Stephan Hawkins worked extensively with theoretical gravitational physics. He is well known for his work with black holes. He passed on March 14th, 2018. It has been observed that a certain star (S2) in the Sagittarius A* (meaning in the direction of the constellation Sagittarius) is orbiting a supermassive black hole. If the period of S2 is 16year and the distance between the two is 1.79e13m. What is the mass of this black hole? (be careful with your units)

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meep26wesley

High School

Stephan Hawkins worked extensively with theoretical gravitational physics. He is well known for his work with black holes. He passed on March 14th, 2018. It has been observed that a certain star (S2) in the Sagittarius A* (meaning in the direction of the constellation Sagittarius) is orbiting a supermassive black hole. If the period of S2 is 16year and the distance between the two is 1.79e13m. What is the mass of this black hole? (be careful with your units)

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cryssatemp

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Answer: $M=1.3(10)^{34}kg$
The equation that relates the period $T$ of a body that orbits a greater body in space with the distance $r$ between both bodies is:
$T^{2}=\frac{4\pi^{2}}{GM}r^{3}$ (1)
Where;
$M$ is the mass of the Black Hole (the value we want to find)
$G$ is the Gravitational Constant and its value is $6.674x10^{-11}\frac{m^{3}}{kgs^{2}}$
$r=1.79(10)^{13}m$ 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).
$T=16y$ is the orbital period of the Star S2
At this point, note we have to transform the units of $T$ from years $y$ to seconds:
$T=16y.\frac{365days}{1y}.\frac{24h}{1day}.\frac{3600s}{1h}$
$T=504796000\approx 5(10)^{8}s$ is the orbital period of the Star S2 in seconds
If we want to find the mass $M$ of the black hole, we have to express equation (1) as written below and substitute all the values:
$M=\frac{4\pi^{2}r^{3}}{G.T^{2}}$ (2)
$M=\frac{4\pi^{2}(1.79(10)^{13}m)^{3}}{(6.674x10^{-11}\frac{m^{3}}{kgs^{2}})(5×10^{8}s)^{2}}$ (3)
$M=\frac{4\pi^{2}5.73(10)^{39}m^{3}}{16675000\frac{m^{3}}{kg}}$ (3)
Finally we have the mass of the black hole:
$M=1.3(10)^{34}kg$
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lizfriend28

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
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