View Full Version : Gravitationally bound

2007-Feb-13, 05:48 PM
It is very common to read in astronomy materials that objects are or are not gravitationally bound to one another. For instance, I just read an article that says the Magellanic Clouds appear not to be gravitationally bound to the Milky Way, but are just cruising past at high speed.

What exactly does that mean?

I spent an hour looking for a clear explanation of this on the Internet last night and couldn't find one.

2007-Feb-13, 06:18 PM
I would assume that it must mean, that the object in question has just the right velocity relative to the attractor, that it could escape and would and will in the long run...
I thought the Magellanic Clouds had an appointment with fate and were to be canabalized by the Milky Way any time soon...

Kwalish Kid
2007-Feb-13, 06:21 PM
The quick and dirty answer is that gravitationally bound objects are orbiting each other or a common centre of mass. Whether or not an object will orbit another is determined by their relative momenta, so for the case of the MCs, if they are moving fast enough, they won't feel enough impact from the gravity of the Milky Way (and vice versa) to form an orbiting system.

Jeff Root
2007-Feb-13, 08:20 PM
Objects which are gravitationally bound are held together by gravity.
I am held to the Earth by gravity. An artificial satellite in low Earth
orbit is held in orbit by gravity. Earth's Moon is held in in orbit by
gravity. Earth is held in orbit around the Sun by gravity. The Solar
system is held in orbit around the galaxy by gravity.

The Cassini spacecraft was gravitationally bound to the Earth until
a few minutes after launch, when it reached escaped speed. It was
still gravitationally bound to the Sun. It flew close by Venus, Earth,
and Jupiter, but it was not gravitationally bound to them. When it
reached Saturn it fired its engine to slow it down enough to be
captured by the planet's gravity.

Pioneers 10 and 11 and the two Voyager spacecraft reached escape
speed from the Sun when they passed Jupiter, which gave them a
major boost in speed.

The New Horizons spacecraft now headed for Pluto was given a fast
enough launch that it was no longer gravitationally bound to the Sun
by the time the engines were done firing, a few minutes after launch.
That was the fastest launch ever.

-- Jeff, in Minneapolis

peter eldergill
2007-Feb-13, 08:45 PM
A graviationally bound object will have an elliptical orbit, a closed path

Otherwise it will have a hyperbolic orbit, which is not a closed path

Theoretically there could be circular or parabolic, but that would be a perfect scenario and any disturbance would change it to elliptical or hyperbolic


2007-Feb-13, 09:32 PM
E=KE-PE<0 is bound.

Translation: If the difference of kinetic energy minus potential energy is less than zero, then the system is bound. If the difference is greater than zero then the system is unbound

2007-Feb-13, 09:40 PM
I thought the Magellanic Clouds had an appointment with fate and were to be canabalized by the Milky Way any time soon...

nobody daring to coment on that?

2007-Feb-13, 10:29 PM
I appreciate the answers, especially the distinction between elliptical and hyperbolic orbits, which I "get." I suppose it must be possible for perturbations from third bodies to cause any two objects to move from a bound to an unbound state, or vice versa.

These answers here, I suppose, are accurate for low-velocity nonrelativistic systems. From a larger perspective, ideas like Mach's Principle, which is said to produce inertia, must mean that, in some ultimate sense, all matter in the universe is gravitationally bound.

I'm just a middle-aged lawyer, not a scientist, so I'm sure the above is fuzzy. But after all, this site IS about "bad astronomy" <grin>. I've been lurking here for years, but I guess my posting rate is about one a year.

2007-Feb-14, 08:32 AM
I suppose it must be possible for perturbations from third bodies to cause any two objects to move from a bound to an unbound state, or vice versa.

This is known as the Hill Sphere. For example, you could compute the circular orbital velocity for an object 2 million kilometers from Earth. Place a real object there and it should have 0 eccentricity. Except at that distance the Sun's gravity will immediately strip the object from Earth.

The maximum value that an object can orbit the Earth is known as the Earth's Hill Sphere. It is about 1.5 million kilometers. But even at the Hill Sphere border, conditions must be just perfect. Only an object in a retrograde orbit will survive for more than a single orbit. And this ignores the Moon's influence. There are no stable prograde orbits around Earth external to the Moon's orbit due to the Moon's pertabutions.

Here's a link to a Hill Sphere calculator. Enter 1 Earth mass for little m, 1 Solar mass for big M, and 1 AU for semi-major axis (a). You should get a value of about 1.5 million km, or .01 AU.


Kaptain K
2007-Feb-14, 09:10 AM
nobody daring to coment on that?
I too, thought that the Magllanic Clouds were gravitationally bound to the Milky Way, as did most of the astronomical community, but new data seems to show that the clouds are moving too fast to be bound to the MW:


However, this is just a press release. I don't know if the accuracy of positional measurements made with Hubble would allow a definitive calculation of transverse velocity of objects that are that for away with only a two year baseline. I would think that the error bars would be huge.

2007-Feb-14, 08:42 PM
thanks Kaptain....
( beytheby : best observatory = internet )
temporarily thawed up