2. How long will it take
for Andromeda to collide with the Milky Way? The time-scale here is the
free-fall time, tff. A
way of finding this is to assume that Andromeda is on a highly elliptical orbit
(e à 1) around the Milky
Way. With this assumption, we can use Kepler’s Third Law \[ P^2 = \frac{4 π^2 a^3}{G(M_{MW} + M_{And})} \]where P is the period of the orbit and a is the semi-major axis. How does tff relate to the period? Estimate it to an order of magnitude.
The key to understanding this problem is the assumption that Andromeda is on a highly elliptical orbit around the Milky Way. A normal elliptical orbit might look something like this:
Because the elliptical orbit of Andromeda is stretched to the point where the eccentricity is nearly one, the orbit can instead essentially be represented by a straight line, with the ellipse's focal points at either end of the line. The Milky Way Galaxy is thus stationery at one end, and Andromeda travels up and down the line, from the Milky Way to the far end of the orbit.
The period, or the time it takes for Andromeda to complete one orbit, is the time it takes Andromeda to travel this line twice - once from the far end to the Milky Way, and once back. The free-fall time, tff, or the time it would take Andromeda to collapse into the Milky Way due to gravity, is simply the time it would take Andromeda to travel from the far end of its orbit to the Milky Way, and thus the the time it would take for Andromeda to travel this line once. The relationship between the period and the free-fall time is therefore \[ t_{\text{ff}} = \frac{P}{2} \]
We can estimate the free-fall time using a combination of this relationship, Kepler's Third Law (above), and given values.
\[ P= (\frac{4 π^2 a^3}{G(M_{MW} + M_{And})})^{\frac{1}{2}} \]
Plug this equation for the period into the previous equation for the free-fall time to yield a new equation that relates the free-fall time to given values.
\[ t_{\text{ff}} = \frac{P}{2} \]
\[ t_{\text{ff}} = \frac{1}{2} (\frac{4 π^2 a^3}{G(M_{MW} + M_{And})})^{\frac{1}{2}} \]
Now this equation can be evaluated using given values to estimate tff to an order of magnitude.
\[ a = (800 \text{ kpc }) (\frac{3 *10^{19} \text{ m }}{1 \text{ kpc }}) ≈ 2 * 10^{22} \text{ m}\]
\[ M_{MW} = M_{And} = (10^{12} \text{ solar masses}) (\frac{10^{33} \text{ g}}{1 \text{ solar mass}})(\frac{1 \text{ kg}}{10^3 \text{ g}}) ≈ 10^{42} \text{ kg}\]
\[ t_{ff} = \frac{1}{2} (\frac{4 π^2 (2 *10^{22} \text{ m})^3}{(6.67 * 10^{-11} \text{ Nm}^2\text{/kg}^2)(2 * 10^{42} \text{ kg})})^{\frac{1}{2}} \]
Canceling the units shows that the answer will have the SI unit of time, seconds.
\[ t_{\text{ff}} = \frac{1}{2} (\frac{4 * 10 * (8 *10^{66})}{(7* 10^{-11})(2 * 10^{42})})^{\frac{1}{2}} \text{ s}\]
\[ t_{\text{ff}}= \frac{1}{2} (\frac{3 *10^{68}}{1 * 10^{32}})^{\frac{1}{2}} \text{ s}\]
\[t_{\text{ff}} = \frac{1}{2} (3 * 10^{36})^{\frac{1}{2}} \text{ s}\]
\[ t_{\text{ff}} = \frac{1}{2} (1.7 * 10^{18}) \text{ s}\]
\[ t_{\text{ff}} = (9 * 10^{17} \text{ s) } (\frac{1 \text{ year}}{3 * 10^7 \text{ s}}) = 3 * 10^{10} \text{ years}\]
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: \[ n(r) \varpropto exp(-r/R_s) \] Rs 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, \(\alpha\) Centauri A and B, Proxima Centauri, and Barnard's Star.
