IPhO 2025 · Theory problem 1

Hydrogen and galaxies

This problem aims to study the peculiar physics of galaxies, such as their dynamics and structure. In particular, we explain how to measure the mass distribution of our galaxy from the inside. For this we will focus on hydrogen, its main constituent.

Throughout this problem we will only use \(\hbar\), defined as \(\hbar=h/2\pi\).

Part A - Introduction

Bohr model

We assume that the hydrogen atom consists of a non-relativistic electron, with mass \(m_e\), orbiting a fixed proton. Throughout this part, we assume its motion is on a circular orbit.

A.1 — 0.2pt

Determine the electron's velocity \(v\) in a circular orbit of radius \(r\).

In the Bohr model, we assume the magnitude of the electron's angular momentum \(L\) is quantized, \(L=n\hbar\) where \(n>0\) is an integer. We define \(\alpha=\dfrac{e^2}{4\pi\varepsilon_0\hbar c}\approx7.27\times10^{-3}\).

A.2 — 0.5pt

Show that the radius of each orbit is given by \(r_n=n^2r_1\), where \(r_1\) is called the Bohr radius. Express \(r_1\) in terms of \(\alpha\), \(m_e\), \(c\) and \(\hbar\) and calculate its numerical value with 3 digits. Express \(v_1\), the velocity on the orbit of radius \(r_1\), in terms of \(\alpha\) and \(c\).

A.3 — 0.5pt

Determine the electron's mechanical energy \(E_n\) on an orbit of radius \(r_n\) in terms of \(e\), \(\varepsilon_0\), \(r_1\) and \(n\). Determine \(E_1\) in the ground state in terms of \(\alpha\), \(m_e\) and \(c\). Compute its numerical value in eV.

Hydrogen fine and hyperfine structures

The rare spontaneous inversion of the electron's spin causes a photon to be emitted on average once per 10 million years per hydrogen atom. This emission serves as a hydrogen tracer in the universe and is thus fundamental in astrophysics. We will study the transition responsible for this emission in two steps.

First, consider the interaction between the electron spin and the relative motion of the electron and the proton. Working in the electron's frame of reference, the proton orbits the electron at a distance \(r_1\). This produces a magnetic field \(\vec B_1\).

A.4 — 0.5pt

Determine the magnitude \(B_1\) of \(\vec B_1\) at the position of the electron in terms of \(\mu_0\), \(e\), \(\alpha\), \(c\) and \(r_1\).

Second, the electron spin creates a magnetic moment \(\vec{\mathcal M}_s\). Its magnitude is roughly \(\mathcal M_s=\dfrac{e}{m_e}\hbar\). The fine (F) structure is related to the energy difference \(\Delta E_{\mathrm F}\) between an electron with a magnetic moment \(\vec{\mathcal M}_s\) parallel to \(\vec B_1\) and that of an electron with \(\vec{\mathcal M}_s\) anti-parallel to \(\vec B_1\). Similarly, the hyperfine (HF) structure is related to the energy difference \(\Delta E_{\mathrm{HF}}\), due to the interaction between parallel and anti-parallel magnetic moments of the electron and the proton. It is known to be approximately \(\Delta E_{\mathrm{HF}}\simeq3.72\dfrac{m_e}{m_p}\Delta E_{\mathrm F}\) where \(m_p\) is the proton mass.

A.5 — 0.5pt

Express \(\Delta E_{\mathrm F}\) as a function of \(\alpha\) and \(E_1\).

Express the wavelength \(\lambda_{\mathrm{HF}}\) of a photon emitted during a transition between the two states of the hyperfine structure and give its numerical value with two digits.

Part B - Rotation curves of galaxies

Data

  • Kiloparsec: \(1\,\mathrm{kpc}=3.09\times10^{19}\,\mathrm m\)
  • Solar mass : \(1\,M_\odot=1.99\times10^{30}\,\mathrm{kg}\)

We consider a spherical galaxy centered around a fixed point \(O\). At any point \(P\), let \(\rho=\rho(P)\) be the volumetric mass density and \(\varphi=\varphi(P)\) the associated gravitational potential (i.e. potential energy per unit mass). Both \(\rho\) and \(\varphi\) depend only on \(r=\|\overrightarrow{OP}\|\). The motion of a mass \(m\) located at \(P\), due to the field \(\varphi\), is restricted to a plane containing \(O\).

B.1 — 0.2pt

In the case of a circular orbit, determine the velocity \(v_c\) of an object on a circular orbit passing through \(P\) in terms of \(r\) and \(\dfrac{d\varphi}{dr}\).

