Black Holes: why is it that...

[illusion88]

Now why is it that, the larger the hole (physical size), the less gravitational pull? Anyone have any clue?

 

[A5]

Originally posted by: illusion88
Now why is it that, the larger the hole (physical size), the less gravitational pull? Anyone have any clue?

Hm...where'd you find this information? I've never heard this before and it sounds interesting.

 

[jeremy806]

--deleted--

 

[heliomphalodon]

Uh, the more massive the black hole, the larger the area of its event horizon. I don't know what you mean by your statement.

 

[illusion88]

Originally posted by: heliomphalodon
Uh, the more massive the black hole, the larger the area of its event horizon. I don't know what you mean by your statement.

all right hang on... I read it in "Black Holes and Time Warps" By Kip Thoran. Ill go check out the book from the library and look it up again.

 

[Shalmanese]

Hmm, the gravitational pull at the event horizon will ALWAYS be the same, the event horizon is DEFINED by the distance away from the singularity where the gravitational pull exactly equals the speed of light.

The larger the black hole, the less the average density is and also, the less difference there is between your head and you foot, I forgot what the term is. With a small balck hole, you would be ripped to shreds trying to enter since thr gravitational differece is so large. With a huge one, you wouldnt be able to feel it.

 

[imgod2u]

I've never heard this myself. Black holes are singularities so its size would be defined not by the physical matter, but by the event horizon. I'm not sure why it'd be less strong (gravitational wise) than a smaller black hole though.

 

[Shalmanese]

The gravitational pull is NOT less, it cant be.

 

[heliomphalodon]

Originally posted by: Shalmanese
The gravitational pull is NOT less, it cant be.

I think what this thread may be trying to discuss is whether the gravitational gradient near the event horizon of a black hole is less for a large black hole than for a small one.

I don't know that this is true -- I think it is, but I haven't checked.

 

[Shalmanese]

Ah, gravitational gradient, thats the word. Thats simple, gravity falls of as the inverse square of distance. Larger black holes mean you are further away from the black hole so the gradient is less.

 

[ant80]

My interpretation of the question is, what is the relationship between the size (radius) of the black hole (or any massive object, for that matter) and its mass? Since the gravitational pull is directly proportional to the mass, we can deduce that part of the equation. I think. I honestly don't think even Stephen Hawkins knows, for that matter.

My guess is it should start out low, go high, and at very high masses, go low. I don't know the how the relationship would be for black holes specefically. Because of the insane distortion to space-time, i don't even know if distance is defined inside the event horizon. Would be interesting to find out. Anyone?

 

[Glytch]

Originally posted by: illusion88
Now why is it that, the larger the hole (physical size), the less gravitational pull? Anyone have any clue?

It was my understanding that a black hole doesn't really have a "physical size". The concept of a black hole is massive amounts of matter in the smallest possible space, most likely Plank's Constant (Plank's length is the minimum length before the general laws of physics no longer apply.).

 

[chsh1ca]

While that's true, you need to separate the Event Horizon (part of the black hole) from the singularity in your conceptualization of it.

A lot of matter can exist outside of the singularity, but within the event horizon. At first this may seem counterintuitive, but remember, a black hole has no 'surface' per se.

The Black Hole is really a combination of the two.

From http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html:

Suppose that, possessing a proper spacecraft and a self-destructive urge, I decide to go black-hole jumping and head for an uncharged, nonrotating ("Schwarzschild") black hole. In this and other kinds of hole, I won't, before I fall in, be able to see anything within the event horizon. But there's nothing locally special about the event horizon; when I get there it won't seem like a particularly unusual place, except that I will see strange optical distortions of the sky around me from all the bending of light that goes on. But as soon as I fall through, I'm doomed. No bungee will help me, since bungees can't keep Sunday from turning into Monday. I have to hit the singularity eventually, and before I get there there will be enormous tidal forces-- forces due to the curvature of spacetime-- which will squash me and my spaceship in some directions and stretch them in another until I look like a piece of spaghetti. At the singularity all of present physics is mute as to what will happen, but I won't care. I'll be dead.

For ordinary black holes of a few solar masses, there are actually large tidal forces well outside the event horizon, so I probably wouldn't even make it into the hole alive and unstretched. For a black hole of 8 solar masses, for instance, the value of r at which tides become fatal is about 400 km, and the Schwarzschild radius is just 24 km. But tidal stresses are proportional to M/r3. Therefore the fatal r goes as the cube root of the mass, whereas the Schwarzschild radius of the black hole is proportional to the mass. So for black holes larger than about 1000 solar masses I could probably fall in alive, and for still larger ones I might not even notice the tidal forces until I'm through the horizon and doomed.

Hopefully that explains it a bit more. If not, then check out the Usenet Physics FAQ, as it has a few more pages on Black Holes.

 

[Smilin]

A possible explanation:

suppose you have a black hole who's event horizon has a radius of 6 feet and you fall into it. The moment your feet touch the event horizon they will be under 4 times as much gravity as your head (assuming you are a 6 foot tall person..and falling in feet first

Now lets say you have one with an event horizon of only 3 feet (radius) and you fall in. The moment you feet touch they will be under 9 times as much gravity as your head.

In either case you are going to look like spaghetti when you hit the horizon.


It all comes from the fact that gravity falls off with the square of distance.

 

[dzt]

gravity is function of mass and distance.
more distance means square less gravity
more mass linearly more gravity
blackholes become smaller as it grows older, it compacting itself and then the size became smaller so the mass density is bigger due to it's own mass or any materials entering it.
that's why smaller black hole have more gravity, since it has more mass and less radius.
am I right ?

 

[Smilin]

Originally posted by: dzt
gravity is function of mass and distance.
more distance means square less gravity
more mass linearly more gravity
blackholes become smaller as it grows older, it compacting itself and then the size became smaller so the mass density is bigger due to it's own mass or any materials entering it.
that's why smaller black hole have more gravity, since it has more mass and less radius.
am I right ?

Hm. Kinda. The event horizon is going to be determined by the mass. More mass, larger event horizon. The density of the black hole can for all practical purposes be considered infinite...somewhere inside that horizon is a singularity. I don't think it's possible to have one black hole be "denser" or "Less dense" than another. As things "fall in" the mass increases and so does the event horizon (not counting black-hole evaporation here). The radius of the horizon will always be proportional to the mass...by definition.

Maybe this is the source of confusion:
Lets say our sun was large enough that it would collapse to a black hole when it died. If you were somehow able to stand on the surface of the sun you would weigh X amount due to gravity. If you were to stay in the same place and let the sun collapse to a black hole under you, you would still weigh X amount due to gravity. The amount of mass under your feet hasn't changed, it's just now in a really, really small area. The kicker is you can move closer to it now than you could before. When you're on the surface of something you can't get any closer...should you go below the surface, gravity will get weaker since there will now be some mass above you as well as below. With a black hole you can get inches from the center of gravity whereas with a star you can only get to within a few thousand miles of the center. The gravity doesn't change a bit, you can just get closer to it.

For the original question in this thread: That statement may apply to stars (Neutron star -vs- blue giant for instance) but not to black holes. The "size" of stars is determined by the amount of matter, and how dense it is. The "size" of a black hole is determined only by the amount of matter.