Does a Black Hole Have a Greater Gravitational Field Than Its Original Star?

[JV]

When a large star goes supernova, it will become a black hole. Is the gravitational field of the black hole larger than that of the original star? At least at a distance, not close to the singularity.
It seems to me that the mass of the black hole is smaller than that of the star, because a lot of matter was ejected with the nova. But then again you always read about the huge gravitational pull of a black hole.

 

[selfAdjoint]

The gravity is as you conjectured. All the mass and energy has to be accounted for, and only what is left after ejection collapses to the black hole. Nothing magical happens to increase that gravity. The singularity doesn't have infinite gravity but "infinite density".

 

[Phobos]

Welcome to Physics Forums, JV!

The mass of the black hole is less than the mass of the original star, so it's overall gravity force (at a distance) is less. HOWEVER! With a black hole, its mass is compressed to a point rather than being spread out over hundreds of thousands of miles. Now recall that gravitational force is based on distance. If you were standing near the surface of the sun (with lots of sunscreen), the gravity from the sun's mass would be spread out...you would still be hundreds of thousands of miles away from the sun's center of mass. But with a black hole, you would be immediately next to all that mass.

In short, the black hole's gravity is more intense (close up).

If our sun were to magically & instantly convert into a black hole, the orbit of the EArth would be largely unaffected because we're so far away. But up-close, things would be very different.

 

[Chronos]

If a star suddenly collapsed to form a black hole, gravitationally bound objects would not get closer to the resulting black hole. They could be pushed farther out, even destroyed, but, never pulled closer. A black hole at a distance behaves the same as ordinary matter.

 

[JV]

ok, thanx. What made me doubt are the animations, you sometimes see, of a binary system where one of the stars has imploded to a black hole. The black hole is "eating" the other star. I presume that the original star (that was the black hole) did not do that. So how is it possible that the black hole is consuming the other star, when the its gravity did not increase?
e.g.:
Binary black hole 1
or
binary black hole 2

 

[Garth]

. So how is it possible that the black hole is consuming the other star, when the its gravity did not increase?

Hi JV! The black hole did not increase, its companion did!

A Black Hole, super-nova remnant can sit there quite happily orbiting with a companion for billions of years. However when the companion eventually runs out of fuel in its nuclear core that core collapses and gets hotter, this causes the rest of the star, its envelope, to expand and get cooler, so the core shrinks and the envelope expands and cools into a Red Giant. [Notice how red the companion is in your first link]

Red Giants are very large (perhaps the size of the Earth's orbit or larger), cool stars, with a small intensely hot core deep inside them.

When the companion goes 'red giant' its envelope may spill out of its Roche* lobe into that of the comanion Black Hole. When that happens material in the red giant's envelope transfers onto the Black Hole's event horizon first forming an accretion disc around the black hole where it gets very hot. This we can see as light/ultra violet/x-ray emission from the disc just before it spirals through the event horizon and its lost from sight. Jets squirting out from the polar positions on the event horizon often form as well - nobody knows quite how - and all these features are illustrated on those pictures you linked us to.
- Garth

* "Roche lobe" around two stars is a 'figure of eight' surface of equal potential with the two stars at the centre of each lobe, inside each of which the central star has the stronger gravitational attraction.
Material that expands out of one lobe into the other then gets attracted into that other lobe