Съдържание

Условие A. Cluster of Galaxies B. Dark Matter in a Galaxy C. Interstellar Gas and Dark Matter Решение

IPhO 2017, Теория — Задача 1. Dark Matter

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Всички материали от IPhO 2017

T1 · Dark Matter · 10 т.

General instructions: Theoretical Examination (30 points)

The theoretical examination lasts for 5 hours and is worth a total of 30 points.

Before the exam

You must not open the envelopes containing the problems before the sound signal indicating the beginning of the examination.

The beginning and end of the examination will be indicated by a sound signal. There will be announcements every hour indicating the elapsed time, as well as fifteen minutes before the end of the examination (before the final sound signal).

During the exam

Dedicated answer sheets are provided for writing your answers. Enter the final answers into the appropriate boxes in the corresponding answer sheet. For every problem, there are extra blank work sheets for carrying out detailed work. Always use the work sheets that belong to the problem you are currently working on (check the problem number in the header). If you have written something on any sheet which you do not want to be graded, cross it out. Only use the front side of every page.
In your answers, try to be as concise as possible: use equations, logical operators and sketches to illustrate your thoughts whenever possible. Avoid the use of long sentences.
Please give an appropriate number of significant digits when stating numbers.
You may often be able to solve later parts of a problem without having solved the previous ones.
A list of physical constants is given on separate page.

You are not allowed to leave your working place without permission. If you need any assistance (need to refill your drinking water bottle, broken calculator, need to visit a restroom, etc.), please draw the attention of a team guide by lifting up one of the three flags into the air ("Refill my water bottle, please", "I need to go to toilet, please", or "I need help, please" in all other cases).

At the end of the exam

At the end of the examination you must stop writing immediately.
For every problem, sort the corresponding sheets in the following order: problem sheets, answer sheets, work sheets and scratch paper.
Put all the sheets belonging to one problem into the same envelope. Also put the general instructions into the remaining separate envelope. Make sure your student code is written on each envelope. Also hand in empty sheets. You are not allowed to take any sheets of paper out of the examination area.
Leave the calculator provided by the organizers on the table.

Editorial source note: these are the shared instructions for all three theory problems. This edition contains only T1 (10 points). The separate physical-constants sheet is not present in the supplied standalone catalogue pack.

Условие

The first formal inference for the existence of dark matter was provided by Fritz Zwicky based on his observation on the dynamics of the Coma galaxy cluster, a cluster of galaxies that consist of about a thousand of galaxies. Zwicky used the Virial theorem to estimate the mass of the galaxy cluster. In a simple sun-planet system, where the planet revolves around the sun in a circular orbit, the Virial theorem states that the kinetic energy of the planet is exactly related to its gravitational potential energy. While in a general case for a system of many particles bounded by some interaction, the Virial theorem, relates the time average total kinetic energy with its time average total potential energy.

In 1933, based on his observation on the velocity of the galaxies near the edge of the Coma galaxy cluster, Zwicky estimated that the cluster has more mass than what was visually observed (i.e. the galaxies). The gravitational attraction from the observable matter (the galaxies) was too small to account for the velocities of the galaxies. Thus there must be some hidden masses that account for such a large velocity. That hidden mass is the dark matter mass. In what follows, assume that the mass of each galaxy is the sum of its visible mass and the mass of the dark matter which moves together with that galaxy, and dark matter interacts with visible matter only gravitationally.

A. Cluster of Galaxies

Consider a cluster of galaxies consist of a large number $N$ of galaxies and dark matter that are distributed homogeneously in a sphere of radius $R$ with the total mass (galaxies and dark matters) of the cluster $M$ . Assume that the average total mass (visible and dark matter) of a galaxy is $m$ .

A.1 Assuming a continuous distribution of matter in the cluster, find the total gravitational potential energy of the cluster, in terms of $M$ and $R$ . [1 т.]

A.2 Due to cosmological expansion, any distance object will be moving away from an observer on Earth with a speed that depends on the distance from the observer to the object. A certain Lyman (a hydrogen emission spectral line) frequency from a type IA supernova on the $i$ -th galaxy in the galaxy cluster is observed to be $f_i$ , with $i=1,\ldots,N$ , while the same corresponding Lyman frequency on Earth is $f_0$ .

