Question

Astronomical observations of our Milky Way galaxy indicate that it has a mass of about 8.0×1011 solar masses. A star orbiting on the galaxy's periphery is about 6.0×104 light years from its center. (a) What should the orbital period of that star be? (b) If its period is 6.0×107 y instead, what is the mass of the galaxy? Such calculations are used to imply the existence of "dark matter" in the universe and have indicated, for example, the existence of very massive black holes at the centers of some galaxies.

Answer 1

Answer:

a) $T = 26.147\times 10^{7}\,y$, b) $1.520\times 10^{13}$ solar masses

Explanation:

Let suppose that galaxy can be treated as a puntual mass.

a) The acceleration experimented by the star is:

$a_{r} = G\cdot \frac{M}{r^{2}}$

$a_{r} = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (8.0\times 10^{11})\cdot\frac{\left(1.989\times 10^{30}\,kg\right)}{\left[(6.0\times 10^{4}\,ly)\cdot \left(9.461\times 10^{15}\,\frac{m}{ly} \right)\right]^{2}}$

$a_{r} = 3.296\times 10^{-10}\,\frac{m}{s^{2}}$

The angular speed of the star is:

$\omega = \sqrt{\frac{a_{r}}{r} }$

$\omega = \sqrt{\frac{3.296\times 10^{-10}\,\frac{m}{s^{2}} }{(6.0\times 10^{4})\cdot (9.461\times 10^{15}\,\frac{m}{ly} )} }$

$\omega = 7.62\times 10^{-16}\,\frac{rad}{s}$

The period is:

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{7.62\times 10^{-16}\,\frac{rad}{s} }$

$T = 8.246\times 10^{15}\,s$

$T = 26.147\times 10^{7}\,y$

b) The period is:

$T = 6\times 10^{7}\,y$

$T = 1.892\times 10^{15}\,s$

The angular speed is:

$\omega = \frac{2\pi}{1.892\times 10^{15}\,s}$

$\omega = 3.321\times 10^{-15}\,\frac{rad}{s}$

The acceleration experimented by the star is:

$a_{r} = \left(3.321\times 10^{-15}\,\frac{rad}{s} \right)^{2}\cdot (6\times 10^{4}\,ly)\cdot \left(9.461\times 10^{15}\,\frac{m}{ly} \right)$

$a_{r} = 6.261\times 10^{-9}\,\frac{m}{s^{2}}$

The mass of the galaxy is:

$M = \frac{a_{r}\cdot r^{2}}{G}$

$M = \frac{\left(6.261\times 10^{-9}\,\frac{m}{s^{2}} \right)\cdot \left[(6.0\times 10^{4}\ly)\cdot (9.461\times 10^{15}\,\frac{m}{ly} )\right]^{2}}{6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} }$

$M = 3.023\times 10^{43}\,kg$

Which is equal to $1.520\times 10^{13}$ solar masses.