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

AI Thread Summary
The discussion centers on the confusion regarding the analysis of orbital motion with a potential U(r) that depends on the radius. The user successfully derived that angular momentum L is conserved and expressed velocity in terms of L and r, but became puzzled when concluding that potential energy is solely a function of radius, which seemed incorrect. They also noted that their method yielded the same force F(r) regardless of the function r(θ) provided in the problem. Upon further reflection, they recognized that their expression for angular momentum only accounted for the angular component of velocity, neglecting the radial component, which clarified their misunderstanding. The discussion highlights the importance of considering all components of motion in orbital dynamics.
Hertz
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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.)
 
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The angular momentum is L=mvr only for special cases. Write it for the general case and you will get a more general result.
 
nasu said:
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 :)
 
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