The equation given above for the distribution of stars tells us that as we travel a distance r away from the center of the galaxy, the density of stars decreases exponentially in proportion with \( e^{-r/R_s} \). Because the equation gives proportionality (using \( \varpropto \) instead of =), in order to determine an actual density n(r), we need to determine the proportionality constant n0 that will give us the correct relation between distance, r, and density, n.
\[ n(r) = n_0e^{-r/R_s} \]
In part A, we are given the information needed to determine the number density of stars near the Sun. Thus, the number density of stars at a distance of \(r = R_{\text{GC}} = 8 \) kpc away from the galactic center can be represented as \( n_\odot \), or the number density of stars at the Sun. If we plug in these values to the above equation for density, we get:
\[ n(R_{\text{GC}}) = n_\odot = n_0e^{-R_{\text{GC}}/R_s} \]
As the scale radius of the galaxy, Rs, is 3.5 kpc, \(R_{\text{GC}} = 8 \) kpc ≈ 2 Rs. We can substitute this approximation into the equation to get:
\[ n_\odot = n_0e^{-2R_s/R_s} \] \[ n_\odot = n_0e^{-2} \] \[ n_0 = e^2 n_\odot \]
To determine \( n_\odot \), we use the fact that there are five stars within a spherical volume around the Sun with a radius of 2 pc.
\[ n_\odot = \frac{ \text{ number of stars}}{\text {volume of sphere}} \] \[ n_\odot = \frac{5 \text{ stars}}{\frac{4}{3}πR_{\text{sun}}^3} \]
Plugging this equation for \( n_\odot \) into the equation for \( n_0\), we get:
\[ n_0 = e^2 (\frac{\text{ number of stars}}{\frac{4}{3}πR_{\text{sun}}^3}) = e^2 (\frac{5 \text{ stars}}{\frac{4}{3}π(2 \text{ pc})^3}) \] \[ n_0 ≈ 1.1 \frac{\text{stars}}{\text{ pc}^3}\] As the large majority of stars in the galaxy are contained within a relatively small volume at the center of the galaxy, the value for \( n_0 \) can be used as an approximation for n.
\[ n = 1 \frac{\text{star}}{\text{ pc}^3}\]
(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\( M_\odot \).
The key to Part B is the fact that most of the Galaxy's stars are contained within the volume defined by the given "scale lengths." As this is where most of the stars are, the average number density of the Galaxy can be estimated by determining the average number density within that volume. To determine this, we calculate the number density within the cylinder as if the density were uniform throughout the cylinder. The problem gives a "scale height" of 330 pc and a "scale radius" of 3.5 kpc. Therefore, the volume at the center of the galaxy will be a cylinder with a height of 330 pc and a radius of 3.5 kpc.
The key to understanding this problem is the assumption that Andromeda is on a highly elliptical orbit around the Milky Way. A normal elliptical orbit might look something like this:
Because the elliptical orbit of Andromeda is stretched to the point where the eccentricity is nearly one, the orbit can instead essentially be represented by a straight line, with the ellipse's focal points at either end of the line. The Milky Way Galaxy is thus stationery at one end, and Andromeda travels up and down the line, from the Milky Way to the far end of the orbit.
We can estimate the free-fall time using a combination of this relationship, Kepler's Third Law (above), and given values.
Kepler's Third Law gives an equation for the square of the period. Take the square root of both sides to give an equation for the period.
\[ P^2 = \frac{4 π^2 a^3}{G(M_{MW} + M_{And})} \]\[ P= (\frac{4 π^2 a^3}{G(M_{MW} + M_{And})})^{\frac{1}{2}} \]
Plug this equation for the period into the previous equation for the free-fall time to yield a new equation that relates the free-fall time to given values.
\[ t_{\text{ff}} = \frac{P}{2} \]
\[ t_{\text{ff}} = \frac{1}{2} (\frac{4 π^2 a^3}{G(M_{MW} + M_{And})})^{\frac{1}{2}} \]
Now this equation can be evaluated using given values to estimate tff to an order of magnitude.
\[ a = (800 \text{ kpc }) (\frac{3 *10^{19} \text{ m }}{1 \text{ kpc }}) ≈ 2 * 10^{22} \text{ m}\]
\[ M_{MW} = M_{And} = (10^{12} \text{ solar masses}) (\frac{10^{33} \text{ g}}{1 \text{ solar mass}})(\frac{1 \text{ kg}}{10^3 \text{ g}}) ≈ 10^{42} \text{ kg}\]
\[ t_{ff} = \frac{1}{2} (\frac{4 π^2 (2 *10^{22} \text{ m})^3}{(6.67 * 10^{-11} \text{ Nm}^2\text{/kg}^2)(2 * 10^{42} \text{ kg})})^{\frac{1}{2}} \]
Canceling the units shows that the answer will have the SI unit of time, seconds.
\[ t_{\text{ff}} = \frac{1}{2} (\frac{4 * 10 * (8 *10^{66})}{(7* 10^{-11})(2 * 10^{42})})^{\frac{1}{2}} \text{ s}\]
\[ t_{\text{ff}}= \frac{1}{2} (\frac{3 *10^{68}}{1 * 10^{32}})^{\frac{1}{2}} \text{ s}\]
\[t_{\text{ff}} = \frac{1}{2} (3 * 10^{36})^{\frac{1}{2}} \text{ s}\]
\[ t_{\text{ff}} = \frac{1}{2} (1.7 * 10^{18}) \text{ s}\]
\[ t_{\text{ff}} = (9 * 10^{17} \text{ s) } (\frac{1 \text{ year}}{3 * 10^7 \text{ s}}) = 3 * 10^{10} \text{ years}\]
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: \[ n(r) \varpropto exp(-r/R_s) \] Rs 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, \(\alpha\) Centauri A and B, Proxima Centauri, and Barnard's Star.