Determine the average speed $V_{cr}$ of the whole galaxy cluster moving away from the Earth in terms of $f_i$ (with $i=1,\ldots,N.$ ), $f_0$ and $N$ . Note that a galaxy speed is very small compared to the speed of light $c$ . [0,5 т.]

A.3 By assuming that the galaxies velocities with respect to the center of the cluster are isotropic (the same in all direction), determine the root-mean square speed $v_{rms}$ of the galaxies with respect to the center of the cluster in terms of $N$ , $f_i$ (with $i=1,\ldots,N.$ ), and $f_0$ . From this result determine the mean kinetic energy of a galaxy with respect to the center of the cluster in terms of $v_{rms}$ and $m$ . [1,5 т.]

A.4 To find the total mass of the cluster, one can use Virial theorem. The theorem stated that for a system of particles bounded by their conservative force,

\langle K\rangle_t=-\gamma\langle U\rangle_t,

where $\langle K\rangle_t$ is the time average total kinetic energy, $\langle U\rangle_t$ is the time average total potential energy, and $\gamma$ is a constant. This theorem can be derived by assuming that for a system of particles bounded by its own interaction, the magnitudes of the position and momentum of each particle are finite, and thus the following quantity

\Gamma=\sum_i\vec p_i\cdot\vec r_i

is finite.

Using the fact that the time average over a long period of time of $d\Gamma/dt$ vanishes, $\langle d\Gamma/dt\rangle_t=0$ , determine $\gamma$ in the Virial theorem above for the case of gravitational interaction. (Hint: Try to do the problem with the summation in $\Gamma$ for a small finite number of galaxies). [1,7 т.]

A.5 From the previous results determine the total dark matter mass of the cluster in terms of $N$ , $m_g$ , $R$ and $v_{rms}$ , where $m_g$ is the average total visible mass of a galaxy. Note that the root-mean square speed of the dark matter is the same as that of the galaxies. [0,5 т.]

B. Dark Matter in a Galaxy

Dark matter also exists inside and around a galaxy. Consider a spherical galaxy with a visible edge radius $R_g$ (an approximate outermost distance where a large number of stars still visible, but note that a very small number of stars may still be distributed in the region beyond $R_g$ ). Assume that the stars in the galaxy are point particles with an average mass $m_s$ . The stars in the galaxy, distributed homogeneously with a number density $n$ , are assumed to move in circular orbits.

B.1 If the galaxy consists only of stars, find the speed $v(r)$ of a star as a function of its distance to the center of the galaxy and sketch $v(r)$ for $r<R_g$ and $r\ge R_g$ . [0,8 т.]

B.2 The existence of dark matter can be inferred from the galaxy rotation curve, which is a plot of $v(r)$ obtained from observations. The figure below shows a common pattern of the galaxy rotation curve. You may assume simplifyingly that $v(r)$ is a linear function for $r\le R_g$ and a constant $v_0$ for $r>R_g$ .

Galaxy rotation curve rises linearly to Rg then remains approximately at v0. — Fig. 1 Plot of galaxy rotation curve in a galaxy.

Find the total mass $m_R$ of that part of the galaxy which lies within the sphere of radius $R_g$ in terms of $v_0$ and $R_g$ . [0,5 т.]

B.3 The discrepancy between the figure in B.2 and the plot obtained in B.1 indicates the existence of dark matter.

Determine the dark matter mass density as a function of $r$ , $R_g$ , $v_0$ , $n$ , and $m_s$ for $r<R_g$ and $r\ge R_g$ . [1,5 т.]

C. Interstellar Gas and Dark Matter

Now consider a young galaxy whose mass is dominated by interstellar gas and dark matters (neglect the mass of the stars). The interstellar gas is assumed to consist of identical particles of mass $m_p$ . The number density $n(r)$ and temperature $T(r)$ of the gas depend on the distance from the center of the galaxy $r$ . Although many physical processes happen in the gas, we can assume the gas is in a hydrostatic equilibrium due to its pressure and the galaxy gravitational attraction.

