Math astronomy problem

卓霖

2013-10-06 11:47 pm

Because the rotation curve for the Milky Way galaxy is approximately flat, a star in a circular orbit 13,000 kpc from the galactic center has the same orbital speed as the Sun, namely 220 km/sec. How much mass is interior to this star's orbit?

更新1:

sorry, it should be 13 kpc instead of 13000 kpc

回答 (1)

天同

2013-10-07 1:45 am

✔ 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

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