# Gravitation potential and gravitational binding energy

by shounakbhatta
Tags: binding, energy, gravitation, gravitational, potential
 P: 266 Hello, The gravitational binding energy of a system is equal to the negative of the total gravitational potential energy. Is this given by the equation: U=-alphaGM^2/R where alpha=3/5 and the equation stands as: U=-3/5GM^2/R? Kindly correct me if I am wrong?
 P: 398 Where did you get the factor of 3/5 from? There is no $\alpha$ = 3/5 in gravitational potential The gravitational potential energy is: $\frac{-GMm}{r^2}$ The gravitational binding energy can be thought of as an internal potential energy. How much grav potential is stored in an object? You must compute the gravitational potential for every particle in the object relative to all the rest of them which simply yields $\frac{-GM^2}{r^2}$ Because m is now M. Both values are negative, by convention.
 P: 266 For a spherical mass of uniform density, the gravitational binding energy U is given by the formula u=3GM^2/5r. The derivation follows from the volume of the sphere, 4 pi r^3 and area of the sphere 4 pi r^2. In Faber Jackson relation the gravitational potential of a mass distribution of radius R and mass M is given by the expression: U=-alphaGM^2/R where alpha=3/5 and the equation stands as: U=-3/5GM^2/R So I asked you the above question.
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P: 38,706

## Gravitation potential and gravitational binding energy

 Quote by soothsayer Where did you get the factor of 3/5 from? There is no $\alpha$ = 3/5 in gravitational potential The gravitational potential energy is: $\frac{-GMm}{r^2}$
Was this a typo? The gravitational force varies as $1/r^2$ but the gravitational potential energy as 1/r, the integral of $1/r^2$.

 The gravitational binding energy can be thought of as an internal potential energy. How much grav potential is stored in an object? You must compute the gravitational potential for every particle in the object relative to all the rest of them which simply yields $\frac{-GM^2}{r^2}$ Because m is now M. Both values are negative, by convention.
 P: 266 If you can please look into the following: http://en.wikipedia.org/wiki/Faber%E...ckson_relation and correct me, what mistake I am making.
 P: 398 Ah, right. I forgot about the factor for the sphere. And yes, the r^-2 factor was a typo.
 P: 266 So, My question still stands: The gravitational binding energy of a system is equal to the negative of the total gravitational potential energy. Is this given by the equation: U=-alphaGM^2/R where alpha=3/5 and the equation stands as: U=-3/5GM^2/R? Thanks.
 Mentor P: 11,048 Yes, this is true if the mass is uniformly distributed in a sphere. This is analogous to the electrostatic potential energy of a sphere with uniform charge density, which also has a factor of 3/5: http://farside.ph.utexas.edu/teachin...es/node56.html (in particular equation 601 and the derivation preceding it)
 P: 398 Sorry, I was confused at to what the question was. Were you asking if your equation was correct? Because yes, it is. Except, it's not equal to the NEGATIVE of the total gravitational potential energy, because gravitational potential is also negative. Both quantities are negative, it's just the gravitational binding the the total internal gravitational potential of the object.
 P: 266 Reading over Wikipedia I found that: For a spherical mass of uniform density, the gravitational binding energy U is given by the formula U=3GM^2/5R Again it states that for a uniform sphere the gravitational binding energy of the sphere is the negative of the gravitational potential energy, which is: U=-3GM^2/5R Are they referring to the same equation? How can one equation have different results?
 Mentor P: 11,048 Binding energy is the energy that we must add to a bound system in order to separate it completely into its constituents. It is always positive. If it were negative, the system could fly apart by itself, releasing energy. For a gravitationally bound sphere with uniform density, it is Eb = 3GM2/5R. When we add energy to a system, we increase its (potential + kinetic) energy. If the kinetic energy of the "pieces" is zero before and after, then only the potential energy changes. Ubound + Eb = Useparated. In the situation that we're dealing with here, we define the potential energy to be zero when the system is completely separated, that is, Useparated = 0. This leads to Ubound = -Eb = -3GM2/5R.
 P: 398 The negative signs are more a thing of convention than anything... I think what jtbell wrote is useful. Often, an object in a bound state, like an atom in a ground state, or a spherical planet, must be at a minimum of potential energy. The easiest way to do this is to say that a potential energy of zero means the system is totally unbound (imagine an ionized atom), and obviously, the bound state would have to have a negative energy, -3GM2/5R, for a planet, star, etc. You can also think of the positive of this as being the energy required to unbind the system, so the energy needed to ionize an atom, or the energy required by the Death Star to disintegrate Aldreraan ;) There is an arbirarity to potential energy. You can set the zero point energy to be anything as long as you are consistent. There are standard conventions for some systems like gravitational binding energy that make everything much nicer, however. I'm guessing this is the wiki page you're referencing? http://en.wikipedia.org/wiki/Gravita...binding_energy I see that they use the positive and negative version of the equation. It makes little difference. It only matters the convention. The top equation is essentially saying that this is the energy stored in the gravitational binding of the object, and the second equation is saying that this is the total binding potential of the spherical object , which is negative if we say that the potential zero point is when the system is completely unbound.
 P: 790 How do you deal with the gravitational self-energy of a point mass? Renormalization?
 P: 266 Thank you jtbell, for your wonderful reply. It just cleared out all my confusions. Thank you soothsayer. Yes, you are very right. I was referring to the wikipedia article only. Thanks for the detailed reply on the topic. Thanks. Thank you all.
P: 398
 Quote by Khashishi How do you deal with the gravitational self-energy of a point mass? Renormalization?
It's just GM2/R, I believe. Integrating over a sphere gives 3GM2/5R.
 P: 790 point mass has no R

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