0
$\begingroup$

We just completed orbital dynamics in my university astrophysics class. The textbook we are using is Carroll and Ostlie. In the textbook, it is mentioned that the average gravitational potential energy is

$$ \langle U \rangle = -\dfrac{GM\mu}{a} $$

where $M$ is the total mass of the system, $\mu$ is the reduced mass of the system and $a$ is the semi-major axis of the elliptical orbit.

My question is: How do we come about this answer?

It is also mentioned in the textbook that

$$ \langle \frac{1}{r} \rangle = \frac{1}{a}. $$

I know that we need to integrate $1/r$ over the period of the orbit but I'm not sure with respect to what we need to integrate it. Would we be required to use the formula

$$r = \dfrac{a(1-e^2)}{1+ecos{\theta}} ?$$

If yes, how can we do this?

Any help would be most appreciated!

$\endgroup$
1

2 Answers 2

1
$\begingroup$

$L_a=\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\theta}^2+\frac{GmM}{r}: $ Lagrangian.

$\frac{d}{dt}(\frac{\partial L_a}{\partial \dot{\theta}})-\frac{\partial L_a}{\partial \theta}=0\implies mr^2\dot{\theta}=L$. Angular momentum is conserved.

$\frac{d}{dt}(\frac{\partial L_a}{\partial \dot{r}})-\frac{\partial L_a}{\partial r}=0\implies m\ddot{r} -mr\dot{\theta}^2+\frac{GMm}{r^2}=0$.

The above by the Euler-Lagrange Equations.

$m\ddot{r}=\frac{L\dot{\theta}}{r}-\frac{GMm^2\dot{\theta}}{L}$ swapping in $L$.

$m\frac{d}{d\theta}(\dot{r})=\frac{L}{r}-\frac{GMm^2}{L}$ dividing by $\dot{\theta}$

$m\dot{r}=m\dot{\theta}\frac{dr}{d\theta}=\frac{L}{r^2}\frac{dr}{d\theta}=\frac{d}{d\theta}(\frac{-L}{r})$ by the chain rule.

Put it all together:

$\frac{d^2}{d\theta^2}(\frac{1}{r})+\frac{1}{r}=\frac{GMm^2}{L^2}$

So $\frac{1}{r}=c_1\cos{\theta}+c_2\sin{\theta}+\frac{GMm^2}{L^2}$

Let's define $\theta =0$ as the aphelion, $p$.This is the sum of the semi-major axis and the distance from the focus to the center of the ellipse, so $p=a+c$. The perihelion, occurring when $\theta=\pi$, has length $a-c$. Since the radial velocity is $0$ at the aphelion, $c_2=0$. So:

$\frac{1}{a+c}=c_1+\frac{GMm^2}{L^2}$ ,$\frac{1}{a-c}=-c_1+\frac{GMm^2}{L^2}$

Then $\frac{2a}{a^2-c^2}=2\frac{GMm^2}{L^2}$ ,And $\frac{-2c}{a^2-c^2}=2c_1.$

By definition, eccentricity $\epsilon=c/a$.

$\frac{GMm^2}{L^2}=\frac{a}{a^2(1-\epsilon^2)}$

$\frac{1}{r}=\frac{-c}{a^2-c^2}\cos \theta +\frac{a}{a^2-c^2}=\frac{-a\epsilon}{a^2(1-\epsilon^2)}\cos \theta + \frac{a}{a^2(1-\epsilon^2)}$

So $r= \frac{a(1-\epsilon^2)}{1-\epsilon \cos \theta}$.

For a given power of $r$, averaging by $\theta$ is $\langle r^n\rangle_\theta =\frac{1}{2\pi}\int_0^{2\pi} \frac{a^n(1-\epsilon^2)^n d\theta}{(1-\epsilon \cos \theta)^n}$

To average by time, you need the period. $\tau =\int_0^{\tau}1 dt =\int_0^{2\pi} \frac{1}{\dot{\theta}} d\theta=\int_0^{2\pi} \frac{mr^2d\theta}{L}$

Then the time average is $\langle r^n\rangle_t=\frac{1}{\tau}\int_0^{2\pi}\frac{mr^2\cdot r^nd\theta}{L}$

I'm not sure how to do it without the Residue Theorem of Complex Analysis, but it can be shown $\int_0^{2\pi} \frac{d\theta}{(1 - \epsilon \cos \theta)^n}=\frac{2\pi}{(1-\epsilon^2)^{n/2}}$.

From there you need to find $V= -GMm\langle 1/r\rangle_t = \frac{-2\pi}{\tau} GMm \langle mr/L \rangle_\theta$.

$\endgroup$
0
$\begingroup$

I assume you can derive the equations for potential & kinetic energy, and angular momentum for a circular orbit. And that you understand that angular momentum and total mechanical energy are conserved. (Conservation of angular momentum is intimately related to Kepler's law of areas, as explained in the answer I linked above in a comment).

You need to integrate with respect to time. However, the angle $\theta$ in that ellipse equation $$r = \frac{a(1-e^2)}{1 + e\cos\theta}$$

is not proportional to time. Kepler named that angle from the ellipse focus to the (reduced) orbiting particle the true anomaly. The angle which is proportional to the time is called the mean anomaly. Linking those two is another important parameter known as the eccentric anomaly.

All three of these angles are measured from the ellipse's major axis, and are equal to zero at the periapsis.

Here's a diagram from Wikipedia which illustrates the eccentric anomaly.

Eccentric anomaly

The eccentric anomaly of point P is the angle E. The center of the ellipse is point C, and the focus is point F.

The blue circle provides the X coordinate of P, and the green circle provides its Y coordinate. The true anomaly is the angle $f$.

The equations relating true & eccentric anomaly are straightforward, but the mean & eccentric anomalies are related via Kepler's equation: $$M = E - e\sin E$$

where $M$ is the mean anomaly, $E$ is the eccentric anomaly, and $e$ is the eccentricity. So it's easy to calculate $M$ if you know $E$, but there's no closed form in elementary functions for the inverse function. This makes analytical calculation of an orbiting body's position from the time somewhat challenging...

A useful stepping-stone to determining the mean gravitational potential energy and mean kinetic energy is to show that the orbital period is purely a function of $a$, the semi-major axis, and is independent of $e$, the eccentricity.

Anyway, I hope that's enough info to get you started.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy