A planet of mass `m` moves around the Sun of mass Min an elliptical orbit. The maximum and minimum distance of the planet from the Sun are `r_(1)` and `r_(2)`, respectively. Find the relation between the time period of the planet in terms of `r_(1)` and `r_(2)`.

Options

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

Explanation

To find the relation between the time period of a planet in an elliptical orbit around the Sun in terms of its maximum and minimum distances from the Sun, we can follow these steps:

### Step 1: Understand the semi-major axis
In an elliptical orbit, the semi-major axis \( A \) is defined as the average of the maximum distance \( r_1 \) (aphelion) and the minimum distance \( r_2 \) (perihelion) from the Sun. The formula for the semi-major axis is:
\[ A = \frac{r_1 + r_2}{2} \]

### Step 2: Apply Kepler's Third Law
Kepler's Third Law states that the square of the time period \( T \) of a planet is directly proportional to the cube of the semi-major axis \( A \) of its orbit. Mathematically, this can be expressed as:
\[ T^2 \propto A^3 \]

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

### Step 4: Simplify the equation
This can be rewritten as:
\[ T^2 \propto \frac{(r_1 + r_2)^3}{8} \]
Thus, we can express the time period \( T \) as:
\[ T \propto \sqrt{\frac{(r_1 + r_2)^3}{8}} = \frac{(r_1 + r_2)^{3/2}}{2\sqrt{2}} \]

### Step 5: Final relation
From the above steps, we conclude that the time period \( T \) of the planet is directly proportional to the \( 3/2 \) power of the sum of the maximum and minimum distances from the Sun:
\[ T \propto (r_1 + r_2)^{3/2} \]

### Summary of the relation
Thus, the relation between the time period \( T \) of the planet and the distances \( r_1 \) and \( r_2 \) is:
\[ T \propto (r_1 + r_2)^{3/2} \]