\[ T \propto (r_1 + r_2)^{\frac{3}{2}} \]

### Conclusion:
Thus, the period of revolution (T) of the planet is proportional to \((r_1 + r_2)^{\frac{3}{2}}\).

### Final Answer:
The period of revolution of the planet will be proportional to \((r_1 + r_2)^{\frac{3}{2}}\).

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