Respuesta :
Answer: [tex]M=1.3(10)^{34}kg[/tex]
The equation that relates the period [tex]T[/tex] of a body that orbits a greater body in space with the distance [tex]r[/tex] between both bodies is:
[tex]T^{2}=\frac{4\pi^{2}}{GM}r^{3}[/tex] (1)
Where;
[tex]M[/tex] is the mass of the Black Hole (the value we want to find)
[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674x10^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
[tex]r=1.79(10)^{13}m[/tex] is the distance from the Black Hole to the Star S2 (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit).
[tex]T=16y[/tex] is the orbital period of the Star S2
At this point, note we have to transform the units of [tex]T [/tex] from years [tex]y[/tex] to seconds:
[tex]T=16y.\frac{365days}{1y}.\frac{24h}{1day}.\frac{3600s}{1h}[/tex]
[tex]T=504796000\approx 5(10)^{8}s[/tex] is the orbital period of the Star S2 in seconds
If we want to find the mass [tex]M[/tex] of the black hole, we have to express equation (1) as written below and substitute all the values:
[tex]M=\frac{4\pi^{2}r^{3}}{G.T^{2}}[/tex] (2)
[tex]M=\frac{4\pi^{2}(1.79(10)^{13}m)^{3}}{(6.674x10^{-11}\frac{m^{3}}{kgs^{2}})(5×10^{8}s)^{2}}[/tex] (3)
[tex]M=\frac{4\pi^{2}5.73(10)^{39}m^{3}}{16675000\frac{m^{3}}{kg}}[/tex] (3)
Finally we have the mass of the black hole:
[tex]M=1.3(10)^{34}kg[/tex]
Answer:Watching the moving images and sounds in the video helped me picture what the text from A Black Hole Is NOT a Hole is describing about Centaurus A. Being able to both see and hear the information helped me gain a better understanding of black holes.
Explanation: bc am now