I would like to show that a particle orbiting another will follow the trajectory

\begin{equation} r = \frac{a(1-e^2)}{1 + e \cos(\theta)}. \end{equation}

I would like to do this with minimal assumptions. Any pointers? I've read that this is nothing but the equation for an ellipse, as defined from the focus of the ellipse. But I'm not satisfied in just taking the result, I mean why should I assume that orbits are elliptical? I guess it should be derivable from energy and angular momentum considerations...

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    $\begingroup$ staff.washington.edu/storm/121C/lecture_notes/ellipse1.pdf $\endgroup$ – John Rennie Mar 12 '13 at 16:09
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    $\begingroup$ In more fancy terms, you are asking about the Kepler orbit solutions to the classical two-body problem. Have you seen the Kepler orbit wiki: en.wikipedia.org/wiki/Kepler_orbit ? If not, that would be a great place to start for a summary of the mathematical and physical details. $\endgroup$ – David H Mar 12 '13 at 19:23
  • $\begingroup$ You can see a detailed derivation here: teletechsvc.net/… $\endgroup$ – neutrino Mar 12 '13 at 20:52
  • $\begingroup$ The equation above is the general equation for a conic section. Based on the value of $e$, the trajectory can be an ellipse, a circle, a parabola and even a hyperbola. I am writing down the derivation as an answer, in case someone needs it further down. $\endgroup$ – Cheeku Mar 13 '13 at 1:43

For the two-body problem:

In any inertial frame, applying Newton's 2nd law, we write: $$ m_1 \ddot{\vec{r_1}} = \frac{-Gm_1m_2}{r^2}\hat{r} $$ $$ m_2 \ddot{\vec{r_2}} = \frac{Gm_1m_2}{r^2}\hat{r} $$

As a side derivation, you can add the above two derivations to obtain the fact that total linear momentum of the system is constant, but we move on for now.

To get relative motion, subtract the equations for $\vec{r_1}$ and $\vec{r_2}$, and obtain: $$ \ddot{(\vec{r_1}-\vec{r_2})} = -(\frac{1}{m_1} + \frac{1}{m_2})\frac{Gm_1m_2}{r^2}\hat{r} $$

Define 'reduced mass' $\mu$ as $$ \frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2} $$

We then have $$ \mu \ddot{\vec{r}} = \frac{-Gm_1m_2}{r^2}\hat{r} $$

Rearranging the above equation, we can rewrite: $$ \ddot{\vec{r}} = \frac{-G(m_1+m_2)}{r^2}\hat{r} = \frac{-\alpha}{r^2}\hat{r} $$

We have our equations ready with us. Now, we solve them. But note that the quantities $J$(angular momentum) and $E$(total energy) are conserved defined as follows. You can easily check this fact. $$\vec{J} = \mu \vec{r} \times \dot{\vec{r}}$$ and $$E = K+V = \frac{1}{2} \mu \dot{\vec{r}}^2 - \frac{Gm_1m_2}{r}$$

How do you know the motion is in plane? Since, $\vec{J}$ is always perpendicular to plane of motion and it is constant, the motion is in a plane. Dividing motion into components, $$\ddot{x} = \frac{-\alpha}{r^3}x \ , \ \ddot{y} = \frac{-\alpha}{r^3}y$$ and $$r^2 = x^2 + y^2$$ and $$J = \mu(x\dot{y}-y\dot{x})$$

Solving the above three equations(I leave that to the reader), we obtain: $$\ddot{r} = \frac{-\alpha}{r^3} + \frac{J^2}{\mu^2 r^3}$$

Our interest is in equation of orbit $r(\theta)$. Note that $$\dot{\vec{r}} = \dot{r}\hat{r} + r \omega \hat{\theta}$$ which implies $$\vec{J} = \mu r^2 \omega \hat{z}$$ or $$J = \mu r^2 \omega$$ Now, the action $$\frac{dr}{dt} = \frac{dr}{d \theta}\frac{d \theta}{dt} = \frac{dr}{d \theta} \omega = \frac{dr}{d \theta}\frac{J}{\mu r^2}$$ and then $$\frac{d^2r}{dt^2} = \frac{d^2 r}{d \theta^2}(\frac{J}{\mu r^2})^2 - \frac{2}{r^3}\frac{J}{\mu}\frac{J}{\mu r^2}(\frac{dr}{d \theta})^2$$ This is a very complicated equation for $r(\theta)$. But the equation is simple in terms of $\rho(\theta) = \frac{1}{r(\theta)}$. Performing change of variables, we get $$\frac{d^2 \rho}{d \theta^2} + \rho = \mu \frac{Gm_1m_2}{J^2}$$ This has a simple solution $$\rho(\theta) = A \cos \theta + \mu \frac{Gm_1m_2}{J^2}$$ It is convenient to write $h = \frac{J}{\mu}$. Finally substituting for $r$, $$r = \frac{\frac{h^2}{\alpha}}{1+\frac{Ah^2}{\alpha}\cos \theta}$$ Naming new variables $a$ and $e$, the above equation becomes $$r = \frac{a(1-e^2)}{1+e \cos \theta}$$ You have been given incomplete information if you've been told this is an equation for ellipse. It is actually the equation for a conic section. Based on values of $e$, the trajectory can be anything. $$e = 0,circle$$ $$0<e<1,ellipse$$ $$e = 1,parabola$$ $$e>1,hyperbola$$ But that doesn't solve your doubt completely. Here's the catch. Note that the values of $a$(semi-major axis) and $e$(eccentricity) of orbit, which are two orbital parameters, depend upon $J$ and $E$ i.e. the angular momentum and energy of the system. You can play with the equations mentioned above to derive various properties of the system.

But, the important thing to keep in mind is the fact that values of $J$ and $E$ dynamically affect the values of $a$ and $e$. It is pure co-incidence that in most of these cases, the trajectory takes the form of an ellipse, but it could be any conic section in theory.

Does this derivation and explanation satisfy you?

P.S. If you want to change the trajectory of say,Earth, you will have to change $J$ and $E$. That's the job of rockets that launch satellites. Scientists, at the base level, take care that $J$ and $E$ fit right and the rest of the physics takes care of itself

  • $\begingroup$ Thanks Cheeku. Nice derivation, yes, it does satisfy the question, which was 'why an ellipse?'. Thanks for the note on $E$ and $L$ but for me at least, this is not a concern, I'm interested in systems with no external influence, so these are constants. $\endgroup$ – user12345 Mar 13 '13 at 10:13

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