# Question on Schwarzschild Geometry

• B
please interpret this observation. There is a specific radius through a given equation that always gives the correct mass to any star or planet, as well a density. What is the logical explanation for this?
Mass = (4π/3) x schwarzschild radius of the star x 4π/3 x (726696460.5 cm.) cube.
For example earth mass is equal to:

M= 4.188786667 x 0.8870085587 cm. x 4.188786667 x (726696460.5 cm.) cube = 5.9726 x (10) +27 gram.

(4π/3) = 4.188786667
schwarzschild radius of the earth = 0.8870085587 cm.
the mentioned radius = 726696460.5 cm.
the density of earth is equal to:
earth density ρ= M/V = (4π/3) x 0.8870085587 cm. x (4π/3) x (726696460.5 cm) cube /(4π/3) x (637758965.3 cm) cube = earth density.
the radius of the earth = 637758965.3 cm.

Ibix
2020 Award
The Schwarzschild radius is the mass multiplied by a constant. You've divided by that constant (or multiplied by its reciprocal). Unsurprisingly, the result is the mass.

Edit: Well, numerically at least. Your units are messed up - a radius times a volume is not a mass.

Last edited:
nabil23 and Dale
I think that schwarzschild radius is equal to:
x C/2G) = M/6.733418682 x (10) +27
any mass divided on schwarzschild radius will give this constant:
M/schwarzschild radius = 6.733418682 x (10) +27
this constant is actually equal to :

(4π/3) x (4π/3) x (726696460.5 cm.) cube = 6.733418682 x (10) +27

c= 29979245800 cm. speed of light in centimeter/second.
726696460.5 is the mentioned radius mentioned above.

Ibix