Right. Let's get this over with. You're here for an explanation of the universe's exit fees. Don't look so confused; it's the only metaphor that fits.

Minimum energy to remove a system from a gravitationally bound state

Before we begin, a simple observation: Galaxy clusters are the largest known gravitationally bound structures in the universe. They are the grand, sprawling cities of the cosmos, held together by a force that couldn't care less about your feelings. This isn't just trivia; it's the scale we're dealing with. Try to keep up.

The gravitational binding energy of a system is, in the simplest terms you might comprehend, the absolute minimum energy you must inject into it to make it fall apart. It’s the price of cosmic divorce. A system that is gravitationally trapped in a bound state has a lower, which means more negative, gravitational potential energy than the sum of its individual components if you were to scatter them across the void. This isn't a coincidence; it's the universe's inherent laziness. Everything settles in accordance with the minimum total potential energy principle, finding the lowest energy state possible to avoid doing any extra work. It's a principle you and your couch are intimately familiar with.

This concept, however, gets complicated depending on whose universe you're living in. In the clockwork cosmos of Newtonian gravity, the binding energy is a simple, almost quaint, calculation: the linear sum of all the little tugs and pulls between every pair of particles in the system. It’s addition. You can handle that.

But then Albert Einstein had to ruin the party with his theory of General Relativity. In his universe, which is inconveniently the one we actually inhabit, the Newtonian approach is merely an approximation that works only when the gravitational fields are pathetically weak. When gravity gets serious, the binding energy becomes a nonlinear property of the entire system. You can't just add up the parts anymore because the energy is woven into the fabric of spacetime itself. In this far more complex reality, the binding energy is the (negative) difference between the system's total mass as seen from a great distance—its ADM mass—and the sum of the rest energies of all its constituent atoms and other elementary particles if you were to painstakingly disassemble it. It's the mass that goes "missing" just by the act of being held together.

For a perfectly spherical body of uniform density—a charming fiction we'll entertain for a moment—the gravitational binding energy U under Newtonian rules is given by this neat little formula:

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

Here, G is the gravitational constant you can't escape, M is the mass of your idealized sphere, and R is its radius.

Let's apply this to the Earth. Assuming it's a uniform sphere, which it absolutely is not, but let's pretend for the sake of an order-of-magnitude estimate, we have a mass M of 5.97×10²⁴ kg and a radius r of 6.37×10⁶ m. Plug those into the formula, and you get a binding energy U of 2.24×10³² J. To put that number into a context your brain might process, that's roughly the total energy the Sun spews out in a week. It's 37.5 MJ/kg, which is 60% of the absolute value of the potential energy per kilogram at the surface. That's the energy required to obliterate the planet. Don't get any ideas.

Of course, reality is more nuanced. The Earth's density changes with depth, a fact we know from tracking seismic travel times (see the Adams–Williamson equation). This more accurate, layered model is called the Preliminary Reference Earth Model (PREM). Using this grown-up model, the actual gravitational binding energy of Earth can be calculated numerically to be U = 2.49×10³² J. A bit higher. Details matter.

This isn't just about planets. According to the virial theorem, the gravitational binding energy of a star must be about twice its internal thermal energy for it to maintain hydrostatic equilibrium. It’s a cosmic balancing act between gravity trying to crush it and thermal pressure trying to blow it apart. When the gas in a star becomes highly relativistic, this delicate balance shatters. The binding energy required for equilibrium approaches zero, and the star becomes terrifyingly unstable. For a high-mass star, this instability, driven by intense radiation pressure, can trigger a supernova. For a dense neutron star, it can lead to collapse into a black hole. The universe does not do half-measures.

Derivation within Newtonian gravity for a uniform sphere

If you insist on seeing the math, fine. To find the gravitational binding energy of a sphere with radius R, we have to imagine pulling it apart, piece by piece. We do this by moving successive spherical shells to infinity, starting with the outermost one, and calculating the total energy required for this thankless task.

Assuming a constant density ρ, the mass of a shell and the mass of the sphere interior to it are:

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

and

$m_{\mathrm{interior}} = \frac{4}{3}\pi r^3 \rho$

The energy needed to remove one shell is the negative of its gravitational potential energy. It’s the work you have to do against gravity's grip.

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

Now, we integrate this process over all the shells, from the center (r=0) to the surface (r=R):

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

Since density ρ for our uniform object is just its total mass divided by its total volume, we have:

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

Finally, we substitute this back into our result, and after the algebraic dust settles, we arrive at the formula I gave you earlier.

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

There. The gravitational binding energy.

Gravitational binding energy

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

Negative mass component

Now we wade into the murky waters where simple models collide with complex reality. You'll notice this section comes with a warning label. Pay attention. It's there for a reason. There appears to be a "serious conceptual inconsistency" between the Newtonian formula we just derived and the relativistic concept of a Schwarzschild radius. The suggestion to delete the section is tempting, but where's the fun in that?

Consider two bodies. When they are close together (at a distance R), they exert a slightly smaller gravitational force on a third body than they would if they were far apart. This deficit can be conceptualized as a negative mass component of the system. For our uniformly spherical solutions, this "binding mass" is:

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

Let's use Earth again. The fact that it is a gravitationally-bound sphere of its current size costs it 2.49421×10¹⁵ kg of mass. That's about a quarter of the mass of Phobos. If you were to take the Earth apart, atom by atom, and spread them across an infinite volume, the total collection would weigh its current mass plus that 2.49421×10¹⁵ kg. This is the practical consequence of mass-energy equivalence, measured in kilograms instead of Joules.

You might wonder if this negative component could ever overwhelm the positive mass of the system. It can't, at least not in any way you could observe. For the negative binding energy to be greater than the system's mass, its radius would have to be smaller than:

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

This is less than $\textstyle{\frac{3}{10}}$ of its Schwarzschild radius:

$R \le \frac{3}{10} r_{\mathrm{s}}$

An object that small would already be well inside its own event horizon, its secrets forever hidden from any external observer. This is a Newtonian approximation straining at its leash, trying to describe a situation that properly belongs to General Relativity, where other factors must be considered. It’s a mathematical curiosity, a warning sign that your map of reality is incomplete.

Non-uniform spheres

As I've mentioned, planets and stars are not uniform. They possess radial density gradients, from their relatively low-density surfaces to their crushingly dense cores. Objects made of degenerate matter, like white dwarfs and neutron star pulsars, not only have these gradients but also require relativistic corrections to be described accurately.

The relativistic equations of state for neutron stars provide a graph of radius versus mass for various models. The most likely radius for a given neutron star mass is typically bracketed by models like AP4 (yielding the smallest radius) and MS2 (the largest). The binding energy, BE, is expressed as the ratio of the gravitational binding energy mass equivalent to the observed gravitational mass M of a neutron star with radius R:

$BE = \frac{0.60\,\beta}{1-\frac{\beta}{2}}$

where

$\beta = \frac{GM}{Rc^2}.$

Given the current, non-negotiable values for the universe's constants:

• $G = 6.6743 \times 10^{-11}\,\mathrm{m^3 \cdot kg^{-1} \cdot s^{-2}}$
• $c^2 = 8.98755 \times 10^{16}\,\mathrm{m^2 \cdot s^{-2}}$
• $M_{\odot} = 1.98844 \times 10^{30}\,\mathrm{kg}$

and expressing the star's mass M relative to the solar mass, $M_x = \frac{M}{M_{\odot}}$, the relativistic fractional binding energy of a neutron star is:

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

This is the kind of formula you end up with when you stop pretending things are simple.

See also

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