C.1 Find the pressure gradient of the gas $dP/dr$ , in terms of $m'(r)$ , $r$ and $n(r)$ . Here, $m'(r)$ is the total mass of gas and dark matters within a sphere of radius $r$ from the galaxy center. [0,5 т.]

C.2 Assuming the interstellar gas as an ideal gas, find $m'(r)$ in terms of $n(r)$ , $T(r)$ and their derivatives with respect to $r$ . [0,5 т.]

C.3 Next for simplicity assume that the gas is in isothermal distribution at temperature $T_0$ and the interstellar gas number density is given by

n(r)=\frac\alpha{r(\beta+r)^2},

where $\alpha$ and $\beta$ are some constants.

Find the dark matter mass density as a function of $r$ inside the galaxy. [1 т.]

Решение

Official solutions and marking scheme. All printed grading subdivisions and alternative derivations are retained below.

A. Cluster of Galaxies

A.1 Potential energy for a system of a spherical object with mass $M(r)=\frac43\pi r^3\rho$ and a test particle with mass $dm$ at a distance $r$ is given by

dU=-G\frac{M(r)}r\,dm.

0.2 pts. Thus for a sphere of radius $R$

\begin{aligned}U&=-\int_0^RG\frac{M(r)}r\,dm=-\int_0^RG\frac{4\pi r^3\rho}{3r}4\pi r^2\rho\,dr\\&=-\frac{16}3G\pi^2\rho^2\int_0^Rr^4dr=-\frac{16}{15}G\pi^2\rho^2R^5.\end{aligned}

0.6 pts. Then using the total mass of the system $M=\frac43\pi R^3\rho$ we have

U=-\frac35\frac{GM^2}R.

0.2 pts. Total 1.0 pts. [1 т.]

A.2 Using the Doppler Effect,

f_i=f_0\frac1{1+\beta}\approx f_0(1-\beta),

where $\beta=v/c$ and $v\ll c$ . Thus the $i$ -th galaxy moving away (radial) speed is

V_{ri}=-\frac{f_i-f_0}{f_0}c.

Alternative without approximation:

f_i=f_0\frac1{1+\beta},\qquad V_{ri}=c\left(\frac{f_0}{f_i}-1\right).

0.2 pts. All the galaxies in the galaxy cluster will be moving away together due to the cosmological expansion. Thus the average moving away speed of the $N$ galaxies in the cluster is

V_{cr}=-\frac c{Nf_0}\sum_{i=1}^N(f_i-f_0)=-\frac cN\sum_{i=1}^N\left(\frac{f_i}{f_0}-1\right).

Alternative without approximation:

V_{cr}=\frac{cf_0}N\sum_{i=1}^N\left(\frac1{f_i}-\frac1{f_0}\right)=\frac cN\sum_{i=1}^N\left(\frac{f_0}{f_i}-1\right).

0.3 pts. Total 0.5 pts.

Editorial source note: “without approximation” is the key's wording for its classical Doppler expression; it is not a relativistic Doppler formula. [0,5 т.]

A.3 The galaxy moving away speed $V_i$ , in part A.2, is only one component of the three component of the galaxy velocity. Thus the average square speed of each galaxy with respect to the center of the cluster is

\frac1N\sum_{i=1}^N(\vec V_i-\vec V_c)^2=\frac1N\sum_{i=1}^N(V_{xi}-V_{xc})^2+(V_{yi}-V_{yc})^2+(V_{zi}-V_{zc})^2.

Due to isotropic assumption

\frac1N\sum_{i=1}^N(\vec V_i-\vec V_c)^2=\frac3N\sum_{i=1}^N(V_{ri}-V_{cr})^2.

