 Blackholes2 Forum Message Forums: Atm · Astrophotography · Blackholes · Blackholes2 · CCD · Celestron · Domes · Education Eyepieces · Meade · Misc. · God and Science · SETI · Software · UFO · XEphem Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place...The Space and Astronomy Agora Re: Imploding Forum List | Follow Ups | Post Message | Back to Thread Topics | In Response ToPosted by Kev on November 5, 1999 16:15:04 UTC

You're right in stating that a black hole does have a gravitational force extending beyond it's horizon and the closer an object is to the black hole, the stronger the force exerted upon an object. The black hole will have an impact upon particles up to a certain radius beyond the event horizon. Without overly complicating the problem, let me run through a few numbers though.

Let's assume a Schwarzschild type black hole with the classic equation for event horizon radius:

R = 2Gm/c^2.

Lets say a baseball has a radius of 6 cm (I don't have one handy to measure, but this is pretty close), so the mass of this black hole is roughly 4e25 kg. Speculation on the core of Jupiter is that it consists of a rocky mass between 10 and 15 earth masses, or 6e25 kg (for the 10 earth mass case). The problem arises however in defining what kind of force needs to be applied in order for particles distant from the black holes event horizon to overcome 'bonding' forces.

Because forces exerted between masses is a function of each individual mass (and radius) once we get beyond space time curvature of the black hole using classical Newtonian physics, we can say the density of the core has increased dramatically, but is it sufficient to destabilize it? If so, this would allow the black hole to consume it, otherwise, a static condition arises where the black hole simply exists in the center of the planet and has 'consumed' what it can.

Curiously, because we're at the center of the planet, the planet exerts it's own force as a summation of incremental forces from each individual mass element surrounding the core. The result of this is that the force of gravity at the center would be zero for a perfectly symmetric planet. However, now we've introduced a black hole that increases the mass at the core and destabilizes the original static conditions.

Let me make some simplifying assumptions here. Lets look at a 1 kg rock, 1000 km (I'm speculating this is the radius of the rock core)from the center of the black hole. The rock is part of a interior surface, i.e. the black hole has hollowed out a 1 km sphere about itself, but the rock is bonded to adjacent rocks. What force would the black hole exert on the rock? Using

Fgravity = G M m/r^2,

where G = 6.672 × 10-11 m^3 /(kg sec^2), the force applied due to the black hole is 5,338 N (I've assumed the black hole has gobbled up the rest of the interior and reached a mass of 8e25 kg) or roughly 545 times the force exerted by gravity on a similar rock on the earth. Is this force sufficient to dislodge the rock from it's surroundings? I really can't answer that because it depends on the density of material, the type of material etc. Also, I've ignored the force applied the the planet itself (since we're offcenter, the large mass of the planet is tugging on the rock which will aid the black hole consuming the rock) but for the sake of simplicity, lets add another 5400 N. Considering that the rock element has experienced tremendous force already in it's formation, it is possible that the the additional contribution from the black hole is insufficient to dislodge it.

So, in the end, after all that rambling, I can't truly say if the black hole reaches a static condition or it does engulf the planet. It depends on too many factors.  