Milkomeda: Blog 2, Worksheet 1.1, Problems 2-4

The Andromeda Galaxy and the Milky Way Galaxy are slowly but surely colliding into the charmingly named Milkomeda Galaxy. In the first worksheet of the semester, we did a few calculations to learn more about Milkomeda.

Source

Problem 2: . How long will it take for Andromeda to collide with the Milky Way? The time-scale here is the free-fall time, tff. One way of finding this is to assume that Andromeda is on a highly elliptical orbit (e -> 1) around the Milky Way. With this assumption, we can use Kepler’s Third Law \[P^2 = \frac{4 \pi^2 a^3}{G (M_{MW} + M_{And} )} \] where P is the period of the orbit and a is the semi-major axis. How does \(t_{ff}\) relate to the period? Estimate it to an order of magnitude.

We can approximate an e --> 1 ellipse as almost a straight line.

Therefore, we can approximate the free fall time of the second galaxy as half of the period of rotation, as it must fall approximately the length of the orbit. \[t_{ff} = \frac{P}{2}\] Plugging this into the given equation for period allows us to approximate free fall time: \[(2t_{ff})^2 \approx \frac{4 \pi^2 a^3}{G (M_{MW} + M_{And} )} \] \[ t_{ff} \approx \left( \frac{ \pi^2 a^3}{G (M_{MW} + M_{And} )} \right)^{\frac{1}{2}} \] Now it is just a matter of plugging in constants, all of which we were given on the worksheet: \[a \approx 400 kpc \approx 1.24 \times 10^{24} cm \] \[M_{\odot} \approx 10^{33} g \] \[M_{MW} + M_{And} \approx 2 \times 10^{12} M_{\odot} \approx 2 \times 10^{45} g  \] \[G \approx 6.68 \times 10^{-8} \] \[ t_{ff} \approx \left( \frac{ \pi^2 (1.24 \times 10^{24})^3}{6.68 \times 10^{-8} (2 \times 10^{45})} \right)^{\frac{1}{2}} \] \[t_{ff} = 3.75 \times 10^{17} \: seconds \] \[1.2 \times 10^{10} years\]

Problem 3: Let’s estimate the average number density of stars throughout the Milky Way, n. First, we need to clarify the distribution of stars. Stars are concentrated in the center of the galaxy, and their density decreases exponentially: \[ n(r) \propto e^{\frac{-r}{R_s}} \] is also known as the “scale radius” of the galaxy. The Milky Way has a scale radius of 3.5 kpc. With this in mind, estimate n in two ways: 
(a) Consider that within a 2 pc radius of the Sun there are five stars: the Sun, α Centauri A and B, Proxima Centauri, and Barnard’s Star. 

We can approach this problem by approximating the galaxy with the 2pc radius around the Sun. The density of stars in the described reason would be: \[ n = density = \frac{ stars}{volume} = \frac{5}{ \frac{4}{3} \pi 2^3 } \approx 0.149 \frac{stars}{pc^3} \] According to google, we are 8 kpc from the center of the universe, which is about 2 scale radii from the center of the galaxy. For the sake of not doing an integral, we can approximate. With every scale radii, the function drop by \( \frac{1}{e} \), so at 2 radii it has dropped by about \( \frac{1}{e^2} \). To account for this, we can just multiply our density by e^2: \[0.149 \frac{stars}{pc^3} \times e^2 \approx 1.10 \frac{stars}{pc^3}  \]

(b) The Galaxy’s “scale height” is 330 pc. Use the galaxy’s scale lengths as the lengths of the volume within the Galaxy containing most of the stars. Assume a typical stellar mass of 0.5\(M_{\odot}\).

To find number density we need, by definition, the number of stars in the galaxy and the volume of the galaxy. To get the number of stars we need to divide the mass of the galaxy by the average mass of a star: \[ \frac{M_{galaxy}}{M_{star}} = \frac{10^{10} M_{\odot} }{ .5 M_{\odot} } = 2 \times 10^{10} stars \] To get volume, we can approximate the galaxy as a cylinder, assuming most stars are within one scale radius of the center.

 \[ V = (3500)^2 \times \pi \times 300 = 1.27 \times 10^{10} pc^3 \] Now that we have both pieces, we can solve for n: \[ n = \frac{N}{V} = \frac{2 \times 10^{10} stars}{1.27 \times 10^{10} pc^3} = 1.57 \frac{ stars}{ pc^3} \]

Problem 4: Determine the collision rate of the stars using the number density of the stars (n), the cross-section for a star \(σ_{\star}\), and the average velocity of Milkomeda’s stars as they collide v. How many stars will collide every year? Is the Sun safe, or likely to collide with another star?

We know collision rate is proportional to velocity, area, and number density. Using dimensional analysis, we can solve for the actual equation \[collision \: rate \: \propto \:v \: σ \: n  \] \[ \frac{stars}{time} = \frac{cm}{s} \times cm^2 \times \frac{stars}{cm^3} \] All the units cancel so the equation would be \[collision \: rate  = v \: σ \: n \] We know all pieces of this equation, but need to do a few calculations. To find velocity we can use the distance and time from problem 2: \[ v = \frac{d}{t_{ff}} =  \frac{ 2.5 \times 10^{24} }{2 \times 10^{18} } = 1.25 \times 10^6 cm/s \] We can convert our n to cgs: \[ 1.6 \times \frac{1}{(3 \times 10^{18})^3} = 6 \times 10^{-56} \frac{stars}{cm^3} \]  If we approximate our sun as an average star in the galaxy, we have the last piece of this puzzle, cross sectional area. Now we can solve: \[ collision \: rate  = v \: σ \: n  = (1.25 \times 10^6 ) \pi (7 \times 10^{10} )^2 (6 \times 10^{-56}) \approx 1 \times 10^{-27} \frac{stars}{s} \] Multiplying that answer by how many seconds there are in a year ( \( 3.2 \times 10^7  \) s ) tells us that there are \(3.4 \times 10^{-20} \) collisions per year. It is pretty unlikely that the sun will hit anything anytime soon. So we can all breathe a sigh of relief.