0.5 pts. And thus the root mean square of the galaxy speed with respect to the cluster center is

\begin{aligned}v_{rms}&=\sqrt{\frac3N\sum_{i=1}^N(V_{ri}-V_{rc})^2}\\&=\sqrt{\frac3N\sum_{i=1}^N(V_{ri}^2-2V_{cr}V_{ri}+V_{cr}^2)}\\&=\sqrt{\frac3N\left(\sum_{i=1}^NV_{ri}^2\right)-3V_{cr}^2},\\v_{rms}&=c\sqrt3\sqrt{\left(\frac1N\sum_{i=1}^N\left(\frac{f_i}{f_0}-1\right)^2\right)-\left(\frac1N\sum_{i=1}^N\left(\frac{f_i}{f_0}-1\right)\right)^2}\\&=\frac{c\sqrt3}{f_0}\sqrt{\left(\frac1N\sum_{i=1}^N(f_i^2-2f_if_0+f_0^2)\right)-\left(\left(\frac1N\sum_{i=1}^Nf_i\right)^2-2\frac{f_0}N\sum_{i=1}^Nf_i+f_0^2\right)}\\&=\frac{c\sqrt3}{f_0N}\sqrt{N\sum_{i=1}^Nf_i^2-\left(\sum_{i=1}^Nf_i\right)^2}.\end{aligned}

Alternative without approximation:

\begin{aligned}v_{rms}&=c\sqrt3\sqrt{\left(\frac1N\sum_{i=1}^N\left(\frac{f_0}{f_i}-1\right)^2\right)-\left(\frac1N\sum_{i=1}^N\left(\frac{f_0}{f_i}-1\right)\right)^2}\\&=\frac{c\sqrt3}{f_0}\sqrt{\left(\frac1N\sum_{i=1}^N\left(\frac1{f_i^2}-2\frac1{f_i}\frac1{f_0}+\frac1{f_0^2}\right)\right)-\left(\left(\frac1N\sum_{i=1}^N\frac1{f_i}\right)^2-2\frac1{f_0}\frac1N\sum_{i=1}^N\frac1{f_i}+\frac1{f_0^2}\right)}\\&=\frac{cf_0\sqrt3}N\sqrt{N\sum_{i=1}^N\left(\frac1{f_i}\right)^2-\left(\sum_{i=1}^N\frac1{f_i}\right)^2}.\end{aligned}

0.7 pts. The mean kinetic energy of the galaxies with respect to the center of the cluster is

K_{ave}=\frac m2\frac1N\sum_{i=1}^N(\vec V_i-\vec V_c)^2=\frac m2v_{rms}^2.

0.3 pts. Total 1.5 pts.

Editorial source note: the key swaps $V_{cr}$ to $V_{rc}$ in one line and prints $c\sqrt3/f_0$ in the intermediate reciprocal-frequency expression, whereas its final expression has $cf_0\sqrt3/N$ . The source's summation grouping in the component formula is also retained. [1,5 т.]

A.4 The time average of $d\Gamma/dt$ vanishes

\left\langle\frac{d\Gamma}{dt}\right\rangle_t=0.

Now

\begin{aligned}\frac{d\Gamma}{dt}&=\frac d{dt}\sum_i\vec p_i\cdot\vec r_i=\sum_i\frac{d\vec p_i}{dt}\cdot\vec r_i+\sum_i\vec p_i\cdot\frac{d\vec r_i}{dt}\\&=\sum_i\vec F_i\cdot\vec r_i+\sum_i m_i\vec v_i\cdot\vec v_i=\sum_i\vec F_i\cdot\vec r_i+2K.\end{aligned}

0.6 pts. Where $K$ is the total kinetic energy of the system. Since the gravitational force on $i$ -th particle comes from its interaction with other particles then

\begin{aligned}\sum_i\vec F_i\cdot\vec r_i&=\sum_{i,j\ne i}\vec F_{ji}\cdot\vec r_i=\sum_{i<j}\vec F_{ji}\cdot\vec r_i-\sum_{i>j}\vec F_{ij}\cdot\vec r_i\\&=\sum_{i<j}\vec F_{ji}\cdot\vec r_i-\sum_{i<j}\vec F_{ji}\cdot\vec r_j=\sum_{i<j}\vec F_{ji}\cdot(\vec r_i-\vec r_j)\\&=-\sum_{i<j}G\frac{m_im_j}{|\vec r_i-\vec r_j|^2}\frac{\vec r_i-\vec r_j}{|\vec r_i-\vec r_j|}\cdot(\vec r_i-\vec r_j)\\&=-\sum_{i<j}G\frac{m_im_j}{|\vec r_i-\vec r_j|}=U_{tot}.\end{aligned}

