USAAAO 2020, II кръг — Задача 7

• (a) T = 334 million years; M = 7.88 × 10¹⁰ solar masses
• (b) $1.08 \times 10^{-15}$ rad/s
• (c) $\sigma = \frac{V_0^2}{2\pi rG}$
• (d) 85 %

Solution:

(a) The period is $T = \frac{2\pi r}{V}$, so we obtain $T = 334$ million years. Equating the centripetal force with the gravitational force (as is the condition for a circular orbit), $M = \frac{V^2r}{G}$, so we obtain $M = 7.88 \times 10^{10}$ solar masses.

(b) $\omega = \frac{V}{r}$ is the angular velocity. Very close to the center, $e^{-r/R} \approx 1 - r/R$, so $\omega \approx V_0/R = 1.08 \times 10^{-15}$ rad/s.

(c) Recall $M = \frac{V^2r}{G}$. Consider an annulus at radius $r$ with small radial thickness $\delta r$. Letting $\sigma$ be the desired mass per unit area, we have that $\sigma 2\pi r\delta r = \delta M$. But note that $\delta M \approx \frac{V^2\delta r}{G}$ very far away from the center (as $V$ is essentially constant). Thus we obtain $\sigma = \frac{V_0^2}{2\pi rG}$ very far away from the center ($r \gg R$).

(d) $M_{bol} = 4.75 - 2.5\log(L/L_{Sun})$, so $L/L_{Sun} = 2.40 \times 10^{10}$. The luminous mass enclosed by an annulus at radius $r$ with small radial thickness $\delta r$ is $2\pi r\delta r\sigma_L = 2\pi k\delta r$, which means we seek $r_{max}$ such that $2\pi kr_{max} = 2.40 \times 10^{10}$. This gives $r_{max} = 14.97$ kpc as the maximum distance from the center of the galaxy for which the galaxy is still visibly defined. The total gravitational mass enclosed at this radius is $M = \frac{V^2r}{G} = 1.63 \times 10^{11}$ solar masses. The fraction of dark matter is approximately $1 - \frac{2.40\times10^{10}}{1.63\times10^{11}} = 85\%$.