Re: Force of Gravity of a 100-solar Mass Lump of Gold


Mark Martin wrote:
> Michael Ejercito wrote:
> > What would be the weight of a 160 kg person standing on the surface
> > a lump of 24-karat gold with a mass one hundred times that of Earth's
> > sun?
>
> The Sun has a mass of 1.99 x 10^30 kg, so the gold lump will be 1.99
> x 10^32 kg. Gold has a mass density of 19,300 kg/m^3. Assuming the gold
> must tend to be spherical, the radius of a sphere is given by
>
> R = [3V/4pi]^1/3
>
> This gives the gold a radius of about 1.3 billion meters. The
> Swarzschild radius for that much mass is a lot less than this, so it'll
> collapse into a black hole. A person, falling into the hole, will
> weight nothin. Instead, it will be stretched to destruction by tidal
> forces.
>
> -Mark Martin

COMMENT:

The Swartzschild radius for 2e32 kg is only GM/c^2 = 1.5e5 m, so the
gold ball of 1.3e9 m is going to have to collapse by compression down
to that size (a factor of almost 9000 in radius), before it disappears
into a black hole. Which it will do, since nothing is stopping it.

How long for that to happen? Not milliseconds. Orbital period for an
object orbiting 1.3e9 m from a mass of 2e32 kg = 2pi srqr(R^3/GM) =
2550 sec = 42.5 min. The Newtonian radial infall time from that orbit
is 1/sqrt(32) of that, which is 7.5 minutes. That's how long until the
gold collapses to a black hole. The actual time is a tad longer due to
the fact that at the end, the collapse rate is limited to c, but (as
you can guess) that doesn't produce significantly longer times.

So, 7.5 minutes. Just one more example of a highschool physics
providing the correct answer to a problem that could never happen in
reality.

SBH

.

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