The movements of Andromeda and the Milky Way

We are bounded in a nutshell of Infinite space:Blog Post #2, Worksheet # 1.1, Problem #2: How long will it take the sky to fall? 

How long will it take for Andromeda to collide with the Milky Way? The time-scale here is the free-fall time, \(t_{ff}\). One way of finding this is to assume that Andromeda is on a highly elliptical orbit \(e \to 1 \) 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.

First off, we establish a few constants that will serve us throughout the problem.
\[G_{(gravitational constant)}=6.674\times10^{-8}\frac{cm^3}{g s^2}\]   and the distance between galaxies (a) is \[D_{Galaxies}=800kpc=2.475\times10^{24}cm\], as well as the mass of both galaxies is \(10^{12}\) solar masses which is equal to \(10^{45} g\), since the mass of the Sun is \(10^{33} grams\).
Starting off, the problem already indicates the use of Kepler’s Third law: \[P^2=\frac{4\pi^2 a^3}{G(M_{MW}+M_{And})}\] and how plugging in for different values once isolated could effectively calculate the time in free fall \(t_{ff}\) for Andromeda to fall and merge with the Milky Way Galaxy.
Having established that the ellipse describing the orbit/period of the fall of Andromeda is \(e \to 1 \), that means the semi-major axis of the ellipse, the line that can be drawn from the furthest two points on an ellipse, is virtually identical to the orbit of Andromeda, so it is traveling in a straight line across \(D_{Galaxies}=800kpc=2.475\times10^{24}cm\). Therefore, if the period of the orbit is the time it takes to return to the point where it began (there and back again), then reaching the other end of \(D_{Galaxies}\) is \(\frac{1}{2}Period\), and concurrently the \(Period=2t_{ff}\).
With these values and reasoning established, we may begin to solve the equation to find \(t_{ff}\). Exchanging values in increments, we start with \[(2t_{ff})^2=\frac{4\pi^2a^3}{G(M_{MW} + M_{And})}\] then \[4t_{ff}^2=\frac{4\pi^2 a^3}{G(M_{MW} + M_{And})} \] and cancel 4 on both sides \[t_{ff}^2 = \frac{\pi^2 a^3}{G(M_{MW} + M_{And})} \] and finally placing the whole equation in terms of \(t_{ff}\), \[t_{ff} = \left(\frac{\pi^2 a^3}{G(M_{MW} + M_{And})}\right)^{\frac{1}{2}}\]
Now  plug in the values and constants we had established earlier and we have:
\[t_{ff} = \left(\frac{\pi^2 (2.475\times10^{24}cm)^3}{6.674\times10^{-8}\frac{cm^3}{g s^2}(2\times10^{45}g)}\right)^\frac{1}{2}\]
And simplifying all values yields the following result:
\[t_{ff} = 1.059\times10^{18}s\] and by dimensional analysis we can find \[t_{ff}=3.36\times10^{10}years\].

We are bounded in a nutshell of Infinite space: Blog Post #2, Worksheet # 1.1, Problem #3: Stars in your backyard

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: \[n(r) \alpha e^{\frac{-r}{R_s}}\]  \(R_s\) 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.
(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 \(0.5M_\odot\).

(a) In the Sun’s vicinity, there are 5 other stars, so in a 2 parsec radius sphere, the amount of stars is 5. From this simple understanding of amount and space which it occupies, a density can be understood. The volume of a sphere is \(V_{sphere} = \frac{4 \pi r^2}{3}\) and if 2 parsecs is plugged in, the volume would be \(33.5 pc^3\). The density, then, would be \(5 suns\) divided by \(33.5 pc^3\), giving the result: \[Density_{stellar}=.149\frac{suns}{pc^3}\]. Now that these values have been established, we can now use the equation: \[n(r)=ke^{\frac{-r}{R_s}}\]. By placing the constant k, the formula becomes an inequality instead of a proportionality. The k in this equation is what we want to find in the first iteration, substituting all the other values to find the constant that describes the density of stars in the galaxy. Here, the substitution would yield: \[n(2)=ke^{\frac{-8 kpc}{3.5 kpc)}}\],  where n(2) is the \(Density_{stellar}\) we just calculated \(\left(.149\frac{suns}{pc^3}\right)\), k is the looked for constant, -r turns into -8 kpc for describing the distance of the sun from the center of the galaxy, and \(R_s\) is the Milky Way Scale radius (3.5 kpc), the distance where largest concentration of stars exists, which was established in the question. This equation solved yields: \[0.149=0.102k\] \[k= 1.529\].
Now this number, k, is plugged into the original equation but solving for n(3.5): \[n(3.5) =1.529e^{\frac{-3.5 kpc}{3.5 kpc}}\] \[n(3.5) =.562\frac{suns}{pc^3}\].

