The acceleration towards the centre of a star is proportional to $x^{-2}$ where $x$ is the distance from the centre. That is, $\ddot{x}=-\frac{k}{x^2}$.
A satellite is orbiting a star with a constant speed, $V$, at a fixed distance $R$ from the centre of the star. Its period of revolution is $T$. If it moves in a uniform circular motion show that $k=\frac{4{\pi}^2R^3}{T^2}$.
I'm confused on how to determine the acceleration. I know that I can get the answer by squaring the speed of rotation, which is $\frac{2\pi}{T}$ and then multiplying it by $-R$ then subbing that into the equation when $x=R$ but I don't understand how squaring the speed of revolution nor multiply it by $-R$ would give the acceleration. Am I on the right train of thought or is it just a coincidence that I can get the answer this way?
For constant revolution in a circle the velocity is given by the following equation
$$ \frac{d}{dt}\vec{r}=\vec{\omega}\times\vec{r} $$
By differentiating once w.r.t. time we get the acceleration equation
$$ \begin{align} \frac{d^{2}}{dt^{2}}\vec{r}&=\vec{\omega}\times\left(\frac{d}{dt}\vec{r}\right)+\left(\frac{d}{dt}\vec{\omega}\right)\times\vec{r}\\ \\ &= \vec{\omega}\times\left(\frac{d}{dt}\vec{r}\right)+0\\ \\ &=\vec{\omega}\times\vec{\omega}\times\vec{r} \end{align} $$