Alternative proof:

\begin{aligned}\sum_i\vec F_i\cdot\vec r_i&=\sum_{i,j\ne i}\vec F_{ji}\cdot\vec r_i\\&=\vec F_{21}\cdot\vec r_1+\vec F_{31}\cdot\vec r_1+\vec F_{41}\cdot\vec r_1+\cdots+\vec F_{N1}\cdot\vec r_1+\\&\quad\vec F_{12}\cdot\vec r_2+\vec F_{32}\cdot\vec r_2+\vec F_{42}\cdot\vec r_2+\cdots+\vec F_{N2}\cdot\vec r_2+\\&\quad\vec F_{13}\cdot\vec r_3+\vec F_{23}\cdot\vec r_3+\vec F_{43}\cdot\vec r_3+\cdots+\vec F_{N3}\cdot\vec r_3+\cdots\\&\quad\vec F_{1N}\cdot\vec r_N+\vec F_{2N}\cdot N_N+\vec F_{3N}\cdot\vec r_N+\cdots+\vec F_{N\,N-1}\cdot\vec r_{N-1}.\end{aligned}

Collecting terms and noting that $\vec F_{ij}=-\vec F_{ji}$ we have

\begin{aligned}&\vec F_{12}\cdot(\vec r_2-\vec r_1)+\vec F_{13}\cdot(\vec r_3-\vec r_1)+\vec F_{14}\cdot(\vec r_4-\vec r_1)+\cdots+\vec F_{23}\cdot(\vec r_3-\vec r_2)\\&\quad+\vec F_{24}\cdot(\vec r_4-\vec r_2)+\cdots+\vec F_{34}\cdot(\vec r_4-\vec r_3)+\cdots=\sum_{i<j}\vec F_{ji}\cdot(\vec r_i-\vec r_j)\\&=-\sum_{i<j}G\frac{m_im_j}{|\vec r_i-\vec r_j|^2}\frac{\vec r_i-\vec r_j}{|\vec r_i-\vec r_j|}\cdot(\vec r_i-\vec r_j)=-\sum_{i<j}G\frac{m_im_j}{|\vec r_i-\vec r_j|}=U_{tot}.\end{aligned}

0.9 pts. Thus we have

\frac{d\Gamma}{dt}=U+2K.

And by taking its time average we obtain $\langle d\Gamma/dt=U+2K\rangle_t=0$ and thus $\langle K\rangle_t=-\frac12\langle U\rangle_t$ . Therefore $\gamma=\frac12$ .

0.2 pts. Total 1.7 pts.

Editorial source note: the alternative expansion prints $N_N$ in place of a position vector and ends with $\vec F_{N\,N-1}\cdot\vec r_{N-1}$ ; these apparent typographical errors are retained. [1,7 т.]

A.5 Using Virial theorem, and since the dark matter has the same root mean square speed as the galaxy, then we have

\langle K\rangle_t=-\frac12\langle U\rangle_t,\qquad \frac M2v_{rms}^2=\frac12\frac35\frac{GM^2}R.

0.3 pts. From which we have

M=\frac{5Rv_{rms}^2}{3G}.

0.1 pts. And the dark matter mass is then

M_{dm}=\frac{5Rv_{rms}^2}{3G}-Nm_g.

0.1 pts. Total 0.5 pts. [0,5 т.]

B. Dark Matter in a Galaxy

B.1 Answer B.1: The gravitational attraction for a particle at a distance $r$ from the center of the sphere comes only from particles inside a spherical volume of radius $r$ . For particle inside the sphere with mass $m_s$ , assuming the particle is orbiting the center of mass in a circular orbit, we have

G\frac{m'(r)m_s}{r^2}=\frac{m_sv_0^2}r.

0.3 pts. With $m'(r)$ is the total mass inside a sphere of radius $r$

m'(r)=\frac43\pi r^3m_sn.

Thus we have

v(r)=\left(\frac{4\pi Gnm_s}3\right)^{1/2}r.

0.2 pts. While for particle outside the sphere, we have

v(r)=\left(\frac{4\pi Gnm_sR^3}{3r}\right)^{1/2}.

