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We will assume that the galaxy is a sphere where stars orbit around the center on different radii. Since there are so many stars, we will not deal with individual stars, but we will assume that the density of the galaxy decreases exponentially from the center outwards as $\rho(r)=\rho_{c}e^{-\frac{r}{r_{0}}}$ We know the characteristic radius $r_{0}$ and the density at the center $\rho_{c}$. Calculate the rotation curve of the galaxy $v(r)$, that is, the dependence of the circumferential velocity of the stars on the dependence from the radius on which they circulate.

In fact, the observations show a rotation curve that is straight. This means that $v(r)= constant$. This can be explained by the concept of dark matter. It is a substance that we cannot detect, we can only see its gravitational effects. In the previous part of the task, we considered only the “visible” mass, now we will add ˇse dark matter. Assume that dark matter is also distributed after galaxy spherically symmetrically and that its density varies with radius. Calculate the mass of the dark matter within a radius as a function of radius for that galaxy.

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  • $\begingroup$ A clear example of a homework question. $\endgroup$
    – ProfRob
    Jan 5, 2021 at 8:28

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The first question seems more complex to me than the second question. An answer to the second question arises naturally out of the work needed to answer the first question.

Imagine the galaxy as the sum of an onion-like collection of infinitesimally thin spherical "shells" of volume, each with a thickness $dr$, surface area $A = 4 \pi r^2$, and thus volume $dV = A \, dr = 4 \pi r^2 \,dr$.

As $m = \rho V$, it follows that the infinitesimally small mass of each shell, $dm$, is $dm = \rho(r) \,dV$. Since the volume of the shall varies with its distance from the centre ($dV = 4\pi r^2 \,dr$), we rephrase $dV$ in terms of $r$ with the intention of integrating both sides - this is especially because our expression for $\rho$ varies with $r$, too.

As such, we find an expression for the mass of the galaxy at a radius $r$, $m(R)$:

$$m(R) = \int_{0}^{R} dm = \int_{0}^{R} \rho(r) dV = \int_{0}^{R} \rho(r) 4 \pi r^2\, dr$$

Where we want to evaluate the integral on the right-hand side of the above equation.

$$4 \pi \rho_c \int^{R}_{0} e^{-\frac{r}{r_0}} r^2\, dr$$

This can be done through integration by parts; take $r^2$ as the expression to differentiate, and $e^{-\frac{r}{r_0}}$ as the expression to integrate. We find that

$$4 \pi \rho_c \int^{R}_{0} e^{-\frac{r}{r_0}} r^2\, dr = 4\pi\rho_c \left[ -e^{-\frac{r}{r_0}} (r^2r_0 + 2r r_0^2 + 2 r_0^3) \right]_{0}^{R} $$

Get rid of the $-1$ factor by reversing the limits:

$$= 4\pi\rho_c \left[ e^{-\frac{r}{r_0}} (r^2r_0 + 2r r_0^2 + 2 r_0^3) \right]^{0}_{R} $$

Factor out one $r_0$ term:

$$= 4\pi\rho_c r_0 \left[ e^{-\frac{r}{r_0}} (r^2 + 2r r_0 + 2 r_0^3) \right]^{0}_{R} $$

Evaluate bounds, and simplify ($e^0 = 1$, et cetera):

$$m(R) = 4\pi\rho_c r_0 \left[ 2r_0^2 - e^{-\frac{R}{r_0}}(R^2 + 2Rr_0 + 2r_0^2) \right]$$

That is the hard part of the problem done. Consider that the mass can be modelled as being concentrated at the centre of the galaxy. Consider also that, as the other shells of the galaxy further out from the centre have graviational force vectors that all cancel out (i.e. the same reason why the gravitational force in a hollow spherical planet is 0), this is the only mass we need to worry about. You can take Newton's Law of Gravitation and the Circular Motion equation to get:

$$G\frac{m(r)}{r^2} = \mathrm{acceleration} = \frac{v(r)^2}{r}$$

$$v(r) = \sqrt{\frac{Gm(r)}{r}}$$

For your final answer:

$$v(r) = \sqrt{\frac{4 G \pi\rho_c r_0 \left[ 2r_0^2 - e^{-\frac{r}{r_0}}(r^2 + 2rr_0 + 2r_0^2) \right]}{r}}$$

The second question is much simpler. Consider the fact that, for $v(r)$ to be constant, $\frac{m(r)}{r}$ must be constant, so that $v(r)$ is always just some multiple of $G$.

For this to be the case, the mass of each shell must therefore be constant, as to integrate those masses with respect to $r$ would then leave you with some $m_{total}(r) = kr$, and $v(r) = \sqrt{kG}$. Suppose then, that $k$ is defined by $v(r) = \sqrt{kG}$, where $v(r)$ is the observed circumferential velocity. As $m_{total}(r) = kr$, and $m_{total}(r) = m(r) + m_{dark\,matter}(r)$; $m_{dark\,matter}(r) = kr - m(r)$.

$$m_{dark\,matter}(r) = kr - \left( 4\pi\rho_c r_0 \left[ 2r_0^2 - e^{-\frac{r}{r_0}}(r^2 + 2rr_0 + 2r_0^2) \right] \right)$$

...is thus the final answer, using the expression for $m(r)$ we derived to answer the first question.

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  • $\begingroup$ Do you mean the first question is harder? :) $\endgroup$
    – Yejus
    Jan 5, 2021 at 4:23
  • $\begingroup$ That's absolutely right - a typo on my part :) $\endgroup$
    – Leo Webb
    Jan 5, 2021 at 5:09
  • $\begingroup$ As a new contributor you should familiarise yourself with the policy on homework questions. This could easily have been set as a student assessment. $\endgroup$
    – ProfRob
    Jan 5, 2021 at 8:30
  • $\begingroup$ I did think to myself that this looks oddly like a kind of assessment but I didn't think too much of it. I will go and familiarise myself with that policy. $\endgroup$
    – Leo Webb
    Jan 5, 2021 at 9:55

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