# The Energy of a Multiparticle System

1. Feb 26, 2012

### aaj92

1. The problem statement, all variables and given/known data

Consider an electron (charge -e and mass m) in a circular orbit of radius r around a fixed proton (charge+e). Remembering that the inward Coulomb force ( ke$^{2}$/r$^{2}$) is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to -$\frac{1}{2}$ times it's PE; that is, T = -$\frac{1}{2}$U and hence E = $\frac{1}{2}$U. (This result is a consequence of the so called virial theorem. Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius r around a fixed proton. Electron 2 approaches from afar with kinetic energy T$_{2}$. When the second electron hits the atom, the first electron is knocked free and the second is captured in a circular orbit of radius r'.

2. Relevant equations

coulomb force : ke$^{2}$/r$^{2}$

virial theorem T = nU/2

3. The attempt at a solution

I'm not really worried about the second part of this problem quite yet. Right now I'm not really sure how to go about proving that the kinetic energy is -$\frac{1}{2}$ times the potential energy... can someone get me started on the right track?

2. Feb 26, 2012

### genericusrnme

You could use the virial theorem! :p

It's been a while since I've done this but I'm pretty sure it's just a case of writing down the kinetic and potential energies and then stating the correlation.

3. Feb 26, 2012

### aaj92

haha sorry I know it probably seems obvious but we haven't even had the virial theorem mentioned in class. So I'm not really sure how to use it? :/ but I'll try to figure it out

4. Feb 26, 2012

### genericusrnme

Don't worry, the virial theorm is part of action based lagrangian mechanics, you probably haven't encountered it.

You should use the fact that the orbit is circular, that is to say that r doesn't change.
So try playing about with the equation for the inwards force and the outwards centripital (centrifugal? I can't never remember which is the outwards pushing one)
$a_{outwards} = \frac{v^2}{r}$
$a_{outwards} + a_{inwards} = 0$

See if you can work out the kinetic energy from that then see what happens when you compare it to the potential energy

5. Feb 27, 2012

### aaj92

I still can't get this. I'm sorry I'm just slightly frustrated with this and now it's late haha can I just get an explanation for this :/ I'm struggling in this class right now

6. Feb 28, 2012

### genericusrnme

I've been given warnings for being too overzealous with my hinting so I'll try my best to help without giving you the answer

Last edited by a moderator: Feb 28, 2012
7. Feb 28, 2012

### aaj92

Thank you so much! This really helps a lot :)

8. Feb 28, 2012

### genericusrnme

no problem buddy!