Mark Oliver wrote:

The current accepted theory of a black hole is NOT consistent with current
accepted physics. Here are the problems;

1) All accepted calculations of gravitational pull are based upon mass and
the distance between two objects. It is not based upon the density or
dimensions of the same "singular" mass. Thus, when a quasar collapses it is
still the same amount of mass, only denser (smaller dimensions). Then why
would we assume that its gravitational pull will change, when its mass does
not change? Here is proof, measure the mass of 100 steel ball bearings on a
scale, then melt all of the steel bearings to a single mass from the 100
smaller bearings. The mass will not change, thus its gravitational pull
does not change either.

From a distance, for any practical purpose, this is correct. But from
up close the two situations will be different. Assuming the
ball-bearings are packed in a spheroidal clump in free space, a test
particle near the surface of the clump will experience a somewhat
weaker pull than when similarly situated WRT a solid cannonball of
the same mass, because it'll be further from the centre of gravity in
the former case. (Moreover a particle located in one of the gaps
between the ball-bearings, and near the centre of the clump, will
experience next to no net gravitational force.) The more concentrated
is a mass, the stronger the gravity at its surface will be. A small
illustration from our solar system: the Jovian satellite Callisto has
a mass about 50% greater than our Moon's, but its surface gravity is
about 25% less, because it's only a little over half as dense.

Likewise, from a great distance a black hole's gravitation will be
pretty much indistinguishable from that of a star of equal mass.
However, from nearby, because of the black hole's small radius the
gravitational gradient (or space-curvature) is extremely great,
making for a qualitative difference in its behaviour -- the formation
of an event horizon -- as compared to ordinary matter.

--
Odysseus