Question

WHAT IS THE MEAN MASS DENSITY FOR A SUPER MASSIVE BLACK HOLE WITH TOTAL MASS OF 1*10^8 th MASS OF SUN INSIDE THE SCHWARZSCHILD RADIUS ?

Answer 1

What is the mean mass density for a super massive black hole with total mass of $1*10^{8}$ mass of sun inside the Schwarzschild radius?

Solution:

Mass density formula looks like this: $\rho = \frac{M}{V}$, where $M$ – mass of a black hole, $V$ – volume of black hole. It's considered, that visible radius of black hole is equal to its Schwarzschild radius, so volume is equal to $V = \frac{4}{3}\pi r_g^3$.

$r_g$ – Schwarzschild's radius, which is equal to $r_g = \frac{2GM}{c^2}$, where $G$ – gravitational constant, $c$ – speed of light, $M$ – mass of the black hole. Substituting $M$ and $V$ in starting formula we get:

$$\rho = \frac {M}{V} = \frac {1 0 ^ {8} * M _ {\odot}}{4 \pi r _ {g} ^ {3}} = \frac {3 * 1 0 ^ {8} * M _ {\odot}}{4 \pi \left(\frac {2 G M}{c ^ {2}}\right) ^ {3}} = \frac {3 * 1 0 ^ {8} * M _ {\odot}}{4 \pi \frac {8 G ^ {3} M ^ {3}}{c ^ {6}}} = \frac {3 * 1 0 ^ {8} * M _ {\odot} * c ^ {6}}{3 2 \pi G ^ {3} M ^ {3}} = \frac {3 * 1 0 ^ {8} * M _ {\odot} * c ^ {6}}{3 2 \pi G ^ {3} (1 0 ^ {8} * M _ {\odot}) ^ {3}} = \frac {3 * c ^ {6}}{3 2 \pi G ^ {3} * 1 0 ^ {1 6} * M _ {\odot} ^ {2}};$$

(Where $M_{\odot}$ is mass of Sun).

Substituting numerical values, we get:

$$\rho = \frac {3 * c ^ {6}}{3 2 \pi G ^ {3} * 1 0 ^ {1 6} * M _ {\odot} ^ {2}} = \frac {3 * (3 * 1 0 ^ {8} m / s) ^ {6}}{3 2 * 3 . 1 4 1 5 9 2 * \left(6 . 6 7 * 1 0 ^ {- 1 1} \frac {m ^ {3}}{s ^ {2} * k g}\right) ^ {3} * 1 0 ^ {1 6} * 3 , 9 2 0 4 * 1 0 ^ {6 0} k g ^ {2}} = \frac {3 * 7 2 9 * 1 0 ^ {4 8} \frac {m ^ {6}}{s ^ {6}}}{3 2 * 3 . 1 4 1 5 9 2 * 2 . 9 6 7 * 1 0 ^ {- 3 1} \frac {m ^ {9}}{s ^ {6} * k g ^ {3}} * 1 0 ^ {1 6} * 3 , 9 2 0 4 * 1 0 ^ {6 0} k g ^ {2}} = \frac {2 . 1 8 7 * 1 0 ^ {5 1} \frac {m ^ {6}}{s ^ {6}}}{2 . 9 8 3 * 1 0 ^ {- 1 3} \frac {m ^ {9}}{s ^ {6} * k g ^ {3}} * 3 , 9 2 0 4 * 1 0 ^ {6 0} k g ^ {2}} = \frac {2 . 1 8 7 * 1 0 ^ {5 1} \frac {m ^ {6}}{s ^ {6}}}{2 . 9 8 3 * 1 0 ^ {- 1 3} \frac {m ^ {9}}{s ^ {6} * k g ^ {3}} * 3 , 9 2 0 4 * 1 0 ^ {6 0} k g ^ {2}} = \frac {2 . 1 8 7 * 1 0 ^ {5 1} \frac {m ^ {6}}{s ^ {6}}}{1 . 1 7 * 1 0 ^ {4 8} \frac {m ^ {9}}{s ^ {6} * k g}} = 1 8 6 9. 2 3 \frac {k g}{m ^ {3}}$$

Answer:

The mean mass density for black hole, mentioned in task, is 1869.23 $\frac{kg}{m^3}$. Answer provided by https://www.AssignmentExpert.com