Consider the equations of motion of an object in an orbit around a central body. We use an inertial frame originating at centre of mass of the central body: $$ \frac{d\mathbf{r}}{dt} = \mathbf{V} \tag{1} $$ $$ \frac{d\mathbf{V}}{dt} = -\frac{\mu}{r^3}\mathbf{r} + \mathbf{A}_\text{ext} \tag{2} $$ where $\mathbf{r}$ is the position vector of the object, $\mathbf{V}$ is the velocity, $\mu$ is the gravitational parameter of the central body and $\mathbf{A}_\text{ext}$ are accelerations due to some other sources (example, thrust from engines, drag, etc.).

I am interested in specifying these with respect to the specific orbital energy $\epsilon$. For this, I need to know $\dfrac{d\epsilon}{dt}$

I know that $$\epsilon = \frac{V^2}{2}-\frac{\mu}{r} \tag{3}$$ So, differentiating: $$ \frac{d\epsilon}{dt} = V\dfrac{dV}{dt}+\dfrac{\mu}{r^2}\dfrac{dr}{dt} \tag{4} $$

If we multiply equation (2) with $\dfrac{d\mathbf{r}}{dt}$, we get: $$ \dfrac{d\mathbf{r}}{dt} \cdot \frac{d\mathbf{V}}{dt} = - \dfrac{d\mathbf{r}}{dt} \cdot \frac{\mu}{r^3}\mathbf{r} + \dfrac{d\mathbf{r}}{dt} \cdot \mathbf{A}_\text{ext} \tag{5} $$

Rearranging it, $$ \mathbf{V} \cdot \frac{d\mathbf{V}}{dt} + \dfrac{d\mathbf{r}}{dt} \cdot \frac{\mu}{r^3}\mathbf{r} = \mathbf{V} \cdot \mathbf{A}_\text{ext} \tag{6} $$

I am stuck here. $\epsilon$ is a scaler, and so is its derivative. However, it appears that the left hand side of the equation 6 is the vector form of equation 4.

Could you help me proceed? If I have the above, then I could formulate the equations of motion. Or, if you already know the equations of motion with respect to energy, that would be great!

Wikipedia article on specific orbital energy states that $\dfrac{d\epsilon}{dt} = \mathbf{V}\cdot \mathbf{A}_\text{ext}$ but doesnt give any reference / steps.



1 Answer 1


As a general rule (regardless of the definition of V) we have:

$$\frac{\mathrm{d}}{\mathrm{d}t}(\bf{V}\cdot\bf{V})=\frac{\mathrm{d}\bf{V}}{\mathrm{d}t}\cdot \bf{V}+\bf{V} \cdot \frac{\mathrm{d}\bf{V}}{\mathrm{d}t}$$ $$\frac{\mathrm{d}}{\mathrm{d}t}V^2=2\left(\bf{V}\cdot \frac{\mathrm{d}\bf{V}}{\mathrm{d}t}\right)$$ $$2V\frac{\mathrm{d}V}{\mathrm{d}t}=2\left(\bf{V}\cdot \frac{\mathrm{d}\bf{V}}{\mathrm{d}t}\right)$$ $$V\frac{\mathrm{d}V}{\mathrm{d}t}=\bf{V}\cdot \frac{\mathrm{d}\bf{V}}{\mathrm{d}t}$$

Using this result, starting at equation (4):

$$ \frac{\mathrm{d}\epsilon}{\mathrm{d}t} = V\dfrac{\mathrm{d}V}{\mathrm{d}t}+\dfrac{\mu}{r^2}\dfrac{\mathrm{d}r}{\mathrm{d}t}=\mathbf{V} \cdot \frac{\mathrm{d}\mathbf{V}}{\mathrm{d}t} + \frac{\mu}{r^3}\mathbf{r}\cdot\dfrac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = \mathbf{V} \cdot \mathbf{A}_\text{ext} $$ Where the last equality comes from equation (6).

  • $\begingroup$ You could be a little more explicit to improve the answer by defining $d\mathbf{r}/dt = \mathbf{V}$ and $d\mathbf{V}/dt = \mathbf{A}_{ext}$. It would also help if you added the general definition of power (i.e., $P = \mathbf{F} \cdot \mathbf{v}$) for this type of situation. $\endgroup$ Oct 28, 2015 at 17:29
  • $\begingroup$ @honeste_vivere I have now edited the answer to give the explicit result that was asked. $\endgroup$
    – Chris
    Oct 28, 2015 at 20:02

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