Black Hole math

I found this math involving a black hole and thought that it might be revealing. In order to do this math we must have a few constants:

The mass of the Sun  Ms = 1.99 x 10^33 g  or 1.99 times 10 to the 33 power grams.

Gravitational constant G = 6.673 x 10^-11 m^3/kg -sec^2 or 6.673 times 10 to the -11 power  meters squared over kilograms seconds squared.

Speed of light c = 300,000 km/sec or 30 billion cm/sec

Normalized mass of the Sun ms = G Ms/c^2 = 1.475 km, which can be approximated to 1.5 km.

Density of nuclear matter Pn = 200 x 10^6 metric-tons/cm^3 or  2 x 10^14 gm/cm^3

This is assumed to be the density of a neutron star. The mass of neutron star is:
Mn = (4π/3) Rn^3 Pn = (4π/3) Rn^3 x 2 x 10^14 gm/cm^3 = 8.38 x 10^14 Rn^3 gm/cm^3

Rn is the radius of the neutron star. If this neutron star is the mass of our sun, we set Mn to our sun's value and this gives us:
Rn^3 = (1.99 x 10^33 g) / (8.38 x 10^14 gm/cm^3) = 2.37 x 10^18 cm^3

The cube root of this is Rn = 1.33 x 10^6 cm = 13.3 km  and this is the radius of a neutron star with the mass of our sun. The normalized mass m of this neutron star is the same as our sun 1.5 km. We can calculate the (2m/r) parameter at the surface of any neutron star is proportional to the square of the neutron star's radius and thus:

2m/r = 2(1.5 km) / (13.3 km) = 0.226

The mass is proportional to the cube of the radius , the (2m/r) parameter at the surface of a neutron star is proportional to the square of the its radius -
2m/r = 0.226 [Rn / (13.3 km)]^2

The 2m/r is unity for the neutron star radius given here
Rn =(13.3 km √ [1/0.226] = 28.0 km

The mass of the neutron star relative to our sun's mass is:
Mn/Msun = [Rn/(13.3 km)]^3 = [(28.0 km)/(13.3km)]^3 =9.33

Since this 2m/r is unity or greater at the surface of the star, the star must collapse into a black hole. Thus a neutron star that exceeds 9.33 solar masses will become a black hole. Actually, this ratio is probably even less than 7.6 because of the uncertainty of the density of an atomic nucleus.

The reason I presented this is to demonstrate why you need math to predict astronomical events. This is a simple example.

Thanks for reading.