What does the mass of a black hole need to be in order for its mass divided by its volume to be equal to the density of water?

Answer 1

Because radius of a black hole varies directly with mass (twice the mass means twice the radius) per r = 2Gm/(c^2), and the volume varies with the cube of the radius (per v = (4/3)*pi*(r^3) ...) (twice the radius is eight times the volume) there will indeed be a theoretical black hole whose density is equal to water (one gram per cubic centimeter, or one thousand kilos per cubic meter). Since density is mass divided by volume, this becomes an exercise in solving mass for that density.

I calculate a mass of about 3.03 x 10^38 kg, or around a hundred and fifty million times the Solar mass.. which would give the black hole a radius of roughly(!) 450 million km, or about 3 times the distance of the Earth from the Sun. Can somebody else check my calculations please :)

Answer 2

For a black hole to have the same density as water, its mass needs to be about the same as that of the Earth, which is approximately 5.972 x 10^24 kg. The density of water is 1000 kg/m^3, so the volume of the black hole would be the mass divided by the density of water, which equals 5.972 x 10^21 m^3.