Why does my energy approach give the same force for any orbital trajectory?

[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 $L=mvr$ is conserved and I will call it L. Thus:
$v=\frac{L}{mr}$

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

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 $r(\theta)$ 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 $r(\theta)$ 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:

$L_{Lagrangian}=\frac{1}{2}mv^2-U(r)$

In polar, $v^2=\dot{r}^2+r^2\dot{\theta}^2$

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

And then in my text I found these equations:

$\frac{\partial L}{\dot{q}_i}=p_i$

$\frac{\partial L}{q_i}=\dot{p}_i$

(Not quite sure why they're true yet)

which implies:
$\frac{d}{dt}(\frac{\partial L}{\dot{q}_i})=\frac{\partial L}{q_i}$

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

-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 :)