Where Did I Go Wrong in My Analysis of Orbital Motion?

[Hertz]

Hi, I stumbled upon this while working on a problem on my physics homework. I still want to solve the problem myself if possible though so I won't post it here, instead, I'll post what is confusing me.

Consider orbital motion with potential U(r), where U(r) is any arbitrary function of r.
I was able to show that the quantity [itex]L=mvr[/itex] is conserved and I will call it L. Thus:
[itex]v=\frac{L}{mr}[/itex]

We know that the system has a total energy that is constant:
[itex]E=T+U[/itex]
[itex]E=\frac{mv^2}{2}+U(r)[/itex]
[itex]U(r)=E-\frac{mv^2}{2}[/itex]
[itex]U(r)=E-\frac{m}{2} \frac{L^2}{(mr)^2}[/itex]

This shows that potential is only dependent on radius. Everything else is a constant. Furthermore, it shows that potential as a function of radius is ALWAYS equal to the same thing... This simply cannot be true... Where am I going wrong?

edit-
The problem that I'm working on gives me a function [itex]r(\theta)[/itex] and asks "What central force is responsible for this motion".

Using the method above... I'm finding that F(r) is the same thing no matter what [itex]r(\theta)[/itex] is... (By taking the negative derivative of U(r) with respect to r.)

 

[nasu]

The angular momentum is L=mvr only for special cases. Write it for the general case and you will get a more general result.

 

[Hertz]

The angular momentum is L=mvr only for special cases. Write it for the general case and you will get a more general result.

Hmm, well this is why I thought it was for general U, maybe you can help me see what I did wrong:

[itex]L_{Lagrangian}=\frac{1}{2}mv^2-U(r)[/itex]

In polar, [itex]v^2=\dot{r}^2+r^2\dot{\theta}^2[/itex]

[itex]L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-U(r)[/itex]

And then in my text I found these equations:

[itex]\frac{\partial L}{\dot{q}_i}=p_i[/itex]

[itex]\frac{\partial L}{q_i}=\dot{p}_i[/itex]

(Not quite sure why they're true yet)

which implies:
[itex]\frac{d}{dt}(\frac{\partial L}{\dot{q}_i})=\frac{\partial L}{q_i}[/itex]

Applied to Lagrangian above:
[itex]\frac{d}{dt}(\frac{\partial L}{\dot{\theta}})=\frac{\partial L}{\theta}[/itex]
[itex]\frac{d}{dt}(mr^2\dot{\theta})=0[/itex]
[itex]mr^2\dot{\theta}=const=mr(r\dot{\theta})=mvr=l[/itex]

-edit
I do see the problem now though! In this expression, this is only velocity in the theta direction. It doesn't account for velocity in the r direction, so I can't use it in the OP. Thanks :)