<p>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.</p>
<p>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.</p>
<p>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.</p>

CC BY-SA 4.0
2022-11-13T15:20:16.317Z
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
CC BY-SA 4.0
2022-11-13T23:42:12.990Z
Can this actually be a full definition for 3+ objects, for reasons given in previous comment? If so, how? If not, what is?
Stilez
CC BY-SA 4.0
2022-11-14T09:28:23.173Z
@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.
Luaan
CC BY-SA 4.0
2022-11-14T09:32:57.447Z
@JamesK Your hypothetical ball at 10 km/s is already exceeding the so called "first cosmological speed" - Which for earth is ~ 7.8 km/s - at this speed the ball will already fly around your athmosphere-less planet on an eliptical path, with the planet in one of the focal points - and without air resistance there is no drag
eagle275
CC BY-SA 4.0
2022-11-14T10:49:52.557Z
@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.
Peter - Reinstate Monica
CC BY-SA 4.0
2022-11-14T12:53:56.207Z
@Stilez if we talk about point-like objects, 3 or more objects are almost impossible to be gravitationally bound. By getting 2 of them closer enough one can always get enough energy to expel the third one. If the objects have finite sizes - 2 of them may collide before they set the third one free.
fraxinus
CC BY-SA 4.0
2022-11-14T20:12:21.483Z
@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  https://en.wikipedia.org/wiki/Specific_orbital_energy / https://en.wikipedia.org/wiki/Vis-viva_equation
PM 2Ring
CC BY-SA 4.0
2022-11-15T06:11:30.263Z
@Luaan: Say you've got a "bound" multi-body system with negative total energy. Then interactions between the bodies cause one of the bodies to escape. The total energy is still negative, so by this definition, the system would still be bound (including the body that escaped). That seems like a failure of the definition.
user2357112
CC BY-SA 4.0
2022-11-15T10:24:16.590Z
@user2357112 Sure, but that's because you consider "a collection of X stars bound to a collection of Y stars" the system in question. That's not how it's actually used, for the exact obvious reason you (and I) have pointed out. When a star gets ejected from the Andromeda galaxy, it may or may not be gravitationally bound to the Milky Way, but it certainly has no impact on whether the Milky Way and Andromeda are gravitationally bound. It just means Andromeda lost a star (just like the rocket in my example). Heck, we don't even consider the individual stars in the first place - just the CoM.
Luaan
CC BY-SA 4.0
2022-11-13T13:35:24.810Z
2022-11-13T13:35:24.810Z
James K
<p>When talking about galaxy clusters there is the frequently used phrase &quot;gravitationally bound&quot;, 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.</p>

<observational-astronomy><galaxy><gravity>
What is the mathematical condition for the statement: "gravitationally bound"?
trynerror
# What is the mathematical condition for the statement: "gravitationally bound"?

(Asked by: username_0 on 2022-11-13)

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.

**Tags:** observational-astronomy, galaxy, gravity


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## 1 answers

### Answer by username_1 on 2022-11-13. Score: 21 | Accepted answer

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.

> **Comments:**
> **username_2**: 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.  
> **username_3**: Can this actually be a full definition for 3+ objects, for reasons given in previous comment? If so, how? If not, what is?  
> **username_4**: @username_3 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.  
> **username_5**: @JamesK Your hypothetical ball at 10 km/s is already exceeding the so called "first cosmological speed" - Which for earth is ~ 7.8 km/s - at this speed the ball will already fly around your athmosphere-less planet on an eliptical path, with the planet in one of the focal points - and without air resistance there is no drag  
> **username_6**: @username_5 *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.  
> **username_7**: @username_3 if we talk about point-like objects, 3 or more objects are almost impossible to be gravitationally bound. By getting 2 of them closer enough one can always get enough energy to expel the third one. If the objects have finite sizes - 2 of them may collide before they set the third one free.  
> **username_8**: @username_5 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  https://en.wikipedia.org/wiki/Specific_orbital_energy / https://en.wikipedia.org/wiki/Vis-viva_equation  
> **username_9**: @username_4: Say you've got a "bound" multi-body system with negative total energy. Then interactions between the bodies cause one of the bodies to escape. The total energy is still negative, so by this definition, the system would still be bound (including the body that escaped). That seems like a failure of the definition.  
> **username_4**: @username_9 Sure, but that's because you consider "a collection of X stars bound to a collection of Y stars" the system in question. That's not how it's actually used, for the exact obvious reason you (and I) have pointed out. When a star gets ejected from the Andromeda galaxy, it may or may not be gravitationally bound to the Milky Way, but it certainly has no impact on whether the Milky Way and Andromeda are gravitationally bound. It just means Andromeda lost a star (just like the rocket in my example). Heck, we don't even consider the individual stars in the first place - just the CoM.  

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