Respuesta :
Answer:
Explanation:
The star is revolving the black hole like earth revolves around the sun .so time period of rotation T is given by the following relation
T² = [tex]\frac{4\pi^2\times R^3}{GM }[/tex] , R is distance between black hole and star , M is mass of black hole
Given T = 4.8 hours
4.8² = [tex]\frac{4\pi^2\times R^3}{GM }[/tex]
Using the same equation for earth sun system
24² = [tex]\frac{4\pi^2\times (50R)^3}{GM_s }[/tex] , Ms is mass of the sun and 50R is distance between the sun and the earth .
Dividing the equation
[tex](\frac{4.8}{24})^2[/tex] = [tex]\frac{M_s}{M}\times\frac{1^3}{50^3}[/tex]
[tex]\frac{M}{M_s}[/tex] = 2x 10⁻⁴
The mass of the black hole is [tex]2\times 10^{-4}[/tex] times the mass of our sun.
Given information:
An X-ray telescope detected a source called Cygnus X-3, whose intensity changed with a period of 4.8 hours (T).
The star is orbiting around the black hole.
The distance between the centers of the star and the black hole is one-fiftieth of the distance between the centers of the Earth and our Sun.
Let R be the center to center distance between the black hole and the sun.
So, the time period T of the sun around the black hole will be,
[tex]T^2=\dfrac{2\pi^2R^3}{GM}\\4.8^2=\dfrac{2\pi^2R^3}{GM}[/tex]
where M is the mass of the black hole.
The distance between the earth and our sun will be 50R.
So, the time period of the earth will be,
[tex]T^2=\dfrac{2\pi^2R^3}{GM}\\24^2=\dfrac{2\pi^2(50R)^3}{GM_s}[/tex]
where [tex]M_s[/tex] is the mass of our sun.
Now, compare the above two relations to get the mass of black hole in terms of mass of our sun as,
[tex]4.8^2=\dfrac{2\pi^2R^3}{GM}\\24^2=\dfrac{2\pi^2(50R)^3}{GM_s}\\\dfrac{24^2}{4.8^2}=\dfrac{50^3}{1}\times \dfrac{M}{M_s}\\\dfrac{M}{M_s}=0.0002\\\dfrac{M}{M_s}=2\times 10^{-4}[/tex]
Therefore, the mass of black hole is [tex]2\times 10^{-4}[/tex] times the mass of our sun.
For more details, refer to the link:
https://brainly.com/question/13717010
