# I Multiplying by dr/dt and integrating with respect to t

1. Jun 20, 2017

### BearY

I came across this step in a derivation:$$m\ddot{r}=\frac{L^2}{mr^3} -V'(r)$$
Multiplying by $\dot{r}$ and integrating with respect to t to get $$\frac{1}{2}m\dot{r}^2+\frac{L^2}{2mr^2}+V(r) = C$$
I am not very clear about how the 1st term came to this. Can some one gives a pointer?

2. Jun 20, 2017

### andrewkirk

Differentiate both sides of the second equation wrt $t$ and we get
$$\frac{1}{2}m\left(2\dot{r}\ddot r\right)+\frac{L^2}{2m}\left(-2r^{-3}\dot r\right)+V'(r)\dot r = 0$$
Re-arranging terms gives us the first equation.

3. Jun 21, 2017

### BearY

It is very clear when it's done this way. Thanks a lot.