The equation given above for the distribution of stars tells us that as we travel a distance r away from the center of the galaxy, the density of stars decreases exponentially in proportion with \( e^{-r/R_s} \). Because the equation gives proportionality (using \( \varpropto \) instead of =), in order to determine an actual density n(r), we need to determine the proportionality constant n0 that will give us the correct relation between distance, r, and density, n.
\[ n(r) = n_0e^{-r/R_s} \]
In part A, we are given the information needed to determine the number density of stars near the Sun. Thus, the number density of stars at a distance of \(r = R_{\text{GC}} = 8 \) kpc away from the galactic center can be represented as \( n_\odot \), or the number density of stars at the Sun. If we plug in these values to the above equation for density, we get:
\[ n(R_{\text{GC}}) = n_\odot = n_0e^{-R_{\text{GC}}/R_s} \]
As the scale radius of the galaxy, Rs, is 3.5 kpc, \(R_{\text{GC}} = 8 \) kpc ≈ 2 Rs. We can substitute this approximation into the equation to get:
\[ n_\odot = n_0e^{-2R_s/R_s} \] \[ n_\odot = n_0e^{-2} \] \[ n_0 = e^2 n_\odot \]
To determine \( n_\odot \), we use the fact that there are five stars within a spherical volume around the Sun with a radius of 2 pc.
\[ n_\odot = \frac{ \text{ number of stars}}{\text {volume of sphere}} \] \[ n_\odot = \frac{5 \text{ stars}}{\frac{4}{3}πR_{\text{sun}}^3} \]
Plugging this equation for \( n_\odot \) into the equation for \( n_0\), we get:
\[ n_0 = e^2 (\frac{\text{ number of stars}}{\frac{4}{3}πR_{\text{sun}}^3}) = e^2 (\frac{5 \text{ stars}}{\frac{4}{3}π(2 \text{ pc})^3}) \] \[ n_0 ≈ 1.1 \frac{\text{stars}}{\text{ pc}^3}\] As the large majority of stars in the galaxy are contained within a relatively small volume at the center of the galaxy, the value for \( n_0 \) can be used as an approximation for n.
\[ n = 1 \frac{\text{star}}{\text{ pc}^3}\]
(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\( M_\odot \).
The key to Part B is the fact that most of the Galaxy's stars are contained within the volume defined by the given "scale lengths." As this is where most of the stars are, the average number density of the Galaxy can be estimated by determining the average number density within that volume. To determine this, we calculate the number density within the cylinder as if the density were uniform throughout the cylinder. The problem gives a "scale height" of 330 pc and a "scale radius" of 3.5 kpc. Therefore, the volume at the center of the galaxy will be a cylinder with a height of 330 pc and a radius of 3.5 kpc.
Using the formula for the volume of a cylinder, \( V = πr^2h \), where r is the radius of the cylinder and h is the height of the cylinder, we get: \[ V_{\text{cylinder}} = πR_{\text{s}}^2H_{\text{s}}\]
We can then use the density formula, \(D = \frac{\text{mass}}{\text{volume}} \), and the given value for the stellar mass of the galaxy, \(10^{10 } M_{\odot}\) to determine the density within the cylinder.
\[ D = \frac{\text{mass of Galaxy}}{\text{volume of cylinder}} = \frac{10^{10 } M_{\odot}}{V_{\text{cylinder}}} = \frac{10^{10 } M_{\odot}}{πR_{\text{s}}^2H_{\text{s}}} \]
To convert the density from its current units of \(\frac{M_{\odot}}{\text{pc}^3}\) to a number density with units of \( \frac{\text{stars}}{\text{pc}^3}\), multiply by the given conversion factor:
\[ D = \frac{10^{10 } M_{\odot}}{πR_{\text{s}}^2H_{\text{s}}}(\frac{1 \text{ star}}{0.5 M_{\odot}}) \]
Now this equation can be evaluated by plugging in the given values for the scale radius and the scale height: \[ D = \frac{10^{10 } M_{\odot}}{π(3.5 * 10^3 \text{pc})^2(330 \text{ pc})}(\frac{1 \text{ star}}{0.5 M_{\odot}}) \] \[ D = \frac{2 * 10^{10 }}{3(4 * 10^3)^2(300)}\frac{\text{ stars}}{ \text{ pc}^3} \] \[D ≈ 1.4 \frac{\text{ stars}}{ \text{ pc}^3} \] As the problem states that the volume of the cylinder given by the scale lengths contains most of the Galaxy's stars, the average number density within the cylinder can be used as an approximation for n, the average number density of stars throughout the Milky Way. \[ n = 1 \frac{\text{star}}{\text{ pc}^3}\]
Both methods of calculating n yield approximately the same value.
