The gravitational force provides the necessary centripetal force for the planet's circular motion around the Sun: $F_g = F_c \Rightarrow \frac{GMm}{r^2} = \frac{mv^2}{r}$. Solving for orbital velocity $v$, we get $v = \sqrt{\frac{GM}{r}}$. The time period of revolution $T$ is given by $T = \frac{2\pi r}{v}$. Substituting $v$: $T = \frac{2\pi r}{\sqrt{GM/r}} = 2\pi \sqrt{\frac{r^3}{GM}}$. Squaring both sides: $T^2 = \left(\frac{4\pi^2}{GM}\right) r^3$. Comparing this with the given Kepler's law equation $T^2 = Kr^3$, we can see that the constant $K$ is: $K = \frac{4\pi^2}{GM}$. Rearranging the terms, we get: $GMK = 4\pi^2$.