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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 Schwarzschild radius, r_s, is greater than the radius of the sphere, r. Since the escape velocity is greater than c, let's assume this object will continue to 'trap' masses which enter it and grow until r=r_s. Then when r \geq r_s, let's assume photons and such will no longer be trapped, and the sphere will appear as a fireball, somewhat like the sun. So now let's ask the question, what is the mass of the sphere at this point?
Lets say this sphere is as dense as a nucleus (\rho \approx 2.3x10^{17} kg/m3), or the core of a neturon star, roughly (\rho \approx 8x10^{17} kg/m3). Let's further assume that \rho is constant.
The Schwarzschild radius is given by r_s=\frac{2Gm}{c^2}. However we can say m=\rho V, and since we have assumed a sphere, V=4/3\pi r^3. Since we are interested in the case r=r_s, this yields
r=\sqrt{\frac{3c^2}{8\pi \rho G}}, and then the mass of this sphere is m=4/3\rho \pi r^3. (An interesting note here is that as \rho increases, r decreases.)
Running the numbers for \rho = 2.3x10^{17} kg/m3 yields m =1.78x10^{31} kg (roughly 8.9x the solar mass),
Using \rho = 8x10^{17} kg/m3 yields m =9.55x10^{30} 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!
Suppose we have a sphere sufficiently dense, so that the Schwarzschild radius, r_s, is greater than the radius of the sphere, r. Since the escape velocity is greater than c, let's assume this object will continue to 'trap' masses which enter it and grow until r=r_s. Then when r \geq r_s, let's assume photons and such will no longer be trapped, and the sphere will appear as a fireball, somewhat like the sun. So now let's ask the question, what is the mass of the sphere at this point?
Lets say this sphere is as dense as a nucleus (\rho \approx 2.3x10^{17} kg/m3), or the core of a neturon star, roughly (\rho \approx 8x10^{17} kg/m3). Let's further assume that \rho is constant.
The Schwarzschild radius is given by r_s=\frac{2Gm}{c^2}. However we can say m=\rho V, and since we have assumed a sphere, V=4/3\pi r^3. Since we are interested in the case r=r_s, this yields
r=\sqrt{\frac{3c^2}{8\pi \rho G}}, and then the mass of this sphere is m=4/3\rho \pi r^3. (An interesting note here is that as \rho increases, r decreases.)
Running the numbers for \rho = 2.3x10^{17} kg/m3 yields m =1.78x10^{31} kg (roughly 8.9x the solar mass),
Using \rho = 8x10^{17} kg/m3 yields m =9.55x10^{30} 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!