4. 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?
To visualize this problem, picture a single star with a cross section of area \( \sigma_*\) traveling through space with a velocity v distance/time. Over the course of one time unit - for example, one second, if measuring velocity in m/s - the star will have traveled a distance of v and crossed through a volume \( \sigma_*v\). If the star is traveling through an area with a number of other stars evenly distributed with number density n, the number of stars the original star collides with will be equal to \( n\sigma_*v\).
To check this formula with units, we can look at the units of the information given and the information wanted. The answer we are looking for, number of stars that will collide every year, has the unit of \( \frac{\text{stars}}{\text{year}}\). The problem gives three quantities to consider - 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.
- The number density of stars, calculated in question 3, has the units \( \frac{\text{stars}}{\text{pc}^3}\).
- The cross-section for a star, or the area of a circle, has units of 2-dimensional area, or \( \text{pc}^2\).
- The average velocity of Milkomeda's stars as they collide, v, has units of distance/time, or \( \frac{\text{pc}}{\text{years}}\).
Multiplying these units together according to the formula above yields the correct units for the answer we are looking for:
\[ (\frac{\text{stars}}{\text{pc}^3})( \frac{\text{pc}^2}{1})(\frac{\text{pc}}{\text{year}})= \frac{\text{stars}}{\text{year}}\]
Now that we know the formula will yield the correct units, we can plug in the values for n,\( \sigma_*\), and v as given or calculated in previous problems:
- From question 3, n = \( 1 \frac{ \text{star}}{\text{pc}^3}\).
- Using the formula for the area of a circle, A = π\(r^2\), with r as the radius of the circle, \( \sigma_*\) = \( π(R_\odot)^2\). \( R_\odot\) is given as \( \frac{1}{200}\) AU, and 1 pc = 2 * \(10^5\) AU. Therefore, \[ R_\odot = (\frac{1 \text{ AU}}{200})(\frac{1 \text{ pc}}{2 * 10^5 \text{AU}}) = 2.5 * 10^{-8} \text{ pc}\]
- Average velocity is the distance traveled / time. The distance between Andromeda and the Milky Way is d = 800 kpc, and the time was calculated in question 2 as \(t_{\text{ff}} = 3 * 10^{10}\) years. Thus the average velocity v = \( \frac{\text{d}}{t_{\text{ff}}}\).
Plugging these values into the formula we get:
\[ \text{collision rate} = n\sigma_*v\]
\[ \text{collision rate} = (n)(πR_\odot^2)(\frac{\text{d}}{t_{\text{ff}}})\]
\[ \text{collision rate} = (\frac{1 \text{star}}{\text{pc}^3})(\frac{π(2.5 * 10^{-8} \text{pc})^2}{1})(\frac{800 * 10^3 \text{ pc}}{3 * 10^{ 10} \text{years}})\]
\[ \text{collision rate} = \frac{3 (2.5 * 10^{-8})^2(800 * 10^3)}{3 * 10^{10}} \frac{\text{stars}}{\text{year}}\]
\[ \text{collision rate} = 5 * 10^{-20} \frac{\text{stars}}{\text{year}}\]
As the collision rate is so extremely slow, the Sun is essentially safe, as it is will never collide with another star within a comprehensible time frame.
Q2:
ReplyDeleteYou seem to have an in-depth understanding of Kepler’s Law and conic sections! Do note that. if we assume the two galaxies are currently at the farthest distance from each other, as you have depicted in your drawing, then semi-major axis should be *half* the current distance separating the two galaxies (800 kpc / 2).
4.5
Q3:
ReplyDeleteVery nice! Just be aware that the scale height extends both above and below the Milky Way mid-plane!
5
Q4:
ReplyDeleteGreat! Just remember to double the stellar density (Milky Way + Andromeda!)
5
Also, very nice graphics! How did you make them?
ReplyDeletePowerpoint! And a little bit of Photoshop.
Delete