Gravitational binding energy

A gravitational binding energy is the minimum energy that must be added to a system for the system to cease being in a gravitationally bound state. A gravitationally bound system has a lower (i.e., more negative) gravitational potential energy than the sum of its parts — this is what keeps the system aggregated in accordance with the minimum total potential energy principle.

For a spherical mass of uniform density, the gravitational binding energy U is given by the formula

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

where G is the gravitational constant, M is the mass of the sphere, and R is its radius.

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.

The actual depth-dependence of density, inferred from seismic travel times (see Adams–Williamson equation), is given in the Preliminary Reference Earth Model (PREM). Using this, the real gravitational binding energy of Earth can be calculated numerically to U = 2.487 · 10³² J

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

Derivation for a uniform sphere

The gravitational binding energy of a sphere with Radius $R$ 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.

Assuming a constant density $\rho$, the masses of a shell and the sphere inside it are:

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

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

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

Integrating over all shells yields:

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

Since $\rho$ is simply equal to the mass of the whole divided by its volume for objects with uniform density, therefore

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

And finally, plugging this into our result leads to

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

Non-uniform spheres

Planets and stars have radial density gradients from their lower density surfaces to their much larger density compressed cores. Degenerate matter objects (white dwarfs; neutron star pulsars) have radial density gradients plus relativistic corrections.

Neutron star relativistic equations of state provided by Jim Lattimer include a graph of radius vs. mass for various models. The most likely radii for a given neutron star mass are bracketed by models AP4 (smallest radius) and MS2 (largest radius). BE is the ratio of gravitational binding energy mass equivalent to observed neutron star gravitational mass of "M" kilograms with radius "R" meters,

$BE = \frac{0.60\,\beta}{1 - \frac{\beta}{2}}$      $\beta \ = G\,M/R\,{c}^{2}$

Given current values

$G = 6.6742\times10^{-11}\, m^3kg^{-1}sec^{-2}$

$c^2 = 8.98755\times10^{16}\, m^2sec^{-2}$

$M_{solar} = 1.98844\times10^{30}\, kg$

and star masses "M" commonly reported as multiples of one solar mass,

$M_x = \frac{M}{M_\odot}$

then the relativistic fractional binding energy of a neutron star is

$BE = \frac{885.975\,M_x}{R - 738.313\,M_x}$

See also

• Nordtvedt effect