As far as I understand it the gravitational binding energy of some distribution of mass is the negative of its gravitational self-potential energy.

I tried to calculate the latter for a solid sphere of radius $R$, mass $M$, and uniform density.

By the shell theorem (or Gauss's law of gravitation), the field strength at a distance $r$ from the center of the sphere is given by


where $M_{enc}=M(r/R)^3$ is the mass enclosed in a sphere of radius $r$.

The gravitational potentiel at a distance $r$ created by this distribution is thus


The self-gravitational potentiel energy is the sum of the gravitational potentiel energies $U \cdot dm$ over all the mass elements $dm$ in the distribution.

Let's proceed by shell integration. The mass contained in the shell of inner radius $r$, outer radius $r+dr$ is simply

$$dm_r=4 \pi r^2\cdot dr\cdot \rho=4 \pi r^2\cdot dr\cdot \frac{M}{4 \pi R^3}=\frac{3Mr^2dr}{R^3}$$

The self-potential energy of the sphere is thus

$$\int^{R}_{0} V(r)dm_r=\int^{R}_{0}\big(\frac{-GMr^2}{2R^3}\big)\big(\frac{3Mr^2dr}{R^3}\big)=\frac{-3GM^2}{2r^6}\int^{R}_{0}r^4dr=-\frac{3GM^2}{10R}$$

which is exactly half of the correct answer.

I checked my work multiple times for simple mistakes but I can't seem to locate the source of the factor of $2$ error. This leads me to believe there is something fundamentally wrong with the way I calculated the energy.

Where is the problem?

  • $\begingroup$ In your MathJax you're using \big for large brackets, which doesn't work. Use matching \left and \right instead. \Big is a fixed size, whereas \left and \right will scale automatically to the size needed for the enclosed contents of the brackets. $\endgroup$ Jul 15, 2017 at 1:12

2 Answers 2


The issue is the manner in which you are forming your shells---whether they are coming from inside or outside the previous shells. For binding energy, this means the amount of energy that it would take to sequentially remove each successive shell to infinity. Thus, the potential needs to be calculated with respect to infinity, not the origin; your expression for potential would suggest that each shell starts at the origin and expands through the existing mass out to a radius $r$, rather than coalescing around an already-existing core from the outside. So, calculate the potential as

$V(r) = \int_{\infty}^r \frac{GM_{enc}(r)}{x^2}\ dx = -\frac{GM_{enc}(r)}{r}.$

This should resolve the factor of two.

Terminology aside, I think we can agree on the concept of what the magnitude of the energy means, so positive or negative doesn't have a huge impact. To get a feel for the integral above, let's imagine a single particle that's being pulled in by the gravitation of the still-forming ball (with radius $r$), rather than a shell. As the particle comes in from infinity, the potential it will feel will be the usual Newtonian gravitational potential, all the way until it hits the surface of the ball. Now, each little bit of mass $dm$ of a shell being added will also feel this same potential; we can think of the shell as being many small particles coming in from all directions at the same time. Every time we add a shell in this way, $r \rightarrow r+dr$, so $M_{enc}$ increases accordingly, which we account for in the integral over $r$. This is in contrast to the integral with bounds $[0,R]$ in the question, because such an integral is more akin to the amount of energy it would take to "inflate" shells of mass outwards from the origin. Such a process would require the ball to be totally permeable as the shells inflate out to the surface, but if this were the case, the whole ball would immediately collapse on itself again due to its lack of rigidity.

  • $\begingroup$ Ok. First I actually don't know what gravitational binding energy. I only know what self-potential energy is. The self-potential energy of a system of mass $m_1,...m_N$ is the sum of $U_{i,j}$ over all pairs $(i,j)$ with $i<j$ where $U_{i,j}=-Gm_im_j/r_{i,j}$, $r_{i,j}$ being the distance between the masses $m_i$ and $m_j$. This is what I tried to calculate. $\endgroup$ Jul 14, 2017 at 1:40
  • $\begingroup$ Second, your integral doesn't make sense to me. $M_{enc}(r)$ should be replaced by $M_{enc}(x)$ no? $\endgroup$ Jul 14, 2017 at 1:41
  • $\begingroup$ Josh is correct: you took the wrong definition of the binding energy. See this Wikipedia article for the full computation: en.m.wikipedia.org/wiki/Gravitational_binding_energy $\endgroup$
    – user154997
    Jul 14, 2017 at 5:56
  • $\begingroup$ @LucJ.Bourhis : In fact, what I calculated is the self-gravitational potentiel energy, which is just the negative of binding energy. I described self-potential energy above, i.e simply the energy of the mass distribution due to its own gravitational field. $\endgroup$ Jul 14, 2017 at 14:50
  • $\begingroup$ I have added clarification in the response, since it wouldn't fit here in the comments. The essential difference in our two quantities is the amount of energy involved in removing all the bits of mass infinitely far away from each other vs. the amount of energy required to keep the ball from collapsing in on itself. The former is the gravitational binding energy (due to the self-potential), and the latter is more a measure of the minimum rigidity of the matter involved. $\endgroup$
    – user138962
    Jul 14, 2017 at 19:55

There are problems with how you are calculating potential and with how you are calculating gravitational binding energy.

The gravitational field inside the sphere is radially inwards and of magnitude $GM_{enc}/r^2= GMr/R^3$. The gravitational field outside the sphere is radially inwards and of magnitude $GM/r^2$.

The gravitational potential is the work done per unit mass bringing that mass from infinity to $r$.

The potential at a radius $r$ inside the sphere is $$V(r) = \int_{\infty}^{R} \frac{GM}{r'^2}\ dr' + \int_{R}^{r} \frac{GMr'}{R^3}\ dr' $$ $$V(r) = -\frac{GM}{R} -\frac{GM}{2R} +\frac{GMr}{2R^3}$$ $$V(r) = \frac{GM}{2R^3}(r^2 - 3R^3)$$

However, this is not needed to calculate the binding energy of a sphere, since the gravitational binding energy is the sum of the energies required to remove mass shells from the surface of a sphere to infinity (imagine peeling off layers from the surface until you reach the centre).

The potential at the surface of a sphere of mass $M'$ is $-GM'/R'$, where the constant density $\rho =3M'/4\pi R'^3$. Thus $$V(R') = -\frac{4\pi G\rho}{3}R'^2$$ and the binding energy is equal to $V(R')$ multiplied by the mass of a shell, $dM = 4\pi R'^2 \rho\ dR'$, integrated over mass shells from zero to the final radius of the star.

$$U = -\int_{0}^{R} \frac{4\pi G\rho}{3}R'^2 \ 4\pi R'^2 \rho \ dR'$$ $$ U = -\frac{16\pi^2 G\rho^2 R^5}{15} = -\frac{3GM^2}{5R}$$


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