0.2 pts. The sketch is given below

Key sketch of rotation speed increasing linearly to radius R then decreasing outside the galaxy. — Sketch of the rotation velocity vs distance from the center of galaxy

0.1 pts. Total 0.8 pts.

Editorial source note: the first force equation prints $v_0$ rather than $v(r)$ , and the exterior formula and sketch use $R$ rather than $R_g$ . [0,8 т.]

B.2 The total mass can be inferred from

G\frac{m'(R_g)m_s}{R_g^2}=\frac{m_sv_0^2}{R_g}.

Thus

m_R=m'(R_g)=\frac{v_0^2R_g}G.

0.5 pts. Total 0.5 pts. [0,5 т.]

B.3 Base on the previous answer in B.1, if the mass of the galaxy comes only from the visible stars, then the galaxy rotation curve should fall proportional to $1/\sqrt r$ on the outside at a distance $r>R_g$ . But in the figure of problem b) the curve remain constant after $r>R_g$ , we can infer from

G\frac{m'(r)m_s}{r^2}=\frac{m_sv_0^2}r.

To make $v(r)$ constant, then $m'(r)$ should be proportional to $r$ for $r>R_g$ , i.e. for $r>R_g$ , $m'(r)=Ar$ with $A$ is a constant.

0.3 pts. While for $r<R_g$ , to obtain a linear plot proportional to $r$ , then $m'(r)$ should be proportional to $r^3$ , i.e. $m'(r)=Br^3$ .

0.3 pts. Thus for $r<R_g$ we have

m'(r)=\int_0^r\rho_t(r)4\pi r'^2\,dr'=Br^3,

dm'(r)=\rho_t(r)4\pi r^2dr=3Br^2dr.

Thus total mass density $\rho_t(r)=3B/(4\pi)$ .

0.2 pts.

m_R=\int_0^{R_g}\frac{3B}{4\pi}4\pi r'^2dr'=BR_g^3,\qquad B=\frac{m_R}{R_g^3}=\frac{v_0^2}{GR_g^2}.

Thus the dark matter mass density

\rho(r)=\frac{3v_0^2}{4\pi GR_g^2}-nm_s.

0.2 pts. While for $r>R_g$ we have

m'(r)=\int_0^{R_g}\rho(r')4\pi r'^2dr'+\int_{R_g}^r\rho(r')4\pi r'^2dr'=Ar,

m'(r)=m_R+\int_{R_g}^r\rho(r')4\pi r'^2dr'=Ar,

\int_R^r\rho(r')4\pi r'^2dr'=Ar-M_0,

\rho(r)4\pi r^2=A,\qquad\rho(r)=\frac A{4\pi r^2}.

0.2 pts. Now to find the constant $A$ .

\int_R^r\frac A{4\pi r'^2}4\pi r'^2dr'=A(r-R_g)=Ar-m_R.

Thus $AR_g=m_R$ and $A=v_0^2/G$ .

We can also find $A$ from the following

G\frac{m'(r)m_s}{r^2}=G\frac{Arm_s}{r^2}=\frac{m_sv_0^2}r,\quad\text{thus } A=\frac{v_0^2}G.

Thus the dark matter mass density (which is also the total mass density since $n\approx0$ for $r\ge R_g$ .

\rho(r)=\frac{v_0^2}{4\pi Gr^2}\quad\text{for }r\ge R_g.

0.3 pts. Total 1.5 pts.

Editorial source note: the source changes $R_g$ to $R$ and $m_R$ to $M_0$ in intermediate lines, and uses $\rho$ both for dark matter and total density. Its unprimed $\rho_t(r)$ inside the first integral and unmatched prose parenthesis are retained. [1,5 т.]

C. Interstellar Gas and Dark Matter

C.1 Consider a very small volume of a disk with area $A$ and thickness $\Delta r$ , see Fig.1

Hydrostatic disk with pressures P(r) and P(r+dr), thickness dr, radial coordinate r and inward acceleration g(r). — Figure 1. Hydrostatic equilibrium

In hydrostatic equilibrium we have

\bigl(P(r)-P(r+\Delta r)\bigr)A-\rho g(r)A\Delta r=0.

