Deduction of Newton's Law of gravitational force from Kepler's Law:
Let us consider:
The mass of a planet = $m$
The radius of the circular path of a planet=$r$
The mass of the sun = $M$
The velocity of the planet = $v$
The time of the revolution=$T$
The attraction force between the planet and the sun is achieved by the centripetal force i.e
$F=\frac{m v^{2}}{r} \qquad(1)$
The orbital velocity of the planet:
$v=\frac{Circumference \: of \: the \: circular \: path}{Time \: period}$
$v=\frac{2\pi r}{T} \qquad(2)$
From equation $(1)$ and equation $(2)$, we get
$F=\frac{m}{r} \left( \frac{2\pi r}{T} \right)^{2}$
$F=\frac{4 \pi^{2} mr}{T^{2}} \qquad(3)$
According to Kepler's third law i.e
$T^{2}=Kr^{3} \qquad(4)$
From equation$(3)$ and equation$(4)$, we get
$F=\frac{4 \pi^{2}mr}{Kr^{3}}$
$F=\frac{4 \pi^{2}}{K}\frac{mr}{r^{3}}$
$F=\frac{4 \pi^{2}}{K}\frac{m}{r^{2}} \qquad(5)$
The source of this force is the sun. So R.H.S of equation $(5)$ should be related to the sun. Since $m$ and $r$ are related to the planet. So the quantity $\frac{4 \pi^{2}}{K}$ should be related to some constant of the sun. Let $\frac{4 \pi^{2}}{K}$ be proportional to the mass of the sun. i.e
$\frac{4 \pi^{2}}{K m} \propto M$
$\frac{4 \pi^{2}}{K m} = G M $
Here $G$ is the proportionality constant and it is called the universal gravitation constant. Now substitute the $\frac{4 \pi^{2}}{K m} = G M$ in equation $(5)$. then we get
$F=G \frac{M m}{r^{2}}$
The above equation represents the force of attraction between the sun and the planet. It is also known as Newton's law of gravitational force.