How much would you have to shrink a baseball to a black hole?

The Schwarzchild radius is . . . R = 2GM/c²  .

Just like you, I'm not completely sure what that means.  But it DOES use the mass of a black hole to calculate a radius associated with it, so with 15 points at stake, it seems like a good-enough formula to use for an answer.

Before I proceed, I really should ask you whether you're talking a softball or a hardball, but again . . . . .

So R = 2GM/c²

G = the  gravitational constant = 6.67 x 10⁻¹¹ N-m²/kg²

M = mass of the baseball = 145 grams = 0.145 kg

c = speed of light = 3 x 10⁸ m/s

R = 2 (6.67 x 10⁻¹¹ m³/kg-s²) (0.145 kg) / (3 x 10⁸ m/s)²

R = (2 x 6.67 x 10⁻¹¹ x 0.145 / 9 x 10¹⁶) (m³-kg-s² / kg-s²-m²)

R = ( 1.9343 x 10⁻¹¹ / 9 x 10¹⁶) (m³-kg-s²/m²-kg-s²)

R = (0.2149 x 10⁻²⁷) meter

R = 2.149 x 10⁻²⁸ meter

For reference: The radius of a Hydrogen atom is 1.2 x 10⁻¹⁰ meter.

So in order to make a black hole out of a baseball, you have to crunch the baseball down to around  0.000000000000000001791 the size of a Hydrogen atom.

Would that be a problem for you ?

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