(b) The other way to calculate the stellar density is by directly finding it in the entire galaxy, by means of conventional geometric analysis. First, we establish the constants: \[Scale Height_{MW} = 330 pc\] \[Mass_{Stars in MW} = 10^{10} M_\odot\] \[Scale Radius_{MW}= 3500 pc\]. The next step would be to decide what geometric figure has height and radius: a cylinder, whose volume equation is \[V_{cylinder} = \pi r^2 h \]. There are now two processes that could be followed, volume in \(pc^3\) or \(cm^3\). First, with \(pc^3\) as our goal, the volume equation previously mentioned would have its values substituted and look like: \[V_{MW} = \pi (3500 pc)^2 (330pc) \]. And solved: \[V_{MW}=1.123\times10^{10} pc^3 \]. Now we know this value, we can find the stellar density (n) by dividing the amount of stars in the galaxy and the volume of the galaxy. We also know that a typical stellar mass is \(0.5 M_\odot\), so if  \(Mass_{Stars in MW} = 10^{10} M_\odot\), then there are \( 2 x 10^{10} stars\) in the Milky Way. Dividing: \[n = \frac {Stars_{MW}} {V_{MW}} \], and substituting the values previously obtained:  \[n = \frac {2 \times 10^{10} stars} {1.123 \times 10^{10} pc^3 } \], then the result is \[n = 1.64 \frac {stars}{pc^3}\].

However, when solving for the stars in a \(cm^3\) (a value that is more useful in the subsequent problem), a few conversion factors must be applied. Still based on the previous constants and the equation \(V_{cylinder} = \pi r^2 h \), the problem must be solved by using the conversion factor of parsecs to cm, \(3.094 \times 10^{18} \frac{cm}{pc}\). Once again, plugging in the values we had seen previously, and adding a conversion factor, we get: \[V_{MW} = \pi (3500 pc \times 3.094 \times 10^{18} \frac{cm}{pc})^2 (330pc \times3.094 \times 10^{18} \frac{cm}{pc}) \], and the result is: \[V_{MW} = 3.761 \times 10^{65} cm^3\]. And again, using the amount of stars in the galaxy \(( 2 \times 10^{10} stars)\), and the volume we just calculated: \[n = \frac {Stars_{MW}} {V_{MW}} \], \[n = \frac {2 \times 10^{10} stars}{3.761 \times 10^{65} cm^3}\], which results in: \[n = 5.318 \times 10^{-56} \frac{stars}{cm^3}\]

We are bounded in a nutshell of Infinite space: Blog Post #2, Worksheet # 1.1, Problem #4: Is the sun safe?

Determine the collision rate of the stars using the number density of the stars (n), the cross-section for a star \(\sigma\), 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?

As always, let us establish the constants, some obtained through queries, and other value we have slowly been accumulating which will help us solve this problem. \[n = 5.318 \times 10^{-56} \frac{stars}{cm^3}\], \[D_{Galaxies} = 800kpc = 2.475 \times 10^{24}cm , \] \[t_{ff} = 1.059 \times 10^{18} s\], and the radius of the sun = 696,300 km = 69,630,000,000 cm. First step, calculate the cross section of a star \(\sigma\), which can be approximated to the Sun’s cross section. The cross section of a sphere is simply the area of the circle with the sphere’s radius, so simply \[\sigma = \pi r^2 \] \[\sigma = \pi (6.963 \times 10^{10} cm)^2 \] \[\sigma = 1.52 \times 10^{22} cm^2 \]. Next, we calculate the speed of the colliding galaxy, by using the average velocity formula: \[v_{Avg} = \frac{distance}{time} \], and in this case, \(D_{Galaxies}\) and \(t_{ff}\) are the time and distance. Plugging these values in, we have: \[v_{Avg} = \frac{2.475\times10^{24}cm }{1.059 \times 10^{18} s } \], which equals: \[v_{Avg} = 2.34 \times 10^6 \frac{cm}{s} \]

Now, this step in the process is to understand how a collision rate can be obtained from n, \(\sigma\) and \(v_{Avg}\) can be turned into a K (rate) of collision. By analyzing the different values, and their units, we see that if all three values were multiplied, the units of \(cm and cm^2\) cancel out \(cm^3\). Thus, the equation and its plug-in procedure becomes: \[K_{Collision} = (n) (\sigma) (v_{Avg}) \] \[K_{Collision} = (5.318 \times 10^{-56} \frac{stars}{cm^3}) (1.52 \times 10^{22} cm^2) (2.34 \times 10^6 \frac{cm}{s}) \] \[K_{Collision} = 1.892 \times 10^{-27} \frac{stars}{s}\]. Using dimensional analysis to turn seconds into years by multiplying it by 60 x 60 x 24 x 365, the value becomes: \[K_{Collision} = 5.965 \times 10^{-20} \frac{stars}{years} \]. Ultimately, the chance of two stars colliding during the combination of the Andromeda and Milky Way galaxies that there is no reason to fret the Sun will collide with another star any time in the foreseeable future.

With resounding joy we exclaim: The Sun is safe!