Black Holes

[Otto]

Riddle me this:  

 If a large star can create a Black Hole when it collapse on itself, after burning all it's fissionable fuel, why can't that same star, in it's normal state with the same mass, accelerate matter it attracts into it's field of gravity, past the Speed of Light and be constantly creating small Black Holes all around it?

[GRUNHERZ]

My guess is because its not as dense.

[ra]

If I understand the question, I've always wondered the same thing.  How come the gravitational pull increases as the star implodes?  Doesn't the mass remain the same?  Is density part of the gravitational equation?  I thought it was just a function mass and distance.

ra

[Holden McGroin]

Fusion (not fission) lessens the mass of the star as energy is emitted because e=mc2. So mass is emmitted in its equivalent energy form.

Weak and strong nuclear forces and electromagnetic radiation are sufficient to counteract the proportionally very weak gravitational forces until the star burns down.

Orbital material is brought into the star and mass increases due to the collision but does it increase faster than the loss of star mass due to the nuclear reaction?  Probably not, as the mass of our sun is like 99% of the mass of our solar system.

The solar impact of Jupiter would be a minor occurance in the life of our sun.

It's not that gravity increases to cause a black hole, it's that the forces associated with the nuclear fusion that lessen due to the running out of fuel.

and ain't nuthin faster than light.

[GRUNHERZ]

Imagine if you have a thin flexible rubber membrane stretched flat and supended in the air,  supported all around its sides.

And you have two spheres.

One has a diamter of three feet.

The other has a diameter of 1/4 inch.

Both spheres weigh 100 lbs.

What will happend when you put the 3 foot sphere on the rubber?  How steep of a depression will it make? How much pressure will it apply to its contact areas on the rubber?

Now do the same thing with the 1/4 inch sphere.  How steep will depression be? How much pressure on the contact area between  the rubber and the sphere.

Remember doth spheres weigh the same, one is just much denser.

The big sphere is a star.
The small one is a black hole.
The rubber membrane is spacetime.

[hawker238]

After the sun has expended so much energy, the electron orbits are no longer as excited and less ionization occurs.  Thusly, the size of the star shrinks, but not its mass.  This causes a higher density, capable of imploding in on itself because its in a tighter area.

Just a shot in the dark.

[AKS\/\/ulfe]

I thought it was because of the rapid rate of collapse when the star runs out of fuel capable of sustaining outer pressure against its own weight to maintain itself. This is why Giant stars produce black holes or neutron stars, while smaller stars like our sun produce brown stars or white dwarfs... the latter doesn't have the mass to collapse at a rate where it falls into itself (black hole) or into a super dense structure (neutron star).
-SW

[lasersailor184]

Gravity is solely reliant on mass and distance from the center of the mass.


The imploded star may seem to have more gravity, but get the same distance away from each's center and the pull will be the same.

[hawker238]

Originally posted by lasersailor184
Gravity is solely reliant on mass and distance from the center of the mass.


The imploded star may seem to have more gravity, but get the same distance away from each's center and the pull will be the same.

Not if you're judging a point within another's radius.

[Otto]

Thanks, many good ideas here.  It wasn't a 'trick' question.

[Octavius]

Both objects have the same gravitational "pull" outside the star's radius in the main sequence state (normal state).  Since the 'normal state' star has the same mass as its final state as a black hole, then anything beyond its 'original radius' behaves as normal.

Now say your normal state star collapses into the black hole, if you were able to go inside the normal's radius (distance from the center to the surface of the star before it collapsed) while in black hole state, then escape speed = 'c' and you're shrecked.    If you were able to hover exactly at the radius, then you'd be squished and torn apart due to tidal forces and event horizon on the inside.  If you orbited the black hole just beyond the normal's radius, then all would be well... You still have a chance to build up enough speed to escape.  

But no, black holes would not 'develop' around the star before it collapsed.

[vorticon]

i personally prefer not to get past the standard rubber membrane simplification of gravity...

[Octavius]

ENTAR TEH WOLRD OF STR1NG THOERY!!11ONEshiftone

[lasersailor184]

Err, yeah, that was assumed.


The imploded star seems to have more pull because the radius gets smaller in the denominator as you go further in, hence the numerator gets larger.  


For the one that hasn't imploded, you can't go inside of it's surface.  So once you hit fat R, that's as close as you get.