# Kepler's First Law

1. Mar 31, 2006

### Rael

Hi folks, my problem is the following one:
Kepler stated that orbits of planets are elliptic. Everytnig's well since Newton obtained the same results, with his formula for gravity

$$F = G(mM)/r^2$$

Now, i tried to write the Lagrangian of the system (L = K-U, K is cinetic energy, U is potential energy) in polar coordinates which (hopefully without errors) should be defined as follows:

$$L = \frac{1}{2}m((\dot{r})^2 + (r\dot{\theta})^2) + G(mM)/r$$

where $$\dot{r}^$$ is radial velocity, $$\dot{\theta}^$$ is angular velocity, r is the distance between the two bodies, one of them is for simplicity considered being still in the origin.

now deriving the differential equations of motion using $$d(\partial{L} / \partial{\dot{x}} )/dt = \partial{L}/\partial{x}$$

we have the following equations :

$$\ddot{\theta} = -2\dot{\theta}\dot{r}/r$$
$$\ddot{r} = r(\dot{\theta})^2 - MG/r^2$$

$$\ddot{r}$$ is the radial acceleration, $$\ddot{\theta}$$ is angular acceleration, M is the mass of the body fixed at the origin.

Now, the equations are by no way linear and easily solvable... and can describe a very rich variety of orbital behaviours, not only the steady Kepler's ellipse (things got even worse when i derived the equations considering the mass M not being fixed but free to move).
Is Kepler's first law an approximation ?? or it's exact ?

Last edited: Mar 31, 2006
2. Mar 31, 2006

### Physics Monkey

Bound orbits in such a 1/r central force field are ellipses. The trouble is that you've made a mess of the equation of motion for $$\theta$$. Notice that $$\theta$$ does not appear explicitly in the Lagrangian. This implies that the canonical momentum conjugate to $$\theta$$ is conserved. What is this momentum?

3. Mar 31, 2006

### Rael

Eghm, well, since the force is central, angular acceleration is 0.
so I have only the equation for radial acceleration left.
It's still not linear, but may be more easily solvable. Is it a brutal approximation to consider the term $$- MG/r^2$$ = 0 due to the little value of G ??
What exactly do you mean by "bound" ? An orbit resulting in collapsing of the body m into M is still bounded ?

Last edited: Mar 31, 2006
4. Mar 31, 2006

### Physics Monkey

A couple of things, Rael.

1) The angular acceleration is not zero.
2) Bound means doesn't escape to infinity. The orbit is finite, in other words, like the orbit of the Earth or Mars. There are unbounded orbits which behave like hyperbolas.
3) It is a horrible approximation to set $$- GM/r^2$$ to zero, you lose all the effects of gravity!

Did you try to figure out what the canonical momentum associated with $$\theta$$ is? The reason why this important is because $$\frac{d p_\theta}{dt} = 0$$ since the Lagrangian is independent of $$\theta$$. In other words, it gives a conservation law.

5. Mar 31, 2006

### Rael

You'r right ! it's an error to consider angular velocity being constant !
Well, angular momentum is derivable from the equation for $$\ddot{\theta}$$, so being h the angular momentum we have

$$h = \dot{\theta}r^2$$

using this in the equation for $$\ddot{r}$$ we obtain

$$\ddot{r} = \frac{h^2}{r^3} - \frac{MG}{r^2}$$

now just matter of solving the nonlinear differential equation for r, and then finding $$\theta$$.
Thank you !
Just a matter of mathematics now .

6. Apr 1, 2006

### Rael

Still have no idea on how to get rid of that nasty NODE (nonlinear differential equation)
it's nonlinear... i wonder if there exists a time dependent solution for the two body problem... (my problem should be a simplified version of that)...

7. Apr 1, 2006

### topsquark

Hmm... You seem to be missing a mass in that angular momentum equation.

You should have:
$$M \ddot r =h^2/(Mr^3)-MG/r^2$$

From here I am going to suggest you write down the total energy of the object in (a bound) orbit. (In polar coordinates!) Recall that E will be a constant of the motion. You can use the angular momentum equation to get rid of the $$\dot \theta^2$$ term and solve for $$\dot r$$. For reference, at this point you should have:
$$\dot r=\sqrt{\frac{2}{M} \left ( E +\frac{GM}{r}-\frac{h^2}{2mr^2} \right ) }$$

You can separate the r and t variables and wind up with an equation for t in terms of an integral over r. Now a trick. Change the variable from r to u = 1/r. The integral is now fairly easy to do (if somewhat long to work out). The only problem is that you wind up with a function t(r) as opposed to r(t).

It is standard not to solve for time, but the angle in the orbit. You can use the angular momentum equation to derive an expression relating d/dt and d/d(theta). The advantage in doing this is that it is pretty straightforward from there to show the bound orbit is an ellipse.

-Dan

8. Apr 6, 2006

### Rael

An ellipse hmmm..., I solved the integral, and winded up with a formula for $$\theta(r)$$ which in polar coordinates is a conic section. Everythings ok until now, but i have simulated the system of differential equations on my PC and I got the following trajectory :

http://img314.imageshack.us/img314/3420/trajectory0lh.jpg [Broken]

the moving body started in the bottom right corner (the big black dot), and moving with a certain initial angular velocity, "converged" to the non-moving body at the center of my system of coordinates.
I suppose two things are possible about the results:
1) numerical errors which lead to "non-conservation" of energy and/or angular momentum (A 4th order Runge-Kutta discretization was used, with adaptive step-size routine).
2) The relation for $$\theta(r)$$ holds in this kind of orbits too, and angular momentum and energy are still conserved...

Don't know what to say, but intuitively I suppose the second possibility is the most realistic.

Last edited by a moderator: May 2, 2017
9. Apr 8, 2006

### topsquark

What you seem to have here is a decaying orbit. This also is predicted by the Mathematics. In order to get an actual conic section for the orbit you need specific boundary conditions on the intial position, velocity, etc.
What the equations imply is that when you DO have those particular boundary conditions, the only stable orbit will be in the form of an ellipse. If the orbit is unbound then (barring running into the central mass) you have a hyperbolic or parabolic path, no matter what your initial conditions may be.

-Dan

Last edited by a moderator: May 2, 2017
10. Apr 8, 2006

### Physics Monkey

Rael,

You shouldn't be getting a decaying orbit. As topsquark said, you need special initial conditions to get an orbit that hits the central mass. In fact, the only way the orbiting body can hit the central mass in this problem is if the orbiting body has exactly zero angular momentum (assuming the central mass is pointlike). This is easy to see since for any finite energy and non-zero angular momentum, the angular momentum barrier $$\frac{\ell^2}{2 m r^2}$$ blows up faster than $$- \frac{G M m}{r}$$ as r goes to zero. Thus you'll always hit a turning point before r = 0. Bound orbits are always ellipses for the Kepler problem unless $$\ell = 0$$. Since it's clear that your orbit has some nonzero angular momentum, you must have made some error in your numerical solution of the differential equation.

Last edited: Apr 8, 2006