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## Main Question or Discussion Point

Yesterday I had a bizarre idea. I supposed that a black hole (which I assumed to be a sufficiently dense sphere) became our sun. I know this is completely wrong, but please humor me and see what I've done.

Suppose we have a sphere sufficiently dense, so that the Schwarzchild radius, [itex]r_s[/itex], is greater than the radius of the sphere, [itex]r[/itex]. Since the escape velocity is greater than c, let's assume this object will continue to 'trap' masses which enter it and grow until [itex]r=r_s[/itex]. Then when [itex]r \geq r_s[/itex], lets assume photons and such will no longer be trapped, and the sphere will appear as a fireball, somewhat like the sun. So now lets ask the question, what is the mass of the sphere at this point?

Lets say this sphere is as dense as a nucleus ([itex]\rho \approx 2.3x10^{17}[/itex] kg/m

The Schwarzchild radius is given by [itex]r_s=\frac{2Gm}{c^2}[/itex]. However we can say [itex]m=\rho V[/itex], and since we have assumed a sphere, [itex]V=4/3\pi r^3[/itex]. Since we are interested in the case [itex]r=r_s[/itex], this yields

[itex]r=\sqrt{\frac{3c^2}{8\pi \rho G}}[/itex], and then the mass of this sphere is [itex]m=4/3\rho \pi r^3[/itex]. (An interesting note here is that as [itex]\rho[/itex] increases, r decreases.)

Running the numbers for [itex]\rho = 2.3x10^{17}[/itex] kg/m

Using [itex]\rho = 8x10^{17}[/itex] kg/m

I was shocked yesterday to find that these numbers were in the ball park of the solar mass. What gives? How could such a rough calculation (not to mention based on a completely wrong foundation) give a relatively accurate answer? What are your thoughts?

Could a star form in such a way?

Thanks!

Suppose we have a sphere sufficiently dense, so that the Schwarzchild radius, [itex]r_s[/itex], is greater than the radius of the sphere, [itex]r[/itex]. Since the escape velocity is greater than c, let's assume this object will continue to 'trap' masses which enter it and grow until [itex]r=r_s[/itex]. Then when [itex]r \geq r_s[/itex], lets assume photons and such will no longer be trapped, and the sphere will appear as a fireball, somewhat like the sun. So now lets ask the question, what is the mass of the sphere at this point?

Lets say this sphere is as dense as a nucleus ([itex]\rho \approx 2.3x10^{17}[/itex] kg/m

^{3}), or the core of a neturon star, roughly ([itex]\rho \approx 8x10^{17}[/itex] kg/m^{3}). Let's further assume that [itex]\rho[/itex] is constant.The Schwarzchild radius is given by [itex]r_s=\frac{2Gm}{c^2}[/itex]. However we can say [itex]m=\rho V[/itex], and since we have assumed a sphere, [itex]V=4/3\pi r^3[/itex]. Since we are interested in the case [itex]r=r_s[/itex], this yields

[itex]r=\sqrt{\frac{3c^2}{8\pi \rho G}}[/itex], and then the mass of this sphere is [itex]m=4/3\rho \pi r^3[/itex]. (An interesting note here is that as [itex]\rho[/itex] increases, r decreases.)

Running the numbers for [itex]\rho = 2.3x10^{17}[/itex] kg/m

^{3}yields [itex]m =1.78x10^{31}[/itex] kg (roughly 8.9x the solar mass),Using [itex]\rho = 8x10^{17}[/itex] kg/m

^{3}yields [itex]m =9.55x10^{30}[/itex] kg (roughly 4.8x the solar mass).I was shocked yesterday to find that these numbers were in the ball park of the solar mass. What gives? How could such a rough calculation (not to mention based on a completely wrong foundation) give a relatively accurate answer? What are your thoughts?

Could a star form in such a way?

Thanks!