0.3 pts.

\frac{\Delta P}{\Delta r}=-\rho\frac{Gm'(r)}{r^2},

\frac{dP}{dr}=-\rho\frac{Gm'(r)}{r^2}=-n(r)m_p\frac{Gm'(r)}{r^2}.

0.2 pts. Total 0.5 pts. [0,5 т.]

C.2 Using the ideal gas law $P=nkT$ where $n=N/V$ where $n$ is the number density, we have

\frac{dP}{dr}=kT\frac{dn(r)}{dr}+kn(r)\frac{dT}{dr}=-n(r)m_p\frac{Gm'(r)}{r^2}.

Thus we have

m'(r)=-\frac{kT}{Gm_p}\left(\frac{r^2}{n(r)}\frac{dn(r)}{dr}+\frac{r^2}{T(r)}\frac{dT(r)}{dr}\right).

0.5 pts. Total 0.5 pts. [0,5 т.]

C.3 If we have isothermal distribution, we have $dT/dr=0$ and

m'(r)=-\frac{kT_0}{Gm_p}\left(\frac{r^2}{n(r)}\frac{dn(r)}{dr}\right).

0.2 pts. From information about interstellar gas number density, we have

\frac1{n(r)}\frac{dn(r)}{dr}=-\frac{3r+\beta}{r(r+\beta)}.

Thus we have

m'(r)=\frac{kT_0r}{Gm_p}\frac{3r+\beta}{r+\beta}.

0.2 pts. Mass density of the interstellar gas is

\rho_g(r)=\frac{\alpha m_p}{r(\beta+r)^2}.

Thus

m'(r)=\int_0^r\bigl(\rho_g(r')+\rho_{dm}(r')\bigr)4\pi r'^2dr'=\frac{kT_0r}{Gm_p}\frac{3r+\beta}{r+\beta},

m'(r)=\int_0^r\left(\frac{\alpha m_p}{r'(\beta+r')^2}+\rho_{dm}(r')\right)4\pi r'^2dr'=\frac{kT_0r}{Gm_p}\frac{3r+\beta}{r+\beta}.

0.3 pts.

\left(\frac{\alpha m_p}{r(\beta+r)^2}+\rho_{dm}(r)\right)4\pi r^2=\frac{kT_0}{Gm_p}\frac{3r^2+6r\beta+\beta^2}{(r+\beta)^2},

\rho_{dm}(r)=\frac{kT_0}{4\pi Gm_p}\frac{3r^2+6r\beta+\beta^2}{(r+\beta)^2r^2}-\frac{\alpha m_p}{r(\beta+r)^2}.

0.3 pts. Total 1.0 pts. [1 т.]

AT1 answer sheet

The official five-page form has six Student Code boxes, blank answer spaces and a blank Marks column. Its A heading reads “Cluster of Galaxie”; B reads “Dark Matter in Galaxy”. The downloadable original preserves the writing space, including the continuation of B.1 on otherwise blank page 3.

Question	Printed answer fields	Points
A.1	$U=$	1.0
A.2	$V_{rc}=$	0.5
A.3	$K_{ave}=$ ; $v_{rms}=$	1.5
A.4	$\gamma=$	1.7
A.5	$M_{dm}=$	0.5
B.1	For $r<R_g$ : $v(r)=$ ; for $r\ge R_g$ : $v(r)=$ ; sketch $v(r)$ for both ranges	0.8
B.2	$m_R=m'(R_g)=$	0.5
B.3	For $r<R_g$ : $\rho(r)=$ ; for $r\ge R_g$ : $\rho(r)=$	1.5
C.1	$dP/dr=$	0.5
C.2	$m'(r)=$	0.5
C.3	$\rho_{dm}(r)=$	1.0

The form prints $V_{rc}$ instead of the question's $V_{cr}$ , and its B.1 boundary inequalities differ from the question.

Page 5: DO NOT WRITE BELOW

SECTION	SCORE
A
B
C
TOTAL

Оригинал в Архива: 1.pdf · официални решения: 1_sol.pdf · допълнителни данни: 1_anssheet.pdf · допълнителни данни: th_instructions.pdf