Fig. 1(A) is a picture of the spiral galaxy NGC 6946 in the visible band (from the 0.8m Schulman Telescope at the Mount Lemmon Sky Center in Arizona). The little ellipses in Fig. 1(B) show experimental measurements of \(v_c\) for this galaxy. The central region (\(r<1\,\mathrm{kpc}\)) is named the bulge. In this region, the mass distribution is roughly homogeneous. The red curve is a prediction for \(v_c\) if the system were homogeneous in the bulge and keplerian (\(\varphi(r)=-\beta/r\) with \(\beta>0\)) outside it, i.e. considering that the total mass of the galaxy is concentrated in the bulge.

Figure 1

Fig. 1: NGC 6946 galaxy: Picture (A) and rotation curve (B).

B.2 — 0.5pt

Deduce the mass \(M_b\) of the bulge of NGC 6946 from the red rotation curve in Fig. 1(B), in solar mass units.

Comparing the keplerian model and the experimental data makes astronomers confident that part of the mass is invisible in the picture. They thus suppose that the galaxy's actual mass density is given by

\[ \rho_m(r)=\frac{C_m}{r_m^2+r^2} \tag{1} \]

where \(C_m>0\) and \(r_m>0\) are constants.

B.3 — 1.8pt

Show that the velocity profile \(v_{c,m}(r)\), corresponding to the mass density in Eq. 1, can be written \[ v_{c,m}(r)=\sqrt{k_1-\frac{k_2\cdot\arctan\!\left(\frac r{r_m}\right)}r}. \] Express \(k_1\) and \(k_2\) in terms of \(C_m\), \(r_m\) and \(G\).

(Hints: \(\displaystyle\int_0^r\frac{x^2}{a^2+x^2}\,dx=r-a\arctan(r/a)\), and: \(\arctan(x)\simeq x-x^3/3\) for \(x\ll1\).)

Simplify \(v_{c,m}(r)\) when \(r\ll r_m\) and when \(r\gg r_m\).

Show that if \(r\gg r_m\), the mass \(M_m(r)\) embedded in a sphere of radius \(r\) with the mass density given by Eq. 1 simplifies and depends only on \(C_m\) and \(r\).

Estimate the mass of the galaxy NGC 6946 actually present in the picture in Fig. 1(A).

Part C - Mass distribution in our galaxy

For a spiral galaxy, the model for Eq. 1 is modified and one usually considers the gravitational potential is given by \[ \varphi_G(r,z)=\varphi_0\ln\!\left(\frac r{r_0}\right) \exp\!\left[-\left(\frac z{z_0}\right)^2\right], \] where \(z\) is the distance to the galactic plane (defined by \(z=0\)), and \(r<r_0\) is now the axial radius and \(\varphi_0>0\) a constant to be determined. \(r_0\) and \(z_0\) are constant values.

C.1 — 0.5pt

Find the equation of motion on \(z\) for the vertical motion of a point mass \(m\) in such a potential, assuming \(r\) is constant. Show that, if \(r<r_0\) the galactic plane is a stable equilibrium state by giving the angular frequency \(\omega_0\) of small oscillations around it.

From here on, we set \(z=0\).

C.2 — 0.6pt

Identify the regime, either \(r\gg r_m\) or \(r\ll r_m\), in which the model of Eq. 1 recovers a potential of the form \(\varphi_G(r,0)\) with a suitable definition of \(\varphi_0\).

Under this condition \(v_c(r)\) no longer depends on \(r\). Express it in terms of \(\varphi_0\).

Therefore, outside the bulge the velocity modulus \(v_c\) does not depend on the distance to the galactic center. We will use this fact, as astronomers do, to measure the galaxy's mass distribution from the inside.

All galactic objects considered here for astronomical observations, such as stars or nebulae, are primarily composed of hydrogen. Outside the bulge, we assume that they rotate on circular orbits around the galactic center \(C\). \(S\) is the sun's position and \(E\) that of a given galactic object emitting in the hydrogen spectrum. In the galactic plane, we consider a line of sight \(SE\) corresponding to the orientation of an observation, on the unit vector \(\hat u_v\) (see Fig. 2).

Figure 2

Fig. 2: Geometry of the measurement

Let \(\ell\) be the galactic longitude, measuring the angle between \(SC\) and the \(SE\). The sun's velocity on its circular orbit of radius \(R_\odot=8.00\,\mathrm{kpc}\) is denoted \(\vec v_\odot\). A galactic object in \(E\) orbits on another circle of radius \(R\) at velocity \(\vec v_E\). Using a Doppler effect on the previously studied 21cm line, one can obtain the relative radial velocity \(v_{rE/S}\) of the emitter \(E\) with respect to the sun \(S\): it is the projection of \(\vec v_E-\vec v_\odot\) on the line of sight.

