✔ 最佳答案
1 pc = 3.086 x 10^16 m
hence, 13 kpc = 13000 x 3.086x10^16 m = 4.012 x 10^20 m
and 220 km/sec = 2.2 x 10^5 m/s
Let m be the mass of the star, and M be the mass of materials inside the orbit of that star.
Gravitational pull on the satr = GmM/(4.012x10^20)^2
where G is the Gravitational constant (= 6.67 x 10^-11 m^3/kg.s^2)
Centripetal force required to perform a circular orbit
= m.(2.2x10^5)^2/(4.012x10^20)
Hence, GmM/(4.012x10^20)^2 = m(2.2x10^5)^2/(4.012x10^20)
i.e. M = (2.2x10^5)^2 x (4.012x10^20)/(6.67x10^-11) kg
M = 2.91 x 10^41 kg