How do Newton’s laws lead to Kepler’s third law for circular orbits?

To derive Kepler’s third law for circular orbits using Newton’s laws of motion, we consider a planet of mass $M_E$ in a uniform circular orbit of radius $r$ around a central body of mass $M_S$ (such as the Sun).

According to Newton’s second law, the gravitational force $F$ provides the necessary centripetal acceleration to keep the planet in orbit. The gravitational force is given by:
$$F=\frac{G{M}_S{M}_E}{r^2}$$
The required centripetal force for an object moving at speed $v$ in a circle of radius $r$ is:
$$F=\frac{M_E{v}^2}{r}$$
Equating these two expressions, we find that the speed $v$ of the planet is independent of its own mass:
$$v^2=\frac{G{M}_S}{r}$$
Since the speed $v$ in a circular orbit is the circumference divided by the orbital period $T$ ($v=2\pi r/T$), we substitute this into the equation:
$${\left(\frac{2\pi r}{T}\right)}^2=\frac{G{M}_S}{r}$$
Rearranging this yields Kepler’s third law:
$$T^2=\left(\frac{4{\pi }^2}{G{M}_S}\right)r^3$$
This demonstrates that the square of the period is directly proportional to the cube of the orbital radius.12