Systems held together by gravity

  • Thread starter Thread starter Wuratch
  • Start date Start date
  • Tags Tags
    Gravity Systems
Wuratch
Messages
3
Reaction score
0

Homework Statement


For any system of particles held together by mutual attraction
U(potential) = -2 U(Kinetic)

Suppose you add energy to such a system and wait for it to equilibrate, does the total kinetic energy increase or decrease? Explain

Homework Equations


The Attempt at a Solution


Apparently the kinetic energy decreases. How can you explain this?
 
Physics news on Phys.org
Well, suppose you add energy to the earth-moon system. That moves the moon to a higher orbit. Does the kinetic energy decrease or increase?
 
Dick said:
Well, suppose you add energy to the earth-moon system. That moves the moon to a higher orbit. Does the kinetic energy decrease or increase?

It would increase and that's what I thought. But the book says that it is a system with a negative heat capacity. It's entropy vs. energy graph is concave-up. When you add energy, all the energy apparently goes into the potential energy and the particles in the system get farther apart and actually slow down. I don't fully understand this and don't know how to to give the explanation without just repeating what the book says
 
Actually the moon moving to a higher orbit means that it must slow down...

Note: mv^2/r=GMm/r^2 (M is mass of Earth, m is mass of moon)=> v=sqrt(GM/r) so that as r increases, v decreases.

Increasing the energy of a system in virial equilibrium goes like this:

E increases, so that T+U increases, but U=-2T, so that T-2T=-T increases. An increase in "negative T" means that T decreases and U increases. By U increases, we mean U becomes less negative.

For example, if E=-2J, T=2J, U=-4J, and I add 1J to this system, E now has to be -1J, and now I have a system of equations:

-1J=T+U, and U=-2T. Solving: -1J=T-2T=-T => T=1J, U=-2J. So you notice that U got less negative (went up) and T went down.
 
Hey, I tried to do this in this way:

I use V for potential energy and T for kinetic energy.

Given equation reads:

V=-2T

V+T=-T

Now you see that the left hand site is a total energy:

E_{tot}=-T

Differentiate the whole thing:

dE_{tot}=-dT

So if the total energy is increasing then dE_{tot}>0 and thus dT<0. This means that the kinetic energy is decreasing.
 
Thanks everyone. I realize my mistake was trying to explain it with just words
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top