Kepler's third law states that square of period revolution `(T)` of a planet around the sun is proportional to third power of average distance `i` between sun and planet i.e. `T^(2)=Kr^(3)`
here `K` is constant
if the mass of sun and planet are `M` and `m` respectively then as per Newton's law of gravitational the force of alteaction between them is `F=(GMm)/(r^(2))`, here `G` is gravitational constant. The relation between `G` and `K` is described as

Options

A. `GK = 4pi^(2)`
B. `GMK = 4pi^(2)`
C. `K = G`
D. `K = (1)/(G)`

Answer

To derive the relationship between the gravitational constant \( G \) and the constant \( K \) in Kepler's third law, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Kepler's Third Law**:
Kepler's third law states that the square of the period of revolution \( T \) of a planet around the sun is proportional to the cube of the average distance \( r \) between the sun and the planet. This can be expressed mathematically as:
\[ T^2 = K r^3 \]
where \( K \) is a constant.

2. **Using Newton's Law of Gravitation**:
According to Newton's law of gravitation, the gravitational force \( F \) between two masses \( M \) (mass of the sun) and \( m \) (mass of the planet) separated by a distance \( r \) is given by:
\[ F = \frac{GMm}{r^2} \]
where \( G \) is the gravitational constant.

3. **Relating Gravitational Force to Circular Motion**:
For a planet in circular orbit, the gravitational force provides the necessary centripetal force for the planet's circular motion. The centripetal force \( F_c \) is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( v \) is the orbital speed of the planet.

4. **Setting Gravitational Force Equal to Centripetal Force**:
Setting the gravitational force equal to the centripetal force gives:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{r^2} = \frac{v^2}{r} \]
Multiplying both sides by \( r \):
\[ \frac{GM}{r} = v^2 \]

5. **Relating Orbital Speed to Period**:
The orbital speed \( v \) can also be expressed in terms of the period \( T \):
\[ v = \frac{2\pi r}{T} \]
Squaring both sides gives:
\[ v^2 = \left(\frac{2\pi r}{T}\right)^2 = \frac{4\pi^2 r^2}{T^2} \]

6. **Substituting for \( v^2 \)**:
Now substitute \( v^2 \) back into the equation:
\[ \frac{GM}{r} = \frac{4\pi^2 r^2}{T^2} \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ T^2 = \frac{4\pi^2 r^3}{GM} \]

8. **Comparing with Kepler's Third Law**:
From Kepler's third law, we have \( T^2 = K r^3 \). By comparing both equations, we find:
\[ K = \frac{4\pi^2}{GM} \]

9. **Final Relationship**:
Rearranging this gives the relationship between \( G \) and \( K \):
\[ GMK = 4\pi^2 \]

### Summary:
The relationship between the gravitational constant \( G \) and the constant \( K \) in Kepler's third law is given by:
\[ GMK = 4\pi^2 \]