Gravitational binding energy

The gravitational binding energy of a sphere with radius ${\displaystyle R}$ 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 ${\displaystyle \rho }$, the masses of a shell and the sphere inside it are:

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

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

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

Integrating over all shells yields:

${\displaystyle U=-G\int _{0}^{R}{\frac {(4\pi r^{2}\rho )({\tfrac {4}{3}}\pi r^{3}\rho )}{r}}dr=-G{\frac {16}{3}}\pi ^{2}\rho ^{2}\int _{0}^{R}{r^{4}}dr=-G{\frac {16}{15}}{\pi }^{2}{\rho }^{2}R^{5}}$

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

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

And finally, plugging this into our result leads to

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

Two bodies, placed at the distance R from each other and reciprocally not moving, exert a gravitational force on a third body slightly smaller when R is small. This can be seen as a negative mass component of the system, equal, for uniformly spherical solutions, to:

${\displaystyle M_{\mathrm {binding} }=-{\frac {3GM^{2}}{5Rc^{2}}}}$

For example, the fact that Earth is a gravitationally-bound sphere of its current size costs 2.49421 × 10¹⁵ kg of mass (roughly one fourth the mass of Phobos – see above for the same value in Joules), and if its atoms were sparse over an arbitrarily large volume the Earth would weigh its current mass plus 2.49421 × 10¹⁵ kilograms (and its gravitational pull over a third body would be accordingly stronger).

It can be easily demonstrated that this negative component can never exceed the positive component of a system. A negative binding energy greater than the mass of the system itself would indeed require that the radius of the system be smaller than:

${\displaystyle R\leq {\frac {3GM}{5c^{2}}}}$

which is smaller than ${\displaystyle {\frac {3}{10}}}$ its Schwarzschild radius:

${\displaystyle R\leq {\frac {3}{10}}r_{\mathrm {s} }}$

and therefore never visible to an external observer. However this is only a Newtonian approximation and in relativistic conditions other factors must be taken into account as well.[5]

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

Neutron star relativistic equations of state include a graph of radius vs. mass for various models.[6] 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 with radius R,

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

Given current values

${\displaystyle G=6.6743\times 10^{-11}\,\mathrm {m^{3}{\cdot }kg^{-1}{\cdot }s^{-2}} }$[7]

${\displaystyle c^{2}=8.98755\times 10^{16}\,\mathrm {m^{2}{\cdot }s^{-2}} }$

${\displaystyle M_{\odot }=1.98844\times 10^{30}\,\mathrm {kg} }$

and the star mass M expressed relative to the solar mass,

${\displaystyle M_{x}={\frac {M}{M_{\odot }}},}$

then the relativistic fractional binding energy of a neutron star is

${\displaystyle BE={\frac {885.975\,M_{x}}{R-738.313\,M_{x}}}}$

• Stress–energy tensor
• Stress–energy–momentum pseudotensor
• Nordtvedt effect