Question

At the center of our galaxy, the Milky Way, there is an object known as Sagittarius A*. Astronomers believe this is a black hole. Evidence for this is that, over several years, astronomers have observed other stars in orbit around Sagittarius A*. One such star has a period of its orbit around Sagittarius A* of 15.2 years and the radius of its orbit has been measured to be 1.42  1014 m. (a) Assuming this orbit is circular, and ignoring the presence of other stars in the neighourhood, calculate the mass of Sagittarius A* in kg. (b) Given that the mass of our Sun is 1.99  1030 kg, how many times more massive is Sagittarius A* than our Sun?

Answer 1

a)

T = time period of star = 15.2 years = 15.2 x 365 x 24 x 3600 sec = 4.8 x 10⁸ sec (Since 1 year = 365 days , 1 day = 24 hours , 1h = 3600 sec)

r = radius of the orbit = 1.42 x 10¹⁴ m

M = mass of Sagittarius A* = ?

Using keplor's third law

T² = 4² r³/(GM)

(4.8 x 10⁸)² = 4(3.14)² (1.42 x 10¹⁴)³ /((6.67 x 10^-11) M)

M = 7.35 x 10³⁶ kg

b)

M_s = mass of sun = 1.99 x 10³⁰ kg

Ratio of mass of Sagittarius A* to our Sun is

ratio = M/M_s = (7.35 x 10³⁶) /(1.99 x 10³⁰) = 3.7 x 10⁶

hence Sagittarius A* 3.7 x 10⁶ times more massive than our Sun