Mechanics Deriving Kepler's Law Knowing that the acceleration of any body towards the centre of a star due to the force of gravity is proprtional to $$x^{-2} $$, where x is the distance of the body form the centre of the star, i.e. $$ x''= \frac k {x^2} $$, where k is a positive constant.
Knowing orbital speed is constant V at distance R from the centre of the star, how do you derive Kepler's 3rd law where $$k = {4\pi R^3\over T^2} $$
 A: The more mathematical solution,
Let $\vec x $ be the two dimensional position vector from mass $M$ to (the center of) the satellite of mass $m$. Then,
$$\vec F=-m\frac{k}{r^2} \frac{\vec{x}}{||\vec x||}=-m\frac{k}{r^3} \vec x$$
(Negative sign because the force is inwards, towards the object of mass $M$)
At the same time $\vec F=m \frac{d^2}{dt^2}(\vec x)$ hence,
$$m \frac{d^2}{dt^2} \vec x=-m\frac{k}{r^3} \vec x$$
Now if we just look at one coordinate of the vector $x$, solving the second order ode:
$$x''=-\frac{k}{r^3}x$$
Gives real solutions of the form,
$$x=A\sin(t\sqrt{\frac{k}{r^3}}+\theta_1)$$
Or,
$$x=B\cos(t\sqrt{\frac{k}{r^3}}+\theta_2)$$
Notice that these solutions have period of $\frac{2\pi}{\sqrt{\frac{k}{r^3}}}$ so,
$$T=\frac{2\pi}{\sqrt{\frac{k}{r^3}}}$$
$$T^2=\frac{4 \pi^2 r^3}{k}$$
Hence,
$$k=\frac{4\pi^2r^3}{T^2}$$

Keplers law describes a body of mass $m$ in orbit with radius $r$ around a body of mass $M$, indeed we have,
$$F=ma=\frac{GMm}{r^2}$$
This is the force that keeps the body in orbit, that is it is what provides the centripetal  force of $m\frac{v^2}{r}$. Where $v$ is the velocity the body is traveling at.
Hence,
$$F=m\frac{v^2}{r}=\frac{GMm}{r^2}$$
This gives,
$$v^2=\frac{GM}{r}$$
But at the same time $v=\frac{2\pi r}{T}$ so,
$$\frac{4\pi^2 r^2}{T^2}=\frac{GM}{r}$$
Giving,
$$k=GM=\frac{4\pi^2 r^3}{T^2}$$
