- #1
Hertz
- 180
- 8
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.)
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.)