Blog #2: Milkomeda

2) How long will it take for Andromeda to collide with the Milky Way? The time-scale here is the free-fall time, $t_{ff}$.  A way of finding this is to assume that Andromeda is on a highly elliptical orbit $(e \rightarrow 1)$ around the Milky Way. With this assumption, we can use Kepler's Third Law:

\begin{align}
P^2 = \frac{4\pi^2 a^3}{G(M_{MW} + M_{And})}
\end{align}

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.

Assuming that Andromeda is on a highly elliptical orbit around the Milky Way since the eccentricity is approaching 1 $(e \rightarrow 1)$, it is approximately a straight line free fall between Andromeda and the Milky Way. We can use the Milky Way as a fixed point on one side of the ellipse because that is our frame of reference. We can use Andromeda on the opposite end of the ellipse because we want to model the free-fall effects of Andromeda onto the Milky Way. It is useful to know that Andromeda is 800 kpc away from the Milky Way. Some other useful values to know are:

\begin{align}
a = 800\text{ kpc} = 2.5 \times 10^{24} \text{ cm}\\
G = 6.67 \times 10^{-8} \frac{cm^3}{g \cdot s^2}\\
M_{MW} + M_{And} = 2 \times 10^{45} g
\end{align}

Using Kepler's Third Law, we know that an an object orbits another object in time $P$. Since Andromeda is half-way through it's highly elliptical orbit, the free fall time, $t_{ff} = \frac{1}{2} P$. In order to find the $t_{ff}$, we can use Kepler's Third Law to solve for $P$.

\begin{align}
t_{ff} &= \frac{1}{2} P\\
t_{ff} &= \frac{1}{2} \left( \frac{4\pi^2 a^3}{G(M_{MW} + M_{And})} \right)^{\frac{1}{2}}\\
t_{ff} &= \frac{1}{2} \left( \frac{4\pi^2 (2.5 \times 10^{24} \text{ cm})^3}{6.67 \times 10^{-8} \frac{cm^3}{g \cdot s^2}(2 \times 10^{45}\text{ g})} \right)^{\frac{1}{2}}\\
t_{ff} &= 1.06 \times 10^{18} \text{ seconds}
\end{align}


The free fall time of the in which Andromeda will collide with the Milky Way is $1.06 \times 10^{18}$ seconds, which is $3 \times 10^{10}$ 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:

\begin{align}
n(r) \propto exp(-\frac{r}{R_s})
\end{align}

$R_s$ is also known as the "scale radius" of the galaxy. The Milky Way has a scale radius of 3.5 kpc. With that in mind, estimate $n$ in two ways:

• Consider that within a 2 pc radius of the Sun there are five stars: the Sun, $\alpha$ Centauri A and B, Proxima Centauri, and Barnard's Star. 
• 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 $0.5M_{\odot}$.

Let's consider the first case, where the frame of reference is the Sun, with 4 other stars within 2 pc: 

Given the equation $n(r) \propto exp(-\frac{r}{R_s})$, where $n(r)$ is the number density ($ n = \frac{\text{# of items}}{\text{volume items occupy}}$) of the number of stars as a function of radius, $r$, and $R_s$ as the scale radius of the galaxy, we can solve for the number density at the center of the galaxy, $n_0$.

But wait, we need more information! Even if we are using the Sun as a frame of reference to other stars, we still need to know how far the Sun is away from the galactic center, $r$. A quick Google search will tell us that the Sun is about 8 kpc away from the galactic center, so $r = 8 kpc$ We also know that the scale radius of the Milky Way is $R_s = 3.5$. Using this information, we can calculate the number density of stars at the galactic center. 

\begin{align}
n &= n_0 e^{-\frac{r}{R_s}} \\
n_0 &= n e^{\frac{r}{R_s}}\\
n_0 &= \frac{\text{# of stars}}{\text{volume stars occupy}}e^{\frac{r}{R_s}}\\
n_0 &= \frac{5 \text{ stars}}{\frac{4}{3}\pi \cdot (2 \text{pc})^3}e^{\frac{8 \text{ kpc}}{3.5\text{ kpc}}}\\
n_0 &\approx 1.5 \frac{\text{ stars}}{\text{ pc}^3}
 \end{align}

Therefore, using the Sun as a frame of reference, we know that the number density at the center of the galaxy is $n_0 = 1.5$ stars for every cubic parsec. 

In this model, we assumed that the majority of stars are located in the center of the galaxy, which is why we can assume that the number density at the center, $n_0$ is the highest, and decreases exponentially the further away from the center you go. So, for example, knowing that the Milky Way Galaxy has a radius, $r = 15$ kpc, we can use the number density function, $n(r)$ to solve for the number density over the entire galaxy:

\begin{align}
n(r) &= n_0 e^{-\frac{r}{R_s}} \\
n(15) &= 1.5 \frac{\text{ stars}}{\text{ pc}^3}  e^{-\frac{15}{3.5}}\\
n(15) &\approx 0.02 \frac{\text{ stars}}{\text{ pc}^3}
\end{align}

This shows that the number density of stars over the entire galaxy is very small, with about 0.02 stars for every cubic parsec

Let's consider the other situation of finding the galactic number density using the "scale height". In order to use the "scale height", we can visualize the Milky Way as a cylindrical shape as follows:

