Blog #2, Worksheet #1.1, Milkomeda Calculations

*Worked with Tim Mcnamara 

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 around the Milky Way (e -> 1). With this assumption we can use Kepler's Third Law where P is the period of the orbit and a is the semi major axis. How does tff relate to the period? Estimate it to an order of magnitude.


We will solve this problem from the Milky Way's frame of reference meaning we will picture the Andromeda Galaxy orbiting the milky way with a highly elliptical shape. The Milky Way will remain stationary and will serve as one of the foci for the Andromeda's orbit. Because it is so highly elliptical, we make the extreme assumption that the Andromeda is more or less moving in a line towards the Milky Way. Since the full period, \(P\), is the time it takes for the Andromeda to get to the Milky Way and then return to its original position, the time it takes to collide with the Milky Way is simply, \(\frac{P}{2}\). The Milky Way and Andromeda have roughly the same mass thus \(M_{Mw}\thickapprox M_{And}\). Thus we can assume Kepler's equation as: \[P^2 = \frac{4\pi^2a^3}{G(2M)}\] If we substitute in \(t_{ff} = \frac{P}{2}\) and simplify we end up with \[t_{ff} = \frac{\pi a^{\frac{3}{2}}}{\sqrt{G(2M)}}\] Substituting in our values for \(M = 10^{12} M_\odot\), \(G = 6.67 x 10^{-11} \frac{N m^2}{kg^2}\), and \(a = 800 kpc\), we end up with \[t_{ff} = 1.1 x 10^{10}\] or 11 billion years!

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. 

Part (a): Consider that within a 2 pc radius of the Sun there are five stars. 

We use the given information to first find our proportionality constant that we can then use to find the average distribution throughout the galaxy. We begin by using the fact that the sun is 8 kpc from the center of the universe so \[n(8000) = \frac{stars}{volume} = \frac{5}{\frac{4}{3} \cdot \pi \cdot 2^3} = \frac{15}{32\pi}\]. We also know that \[n(8000) = C \cdot e^{\frac{-8000}{R_s}}\] where \(R_s = 3.5 kpc\). Which gives us constant, C: \[C = 1.47 \frac{stars}{pc^3}\] This is the central density of our galaxy and can be taken as the average number density of the stars throughout our Milky Way. 

Part (b): Use the galaxy's "Scale height" of 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 \(0.5_\odot\). 

To begin we will assume that our galaxy takes the rough shape of a cylinder with the scale lengths given: \[h = 330 pc \qquad r = 3.5 kpc\] We then solve for the volume of the galaxy using \[v = \pi \cdot r^2 \cdot h = 1.27x10^{10} pc^3\] We also know that the total mass of the Milky Way is \(10^10 M_\odot\) and the avg. stellar mass is \(0.5_\odot\) which gives us an approximation of \(2x10^{10}\) stars in the galaxy. We then solve for the density using \[\rho = \frac{m}{v}\] which leads to \[\rho = \frac{2x10^{10}}{1.27x10^{10}}\] which gives us a similar final answer of \[\rho_{avg} = 1.57 \frac{stars}{pc^3}\]

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

We begin with finding a good estimate for the cross-sectional area of a star. Because our sun is an average sized star, we will use its radius to approximate a circular cross-section of an average star. The sun has a radius \(R_s \thickapprox 10^9 m\). When converting this to parsecs we end up with \(R_s = 3.24 x 10^{-8} pc\). We then solve for the cross-sectional area \(\sigma_\divideontimes = \pi \cdot {R_s}^2 = 1x10^{-15}pc^2\).

For the next step we must solve for the velocity, \(\overline{v}\), using our information from part two. We know that the time, t, that it will take for stars in the Milkomeda to collide is \(t = 1.1 x 10^{10} years\), and the distance between the two galaxies is 800 kpc. Using \(v = \frac{d}{t}\), we find that \(\overline{v} = 7.2x10^{-5} \frac{pc}{year}\). Generalizing our problem, we can assume that the number of collisions is equal to the number of stars occupying the set volume in a set amount of time. To find the volume occupied we multiply our area, \(\sigma_\divideontimes\) by the distance it will travel in a certain time. Thus \(Volume = \sigma_\divideontimes \cdot \overline{v} \cdot t\). Then our collisions will be this volume multiplied with the density of stars in the Milkomeda, which we found in part 3 (though we must use twice the density to approx. the density of both galaxies) We are left with \(collisions = Volume \cdot n = \sigma_\divideontimes \cdot \overline{v} \cdot t \cdot n\). But we are looking for \(\frac{collisions}{year}\) and thus must divide by time, t, leaving us with \(\frac{collisions}{year} = \sigma_\divideontimes \cdot \overline{v} \cdot 2n\). We then put in our constants \[\sigma_\divideontimes = 1x10^{-15}pc^2 \qquad \overline{v} = 7.2x10^{-5}\frac{pc}{year} \qquad n = 1.47 \frac{stars}{pc^3}\] Which gives us a final answer of \[2.26x10^{-19} \frac{collisions}{year}\]This is an incredibly low number of collisions and thus it would be expected that our sun is relatively safe and would not collide with another star once the two galaxies begin to collide.