# Formulation of orbital kinetic energy

1. Jul 28, 2012

### Woodles

For a radial potential proportional to r^k, the virial theorem says $\bar{T}=c(k)\bar{V}$. My problem says to show that c(k)=k/2 for a circular orbit.

I actually solved the problem all ready (by setting the gradiant of the potential equal to the centripetal force and solving for 1/2*m*v^2.)

However, while I was looking around, I came across something.

I found a formulation of kinetic energy saying that $\bar{T}= 1/2 \vec{∇}(V) .\vec{r}$.

How could you derive this? Is it true? Why? The original post said this came from the virial theorem.

2. Jul 28, 2012

### dextercioby

Just integrate with respect to time in the LHS and you'll get it.

$$\frac{m}{2} \int_{t_1}^{t_2} \frac{dr}{dt}\left(\frac{dr}{dt} dt\right) = ...$$