# Given the following information, what is the density of the black hole in g/cm^3?

## Scientists discovered a new class of black holes with masses 100 to 10000 times the mass of our sun, but occupying less space than our moon. Suppose that one of these black holes has a mass of $1 x {10}^{3}$ suns and a radius equal to one half the radius of our moon. The radius of our sun is 7.0 x 10^5 km with an average density of $1.4 x {10}^{3} k \frac{g}{m} ^ 3$. The diameter of the moon is $2.16 x {10}^{3}$miles.

Oct 26, 2016

I found: $1 \times {10}^{12} \frac{g}{c {m}^{3}}$
But...I strongly suggest you to check my maths!!!

#### Explanation:

We know that:
$\text{density"="mass"/"volume}$

1) Mass of the Black Hole:
From density we can evaluate the mass of our Sun:
Rearranging:
$\text{mass"="density"xx"volume}$
$\text{mass"_"(Sun)"=4/3pir^3"density(Sun)} = 2 \times {10}^{30} k g$

The Black Hole will have:
${\text{mass}}_{B H} = 1 \times {10}^{3} \cdot 2 \times {10}^{30} = 3 \times {10}^{33} k g$

2) Volume of the Black Hole;
assuming a sperical shape with half the radius or our Moon we have:
${V}_{B H} = \frac{4}{3} \pi {r}_{\text{moon}}^{3} = \frac{4}{3} \pi {\left(\frac{3.5 \times {10}^{6}}{4}\right)}^{3} = 2.7 \times {10}^{18} {m}^{3}$
I changed into meters to be consistent and reduced from diameter to radius (divide by 2) and I halved it as required by the question.

3) Density of Black Hole:
from: $\text{density"="mass"/"volume}$
We get:
$\text{density} = \frac{\left(3 \times {10}^{33} \cdot 1 \times {10}^{3}\right) g}{\left(2.7 \times {10}^{18} \cdot 1 \times {10}^{6}\right) c {m}^{3}} = 1 \times {10}^{12} \frac{g}{c {m}^{3}}$