Density of Black Hole

Sep 2018
7
0
So when I tested the sample question I got the right answer. But doing a practice one I feel like I got the wrong answer - it seems way too high:

Black hole mass - 4.31 x 10^6
I calculated Rs to be = 1.27 x 10^7
Average density:
P = m/v

m = (4.31x10^6) x (1.989x10^30)
m = 8.57x10^36 km

v=4/3 x π x r^3
v= 4.3 x π x (4.127x10^7)^3
v= 8.22 x 10^21

P = (8.57x10^26) / (8.22 x 10^21)
P = 1.04 x 10^15 km m^-3

Have I calculated this correctly or have I done something wrong?
When I go on to look a Roche Limit:
1400 kg m^-1 = Sun's average density

r = 2.456(((1.04x10^15)/(1400))^1/3)
r = 2243.17

Which seems way too high?
 
Oct 2018
1
0
A black hole contains about 3.7M (3.7 million) solar masses (a solar mass is 1.98892 × 1033 grams) and is assumed to be a sphere with a Schwarzschile radius of 3x105 cm) x (M / Msun)where M is the mass of the black hole and Msun is the mass of the Sun. Typically, M for a black hole in our galaxy is around 10 times the mass of the Sun, but for supermassive black holes at the centers of galaxies it can be millions or even billions [3]. What is its density?


Solution You can use the Schwarzschild radius to calculate the "density" of the black hole - i.e., the mass divided by the volume enclosed within the Schwarzschild radius.

The radius of the sphere must be calculated first, using the formula 3x105 cm) x (M / Msun) we get 3x105 cm) x 3.7 × 106 = 1.125 × 1012

V=43πr3=43×3.142×(1.125×1012cm)3=5.964×1036cm3


The mass of the sphere is 3.7 x 106 Msun × 1.98892 × 1033 g/Msun = 7.359 × 1039 g

Density=ρ=mV
= 7.359×10395.964×1036=1.31×103gcc


You might have recognized a shortcut: the density is

(1.8×1016gcm3)×(MsunM)2
or

1.8×1016gcm3×(13.7×106)2=1.31×103gcc
From the point of view of an outside observer, this might as well be the actual black hole density, since the distribution of matter within the Schwarzschild radius has no effect on the outside.