0
$\begingroup$

Given an inverse square law $\ddot{\vec{r}}=-\frac{\mu}{r^2}\hat{r}$, I define the Angular momentum per unit mass as $\vec{H}=\vec{r}\times\dot{\vec{r}}$. Showing it's constant is strightfoward. Then I defined the Laplace–Runge–Lenz vector per unit mass

$\vec{c}=\dot{\vec{r}}\times\vec{H}-\mu\hat{r}$

I think (yet not sure) this is the normal one scaled down by $m^2$. I now try to take the derivative and get

$\frac{d\vec{c}}{dt}=\ddot{\vec{r}}\times\vec{H}+\dot{\vec{r}}\times\dot{\vec{H}}-\mu\dot{\hat{r}}=\ddot{\vec{r}}\times\vec{H}-\mu\dot{\hat{r}}=-\frac{\mu}{r^2}\hat{r}\times\vec{H}-\mu\dot{\hat{r}}$

I'm not sure how to get to $0$. I thought of substituting $\vec{H}$'s definition and using a triple cross identity but that wont yield anything.

Improve this question
$\endgroup$
7
  • $\begingroup$ This looks like homework, so just a hint: Start with $\vec r = r \hat r$. What is it's time derivative? To answer that, you'll need to answer the auxiliary question, what is the time derivative of $\hat r$? $\endgroup$
    – David Hammen
    Commented Jun 16, 2017 at 19:14
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/18088/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Jun 16, 2017 at 19:30
  • $\begingroup$ @Qmechanic the maths there is over the top for me and it doesn't revole around the one per unit mass. $\endgroup$
    – Theorem
    Commented Jun 17, 2017 at 7:28
  • $\begingroup$ @David $\dot{\hat{r}}=\dot{\theta}\hat{\theta}$ so $\dot{\vec{r}}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}$ but what does that give me? $\endgroup$
    – Theorem
    Commented Jun 17, 2017 at 7:28
  • 1
    $\begingroup$ @Theorem - Use that to calculate $\vec H$ and then $\hat r \times \vec H$. Everything will cancel. $\endgroup$
    – David Hammen
    Commented Jun 17, 2017 at 8:33

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.