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Assume that a mass $m$ is in a gravitational orbit around a much larger mass $M$, as in the case of the earth revolving around the sun. Also, assume the motion is constrained to a single horizontal $xy$ plane.

I set up the Lagrangian equation of motion in polar coordinates as follows:

$\large L = T - V$

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

Euler-Langrange equation for polar coordinates:

$\large\frac{\partial{L}}{\partial{\theta}}-\frac{d}{dt}(\frac{\partial{L}}{\partial{\dot{\theta}}}) = 0$ $\rightarrow$

$\large\frac{d}{dt}(mr^2\dot{\theta}) = m(2r\dot{r}\dot{\theta}+r^2\ddot{\theta}) = 0$ $\rightarrow$

(assuming the trivial solution $r=0$ is not the answer, i.e. the orbiting has a non-zero potential)

$\large\mathbf{2\dot{r}\dot{\theta}+r\ddot{\theta} = 0}$

Here is where I am stuck.

Assuming that I performed the above steps correctly, how can I solve this differential equation in $r$ and $\theta$ to come up with some generalized equation of motion? I know that Kepler's Law dictates that a stable orbit must be an ellipse. But here is where my skill with differential equations and conical sections fails me.

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  • $\begingroup$ Related: physics.stackexchange.com/q/353239 $\endgroup$
    – PM 2Ring
    Dec 28, 2018 at 7:25
  • $\begingroup$ Factor 2 is wrong, you can write your equation like this $\dfrac {d}{dt}\cdot \left( \dfrac {d\theta }{dt}r\right) =0$ $\endgroup$
    – Eli
    Dec 28, 2018 at 8:06
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    $\begingroup$ Don't forget that there's an E-L equation for $r$ too! $\endgroup$ Dec 28, 2018 at 8:24
  • $\begingroup$ What you derived is the conservation of angular momentum. You can set that to a constant L and use that to solve the equation for r. $\endgroup$
    – my2cts
    Dec 28, 2018 at 11:34

1 Answer 1

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Let's start from

$$ L = \frac{1}{2}(\dot{r}^2 + r^2\dot{\theta}^2) - V(r) $$

There are two degrees of freedom: $r$ and $\theta$. Start with $\theta$

$\theta$

$$ \frac{{\rm d}}{{\rm d}t}\frac{\partial L}{\partial \dot{\theta}} = 0 = \frac{{\rm d}}{{\rm d}t}(r^2\dot{\theta})~~~\Rightarrow~~~ r^2\dot{\theta} = l = {\rm const} \tag{1} $$

which means that the number $l = r^2\dot{\theta}$ is a constant (angular momentum!).

$r$

$$ \frac{{\rm d}}{{\rm d}t}\frac{\partial L}{\partial \dot{r}} - \frac{\partial L}{\partial r} = 0 = \ddot{r} - r\dot{\theta}^2 + \frac{{\rm d}V}{{\rm d}r} \tag{2} $$

Use (1) to write

$$ \frac{{\rm d}}{{\rm d}t} = \frac{l}{r^2}\frac{{\rm d}}{{\rm d}\theta} $$

and define the variable $u = 1/r$, if you replace both into (2) you will get

$$ \frac{{\rm d}^2u}{{\rm d}t^2} + u = \frac{1}{l^2 u^2}\frac{{\rm d}V(1/u)}{{\rm d}r} \tag{3} $$

For a Keppler potential

$$ V(r) = -\frac{GM}{r} = -GMu $$

Replace that in (3) and you get

$$ \frac{{\rm d}^2u}{{\rm d}t^2} + u = \frac{GM}{l^2} $$

whose solution is

$$ u(\theta) = C\cos(\theta - \theta_0) + \frac{GM}{l^2} $$

Now define

$$ e = \frac{Cl^2}{GM} ~~~\mbox{and}~~~ a = \frac{l^2}{GM(1 - e^2)} $$

such that the equation above becomes

$$ \bbox[5px,border:2px solid blue] { r(\theta) = \frac{a(1 - e^2)}{1 + e\cos(\theta - \theta_0)} } \tag{4} $$

And there you have it, the solution are conics

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  • $\begingroup$ That is awesome. Yes, I did miss the fact that I had simply derived conservation of angular momentum, and also I forgot about the second E-L equation. Thank you so much! $\endgroup$
    – J.T.
    Dec 28, 2018 at 16:52

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