Answer :
To calculate the diameter of the supermassive black hole at the center of the Milky Way galaxy, we need to follow these steps:
1. Determine the mass of the black hole:
[tex]\[ \text{mass of the black hole} = 4.3 \times 10^6 \times \text{mass of the Sun} \][/tex]
Given the mass of the Sun [tex]\( = 2 \times 10^{30} \, \text{kg} \)[/tex],
[tex]\[ \text{mass of the black hole} = 4.3 \times 10^6 \times 2 \times 10^{30} \, \text{kg} \][/tex]
[tex]\[ \text{mass of the black hole} = 8.6 \times 10^{36} \, \text{kg} \][/tex]
2. Determine the density of the black hole:
The density of the black hole is given as 1000 times the density of water.
[tex]\[ \text{density of the black hole} = 1000 \times \text{density of water} \][/tex]
Given the density of water [tex]\( = 1000 \, \text{kg/m}^3 \)[/tex],
[tex]\[ \text{density of the black hole} = 1000 \times 1000 \, \text{kg/m}^3 \][/tex]
[tex]\[ \text{density of the black hole} = 10^6 \, \text{kg/m}^3 \][/tex]
3. Calculate the volume of the black hole:
Using the formula for the volume of a sphere [tex]\(V = \frac{4}{3} \pi r^3\)[/tex], and knowing that density [tex]\(\rho = \frac{mass}{volume}\)[/tex],
[tex]\[ \text{volume of the black hole} = \frac{\text{mass of the black hole}}{\text{density of the black hole}} \][/tex]
[tex]\[ \text{volume of the black hole} = \frac{8.6 \times 10^{36} \, \text{kg}}{10^6 \, \text{kg/m}^3} \][/tex]
[tex]\[ \text{volume of the black hole} = 8.6 \times 10^{30} \, \text{m}^3 \][/tex]
4. Calculate the radius of the black hole:
Given [tex]\(V = \frac{4}{3} \pi r^3\)[/tex], we need to solve for [tex]\(r\)[/tex]:
[tex]\[ r^3 = \frac{3V}{4\pi} \][/tex]
Plugging in the volume:
[tex]\[ r^3 = \frac{3 \times 8.6 \times 10^{30}}{4\pi} \][/tex]
Evaluating this,
[tex]\[ r^3 \approx 2.049 \times 10^{30} \, \text{m}^3 \][/tex]
Taking the cube root to find [tex]\(r\)[/tex],
[tex]\[ r \approx 1.2709738466 \times 10^{10} \, \text{m} \][/tex]
5. Calculate the diameter of the black hole:
The diameter [tex]\(d\)[/tex] is twice the radius:
[tex]\[ d = 2r \][/tex]
[tex]\[ d \approx 2 \times 1.2709738466 \times 10^{10} \, \text{m} \][/tex]
[tex]\[ d \approx 2.5419476932 \times 10^{10} \, \text{m} \][/tex]
6. Round to 3 significant figures in standard form:
[tex]\[ d \approx 2.54 \times 10^{10} \, \text{m} \][/tex]
So, the diameter of the supermassive black hole at the center of the Milky Way galaxy is approximately [tex]\(2.54 \times 10^{10} \, \text{m}\)[/tex] in standard form to 3 significant figures.
1. Determine the mass of the black hole:
[tex]\[ \text{mass of the black hole} = 4.3 \times 10^6 \times \text{mass of the Sun} \][/tex]
Given the mass of the Sun [tex]\( = 2 \times 10^{30} \, \text{kg} \)[/tex],
[tex]\[ \text{mass of the black hole} = 4.3 \times 10^6 \times 2 \times 10^{30} \, \text{kg} \][/tex]
[tex]\[ \text{mass of the black hole} = 8.6 \times 10^{36} \, \text{kg} \][/tex]
2. Determine the density of the black hole:
The density of the black hole is given as 1000 times the density of water.
[tex]\[ \text{density of the black hole} = 1000 \times \text{density of water} \][/tex]
Given the density of water [tex]\( = 1000 \, \text{kg/m}^3 \)[/tex],
[tex]\[ \text{density of the black hole} = 1000 \times 1000 \, \text{kg/m}^3 \][/tex]
[tex]\[ \text{density of the black hole} = 10^6 \, \text{kg/m}^3 \][/tex]
3. Calculate the volume of the black hole:
Using the formula for the volume of a sphere [tex]\(V = \frac{4}{3} \pi r^3\)[/tex], and knowing that density [tex]\(\rho = \frac{mass}{volume}\)[/tex],
[tex]\[ \text{volume of the black hole} = \frac{\text{mass of the black hole}}{\text{density of the black hole}} \][/tex]
[tex]\[ \text{volume of the black hole} = \frac{8.6 \times 10^{36} \, \text{kg}}{10^6 \, \text{kg/m}^3} \][/tex]
[tex]\[ \text{volume of the black hole} = 8.6 \times 10^{30} \, \text{m}^3 \][/tex]
4. Calculate the radius of the black hole:
Given [tex]\(V = \frac{4}{3} \pi r^3\)[/tex], we need to solve for [tex]\(r\)[/tex]:
[tex]\[ r^3 = \frac{3V}{4\pi} \][/tex]
Plugging in the volume:
[tex]\[ r^3 = \frac{3 \times 8.6 \times 10^{30}}{4\pi} \][/tex]
Evaluating this,
[tex]\[ r^3 \approx 2.049 \times 10^{30} \, \text{m}^3 \][/tex]
Taking the cube root to find [tex]\(r\)[/tex],
[tex]\[ r \approx 1.2709738466 \times 10^{10} \, \text{m} \][/tex]
5. Calculate the diameter of the black hole:
The diameter [tex]\(d\)[/tex] is twice the radius:
[tex]\[ d = 2r \][/tex]
[tex]\[ d \approx 2 \times 1.2709738466 \times 10^{10} \, \text{m} \][/tex]
[tex]\[ d \approx 2.5419476932 \times 10^{10} \, \text{m} \][/tex]
6. Round to 3 significant figures in standard form:
[tex]\[ d \approx 2.54 \times 10^{10} \, \text{m} \][/tex]
So, the diameter of the supermassive black hole at the center of the Milky Way galaxy is approximately [tex]\(2.54 \times 10^{10} \, \text{m}\)[/tex] in standard form to 3 significant figures.
