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The gravitational binding energy of an object consisting of loose material, held together by gravity alone, is the amount of energy required to pull all of the material apart, to infinity. It is also the amount of energy that is liberated (usually in the form of heat) during the accretion of such an object from material falling from infinity.

The gravitational binding energy of a system is equal to the negative of the total gravitational potential energy, considering the system as a set of small particles. For a system consisting of a celestial body and a satellite, the gravitational binding energy will have a larger absolute value than the potential energy of the satellite with respect to the celestial body, because for the latter quantity, only the separation of the two components is taken into account, keeping each intact.

For a spherical mass of uniform density, the gravitational binding energy U is given by the formula^[1]^[2]

$U = \frac{3GM^2}{5r}$

where G is the gravitational constant, M is the mass of the sphere, and r is its radius. This is 80% greater than the energy required to separate to infinity the two half-spheres touching each other got by a cut through the center of a sphere of radius r.

Assuming that the Earth is a uniform sphere (which is not correct, but is close enough to get an order-of-magnitude estimate) with M = 5.97×10²⁴kg and r = 6.37×10⁶m, U is 2.24×10³²J. This is roughly equal to one week of the Sun's total energy output. It is 37.5 MJ/kg, 60% of the absolute value of the potential energy per kilogram at the surface.

According to the virial theorem, the gravitational binding energy of a star is about two times its internal thermal energy.^[1]

Derivation for a uniform sphere

The gravitational binding energy of a sphere is found by imagining that it is pulled apart by successively moving spherical shells to infinity, the outermost first, and finding the total energy needed for that.

If we assume a constant density $ρ$ then the masses of a shell and the sphere inside it are:

$m_{{shell}}=4\,\pi \,{r}^{2}\rho\,dr$ and $m_{{interior}}=4/3\,\pi \,{r}^{3}\rho$

The required energy for a shell is the negative of the gravitational potential energy:

${\it dU}= G\frac{\,m_{shell}\,m_{interior}} {r}$

Integrating over all shells we get:

$U=G\,\int_0^{R} {\frac {(4\,\pi \,{r}^{2}\rho)\,\,(\frac{4}3\,\pi \,{r}^{3}\rho)}{r}}\,dr=G{\frac {16}{15}}\,{\pi }^{2}{\rho}^{2}{R}^{5}$

Remembering that $ρ$ is simply equal to the mass of the whole divided by its volume for objects with uniform density we get:

$\rho={\frac {M}{\frac{4}{3}\pi \,{R}^{3}}}$

And finally, plugging this in to our result we get:

$U=G\frac{16}{15} \pi^2 R^5 \left(\frac {M}{\frac{4}{3}\pi R^3}\right)^2= \frac{3GM^2}{5R}$

References

^ ^a ^b Chandrasekhar, S. 1939, An Introduction to the Study of Stellar Structure (Chicago: U. of Chicago; reprinted in New York: Dover), section 9, eqs. 90-92, p. 51 (Dover edition)
^ Lang, K. R. 1980, Astrophysical Formulae (Berlin: Springer Verlag), p. 272

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