If a graph is plotted between `T^(2) and r^(3)` for a planet, then its slope will be be (where `M_(S)` is the mass of the sun)

Options

A. A. `(4pi^(2))/(GM_(S))`
B. B.`(GM_(S))/(4pi)`
C. C. `4piGM_(S)`
D. `D. GM_(S)`

Explanation

To solve the problem, we need to determine the slope of the graph plotted between \( T^2 \) (the square of the orbital period of a planet) and \( r^3 \) (the cube of the radius of the orbit) for a planet orbiting the Sun. We will use Kepler's Third Law of planetary motion, which states that the square of the period of orbit \( T \) is directly proportional to the cube of the semi-major axis of its orbit \( r \).

### Step-by-Step Solution:

1. **Understanding the Relationship**: According to Kepler's Third Law, we have the relation:
\[ T^2 \propto r^3 \]
This implies that:
\[ T^2 = k \cdot r^3 \]
where \( k \) is a constant.

2. **Expressing the Constant \( k \)**: For a planet orbiting the Sun, we can express \( k \) in terms of the gravitational constant \( G \) and the mass of the Sun \( M_S \):
\[ k = \frac{4 \pi^2}{G M_S} \]
Thus, we can rewrite the equation as:
\[ T^2 = \frac{4 \pi^2}{G M_S} \cdot r^3 \]

3. **Identifying the Slope**: From the equation \( T^2 = \frac{4 \pi^2}{G M_S} \cdot r^3 \), we can see that if we plot \( T^2 \) on the y-axis and \( r^3 \) on the x-axis, the slope of the line will be:
\[ \text{slope} = \frac{4 \pi^2}{G M_S} \]

4. **Conclusion**: Therefore, the slope of the graph plotted between \( T^2 \) and \( r^3 \) is:
\[ \text{slope} = \frac{4 \pi^2}{G M_S} \]
This corresponds to option A.

### Final Answer:
The slope of the graph between \( T^2 \) and \( r^3 \) is \( \frac{4 \pi^2}{G M_S} \) (Option A).

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