Perturbation of the energy and the angular momentum

The energy for the Kepler motion is

$$E = \frac{1}{2}({\bf v}\cdot {\bf v}) -\frac{\mu}{r}$$

Then,

$$\dot{E} = \frac{1}{2}(\dot{{\bf v}}\cdot{\bf v}+{\bf v}\cdot\dot{{\bf v}})+\frac{\mu}{r^2}\dot{r}$$

or

$$\dot{E} = \frac{1}{2}(\dot{{\bf v}}\cdot{\bf v})+\frac{\mu}{r^2}\dot{r}$$

After substituting the equations of motion for $\dot{{\bf v}}$ and rearranging terms, we get

$$\dot{E} = \left( -\frac{\mu}{r^3}{\bf r} + {\bf F}\right)\cdot{\bf v} + \frac{\mu}{r^2}\dot{r}$$

We need now $r\dot{r} = {\bf r}\cdot {\bf v}$, and we get

$$\dot{E} = \left( -\frac{\mu}{r^r}{\bf r} + {\bf F}\right)\cdot{\bf v} + \frac{\mu}{r^3}({\bf r}\cdot {\bf v})$$

which simplifies to

$$\dot{E} = {\bf F}\cdot {\bf v}$$

This is what we expected, i.e. the change of the energy is proportional to the perturbation. In particular, since ${\bf v}$ is tangential to the orbit, only the component of ${\bf F}$ tangential to the orbit changes the energy of the orbit. Also note that when ${\bf F}= 0$, $E$ is conserved. Conserved quantities are often called first inetrals of the motion.

Next, recall that the angular momentum is defined as

$${\bf A}= {\bf r}\times {\bf v},$$

and was also a conserved quantity (or first integral) of the unperturbed Kepler motion. To investigate the effect of the perturbation we compute its time derivative

$$\dot{{\bf A}} = \dot{{\bf r}} \times {\bf v}+ {{\bf r}} \times \dot{{\bf v}}$$

which after substituting $\dot{\bf v}$ again yields

$$\dot{{\bf A}} = {\bf v}\times {\bf v}+ {{\bf r}} \times \left( -\frac{\mu}{r^3}{\bf r} + {\bf F}\right).$$

Here ${\bf v}\times {\bf v}= 0$ and ${\bf r}\times {\bf r}= 0$, leading to

$$\dot{{\bf A}} = {\bf r}\times {\bf F}$$

Note that only components of ${\bf F}$ which are perpendicular to ${\bf r}$ contribute to the change of the angular momentum. The quantity ${\bf r}\times{\bf F}$ is known as the torque in mechanics.