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標題:
Math astronomy problem
發問:
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: sorry, it should be 13 kpc instead of 13000 kpc
最佳解答:
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
其他解答:C8D74AB62542840B
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