# What is the mathematical condition for the statement: "gravitationally bound"?

When talking about galaxy clusters there is the frequently used phrase "gravitationally bound", f.e. we are gravitationally bound to our neighbor galaxy the Andromeda Galaxy. But how is this statement mathematically defined since gravitation has an infinite reach, we (the Milkyway) have a tiny gravitational influence on any other galaxy in the universe.

It means the total energy (kinetic + gravitational potential) is negative. This assumes the convention that gravitational potential energy approaches zero as the distance tends to infinity, and is negative for all finite distances.

So if I imagine an airless planet the same size and mass as the Earth, A ball that is moving at 10 km/s (away from the planet, at the planet's surface) won't have enough kinetic energy to keep moving away from the planet forever. It will move away and slow down as kinetic energy is converted to potential and eventually halt and fall back. Its total energy is negative, and it is gravitationally bound to the planet. On the other hand, a ball moving at 12 km/s will continue to move away forever, gravity will slow it down but never completely stop it. The ball moving at 12km/s has positive total energy.

You can do the same thing, but on a much larger scale with galaxies. The Milky-way Andromeda system has negative total energy, and so is gravitationally bound. But more distant galaxies have positive (or very close to zero) total energy.

• As a footnote, when there are two distinct, compact objects bound can be defined, but it gets messy for 3-body and n-body situation; sometimes several objects have enough energy to promote one object to unbound, it's just a question of if/when it happens.
– uhoh
Commented Nov 13, 2022 at 15:20
• Can this actually be a full definition for 3+ objects, for reasons given in previous comment? If so, how? If not, what is? Commented Nov 13, 2022 at 23:42
• @Stilez Of course. Just because something is gravitationally bound now doesn't mean that it will always be gravitationally bound. A rocket on a launchpad is gravitationally bound to Earth - but a few minutes later (parts of) it very well might not be. Gravitationally bound doesn't imply stable - indeed, mass and energy is continuously lost from gravitationally bound objects. We use what is useful, not what is "technically true" or "absolutely true"; nature doesn't care about our ability to comprehend reality. "Gravitationally bound" is a tool, and it works pretty well. Commented Nov 14, 2022 at 9:28
• @eagle275 All objects free-falling in the vicinity of a body to which they are gravitationally bound and which is spherically symmetrical follow an approximate (because of other bodies) elliptic trajectory (the high-school parabola is of course only a convenient approximation). Sometimes the ellipse is degenerated to a straight line or a circle. And whenever you throw something from the surface of a planet the ellipse will intersect the surface at that point even if the speed exceeds the first cosmological speed -- they will not magically enter orbit. Commented Nov 14, 2022 at 10:49
• @eagle275 The first cosmic velocity 7.9 km/s is the minimum orbit speed. It's the speed at the Earth's equatorial radius (including the Earth's rotation speed) for a circular orbit at ground level for a circular orbit (ignoring air resistance). But if we multiply that speed by $\sqrt2$, giving 11.18 km/s, we get the escape velocity at ground level. Those speeds can be derived from en.wikipedia.org/wiki/Specific_orbital_energy / en.wikipedia.org/wiki/Vis-viva_equation Commented Nov 14, 2022 at 20:12