In this case, the majority of the stars are in this cylinder. We can use the equation we derived in the first part to get the number density of stars in this cylinder using the equation for a volume of a cylinder, $V = \pi r^2 h$. In this case, the height of the cylinder, $h$ can be represented by the scale height of the galaxy, $h = H_s$, and the radius of the cylinder, $r$ can be represented by the scale radius, $r = R_s$. Also, the number of stars in the entire galaxy can be obtained by using the total mass of the Milky Way, $M_{MW} = 10^{10} M_{\odot}$ and the mass of the average star in the Milky Way, $M_* = 0.5 M_{\odot}$. Finally, since we are using the cylindar as the boundary for stars, the radius, $r = R_s$. Using this information, we can get the number density of stars in the center of this cylindrical galaxy as follows:

\begin{align}
n_0 &= n e^{\frac{r}{R_s}}\\
n_0 &= \frac{\text{# of stars}}{\text{volume stars occupy}}e^{\frac{r}{R_s}}\\
n_0 &= \frac{\frac{M_{MW}}{M_*}}{\pi R_s^2 H_s}e^{\frac{R_s}{R_s}}\\
n_0 &= \frac{\frac{10^{10} M_{\odot}}{0.5 M_{\odot}}}{\pi (3500 \text{ pc})^2 (330 \text{ pc})}e\\
n_0 &= \frac{2 \times 10^{10} \text{ stars}}{1.3 \times 10^{10}\text{ pc}^3}e\\
n_0 &\approx 4.3 \frac{\text{ stars}}{\text{ pc}^3}
\end{align}

Based on the above calculation, the number density of the stars in this cylinder of scale radius 3500 pc of the galaxy is about 4.3 stars per cubic parsec. 

In order to determine the number density of stars over the entire galaxy, we can use the number density function, $n(r)$ to calculate the number density of the Milky Way over its entire radius, $r = 15 kpc$ as follows:

\begin{align}
n(r) &= n_0 e^{-\frac{r}{R_s}} \\
n(15) &= 4.3 \frac{\text{ stars}}{\text{ pc}^3}  e^{-\frac{15}{3.5}}\\
n(15) &\approx 0.03 \frac{\text{ stars}}{\text{ pc}^3}
\end{align}

This cylindrical model shows that at over the entire galaxy, the number density of stars is 0.03 stars per cubic parsec.

4) Determine the collision 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 $\overline{v}$.

How many stars will collide every year? Is the Sun safe, or likely to collide with another star?

In order to figure out the collision rate of stars, $K_{collision}$, we can use the number density, $n = 0.03 \frac{\text{ stars}}{\text{ pc}^3}$, the cross-sectional area of a star, such as the Sun, which is $\sigma_{*}$, and the average velocity of Milkomeda's stars, $\overline{v}$. The relationship between $K_{collision}, n, \sigma_{*}$ and $\overline{v}$ can be modeled by the equation:

\begin{align}
K_{collision} = n \times \sigma_{*} \times \overline{v}
\end{align}

$ \sigma_{*}$ is the cross-sectional area of a star, which can be calculated using $A = \pi R_{\odot}^2$. We know that the radius of of the Sun is$R_{\odot} = 6.96 \times 10^{10}$, so we can calculate $\sigma_{*}$ as follows:

\begin{align}
\sigma_* &= \pi R_{\odot}^2\\
\sigma_* &= \pi (6.96 \times 10^{10})^2\\
\sigma_* &= 1.52 \times 10^{22} \text{ cm}^2
\end{align}

We can also calculate the average velocity of stars as they collide using $\overline{v} = \frac{\text{distance}}{\text{time}}$, where the distance between the Milky Way and Andromeda is $800 \text{ kpc} = 2.5 \times 10^{24} \text{ cm}$, and the time to the collision is $t_{ff}$ calculated earlier.

\begin{align}
\overline{v} &= \frac{\text{distance}}{\text{time}}\\
\overline{v} &= \frac{2.5 \times 10^{24} \text{ cm}}{1.06 \times 10^{18} \text{ seconds}}\\
\overline{v} &= 2.36 \times 10^{6} \frac{\text{ cm}}{\text{ seconds}}
\end{align}

Now that we know both $\overline{v}$ and $\sigma_*$, we can calculate $K_{collisions}$ as follows:

\begin{align}
K_{collision} &= n \times \sigma_{*} \times \overline{v}\\
K_{collision} &= (0.03 \frac{\text{ stars}}{\text{ pc}^3}) \times (1.52 \times 10^{22} \text{ cm}^2) \times (2.36 \times 10^{6} \frac{\text{ cm}}{\text{ seconds}})\\
K_{collision} &= (1.01 \times 10^{-57 } \frac{\text{ stars}}{\text{ cm}^3}) \times (1.52 \times 10^{22} \text{ cm}^2) \times (2.36 \times 10^{6} \frac{\text{ cm}}{\text{ seconds}})\\
K_{collision} &= 3.7 \times 10^{-29}\frac{\text{ stars}}{\text{ second}}
\end{align}

This shows that during the collision of Milkomeda, the chance of stars colliding is on the scale of $3.7 \times 10^{-29}\frac{\text{ stars}}{\text{ second}}$, or $1.17 \times 10^{-21}$ collisions per year!!! That is effectively zero, so the chances of the Sun colliding with another star are close to 0.