Derivation of velocity dispersion from virial theorem?

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SUMMARY

The discussion focuses on deriving velocity dispersion from the virial theorem, specifically using the equation 2K + U = 0, where K represents kinetic energy and U represents potential energy. The user correctly identifies that K/m is approximately equal to sigma^2, and U/m is expressed as -GM/R. The conclusion drawn is that the velocity dispersion, sigma^2, is equal to GM/R, emphasizing the importance of the negative sign in gravitational potential energy. The factor of 2 mentioned is not ignored but is accounted for in the derivation.

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Derivation of velocity dispersion from virial theorem??

Hey,

Im probably being a bit dim here but could anyone help me derive the velocity dispersion from the virial theorem. I've got 2K+U=0, K/m~sigma^2 and U/m~GM/R.

From rearranging I get a negative velocity^2? Or maybe its the gravitational force that's negative. Oh and a factor of 2. Is that just ignored??

My notes say the answer is sigma^2~GM/R

Just a bit confused, anyone got any clues?
 
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The virial theorem says that 2T - |U| = 0, where U = -GM/R (the negative sign being important, since gravity is an attractive force).

2T = |U|
 

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