Solution:

To find the relation between \( G \) and \( K \), we can use both Kepler’s third law and Newton’s law of gravitation.

1. **Gravitational Force**:
From Newton's law of gravitation, the gravitational force between the Sun and a planet is:
\[
F = \frac{GMm}{r^2}
\]

2. **Centripetal Force**:
For a planet moving in a circular orbit, the gravitational force provides the necessary centripetal force. The centripetal force for a planet with mass \( m \) and orbital speed \( v \) is:
\[
F = \frac{mv^2}{r}
\]

Equating the two expressions for force:
\[
\frac{GMm}{r^2} = \frac{mv^2}{r}
\]
Simplifying, we get:
\[
v^2 = \frac{GM}{r}
\]

3. **Orbital Period**:
The orbital speed \( v \) is related to the period \( T \) by:
\[
v = \frac{2 \pi r}{T}
\]
Substituting into \( v^2 = \frac{GM}{r} \), we get:
\[
\left( \frac{2 \pi r}{T} \right)^2 = \frac{GM}{r}
\]
Simplifying:
\[
\frac{4 \pi^2 r^2}{T^2} = \frac{GM}{r}
\]
\[
T^2 = \frac{4 \pi^2 r^3}{GM}
\]

4. **Kepler’s Third Law**:
From Kepler's third law, we know:
\[
T^2 = Kr^3
\]

Comparing both expressions for \( T^2 \):
\[
K = \frac{4 \pi^2}{GM}
\]

### Conclusion:
The relation between \( G \) and \( K \) is:
\[
K = \frac{4 \pi^2}{GM}
\]