Lagrangian Equation of Motion for Planetary Orbit in a Single Plane

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.

• – PM 2Ring Dec 28 '18 at 7:25
• Factor 2 is wrong, you can write your equation like this $\dfrac {d}{dt}\cdot \left( \dfrac {d\theta }{dt}r\right) =0$ – Eli Dec 28 '18 at 8:06
• Don't forget that there's an E-L equation for $r$ too! – GodotMisogi Dec 28 '18 at 8:24
• 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. – my2cts Dec 28 '18 at 11:34

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

• 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! – J. Tungsten Dec 28 '18 at 16:52