C.3 — 0.7pt

Determine \(v_{rE/S}\) in terms of \(\ell\), \(R\), \(R_\odot\) and \(v_\odot\). Then, express \(R\) in terms of \(R_\odot\), \(v_\odot\), \(\ell\) and \(v_{rE/S}\).

Using a radio telescope, we make observations in the plane of our galaxy toward a longitude \(\ell=30^\circ\). The frequency band used contains the 21cm line, whose frequency is \(f_0=1.42\,\mathrm{GHz}\). The results are reported in Fig. 3.

Figure 3

Fig. 3: Electromagnetic signal as a function of the frequency shift, measured in the radio frequency band at \(\ell=30^\circ\) using EU-HOU RadioAstronomy

C.4 — 0.6pt

In our galaxy, \(v_\odot=220\,\mathrm{km\cdot s^{-1}}\). Determine the values of the relative radial velocity (with 3 significant digits) and the distance from the galactic center (with 2 significant digits) of the 3 sources observed in Fig. 3. Distances should be expressed as multiples of \(R_\odot\).

C.5 — 0.6pt

On the top view of our galaxy (in the answer box), indicate the positions of the sources observed in Fig. 3.

What could be deduced from repeated measurements changing \(\ell\)?

Figure 4

Part D - Tully-Fisher relation and MOND theory

The flat external velocity curve of NGC 6946 in Fig. 1 is a common property of spiral galaxies, as can be seen in Fig. 4 (left). Plotting the external constant velocity value \(v_{c,\infty}\) as a function of the measured total mass \(M_{\mathrm{tot}}\) of each galaxy gives an interesting correlation called the Tully-Fischer relation, see Fig. 4 (right).

Figure 5

Fig. 4. Left: Rotation curves for typical spiral galaxies - Right: \(\log_{10}(M_{\mathrm{tot}})\) as a function of \(\log_{10}(v_{c,\infty})\) on linear scales. Colored dots correspond to different galaxies and different surveys. The green line is the Tully-Fischer relation which is in very good agreement with the best fit line of the data (in black).

D.1 — 0.4pt

Assuming that the radius \(R\) of a galaxy doesn't depend on its mass, show that the model of Eq. 1 (part B) gives a relation of the form \(M_{\mathrm{tot}}=\eta v_{c,\infty}^{\gamma}\) where \(\gamma\) and \(\eta\) should be specified.

Compare this expression to the Tully-Fischer relation by computing \(\gamma_{TF}\).

In the extremely low acceleration regime, of the order of \(a_0=10^{-10}\,\mathrm{m\cdot s^{-2}}\), the MOdified Newtonian Dynamics (MOND) theory suggests that one can modify Newton's second law using \[ \vec F=m\mu\!\left(\frac a{a_0}\right)\vec a \] where \(a=\|\vec a\|\) is the modulus of the acceleration and the \(\mu\) function is defined by \(\mu(x)=\dfrac{x}{1+x}\).

D.2 — 0.2pt

Using data for NGC 6946 in Fig. 1, estimate, within Newton's theory, the modulus of the acceleration \(a_m\) of a mass in the outer regions of NGC 6946.

D.3 — 0.8pt

Let \(m\) be a mass on a circular orbit of radius \(r\) with velocity \(v_{c,\infty}\) in the gravity field of a fixed mass \(M\).

Within the MOND theory, with \(a\ll a_0\), determine the Tully-Fischer exponent.

Using data for NGC 6946 and/or Tully-Fischer law, calculate \(a_0\) to show that MOND operates in the correct regime.

D.4 — 0.9pt

Considering relevant cases, determine \(v_c(r)\) for all values of \(r\) in the MOND theory in the case of a gravitational field due to a homogeneously distributed mass \(M\) with radius \(R_b\).

General Data Sheet

Constant Notation Value Unit
Speed of light in vacuum \(c\) \(2.997\,924\,58\times10^8\) \(\mathrm{m\cdot s^{-1}}\)
Planck constant \(h\) \(6.626\,070\,15\times10^{-34}\) \(\mathrm{kg\cdot m^2\cdot s^{-1}}\)
Reduced Planck constant \(\hbar=\dfrac h{2\pi}\) \(1.054\,571\,818\times10^{-34}\) \(\mathrm{kg\cdot m^2\cdot s^{-1}}\)
Boltzmann constant \(k_{\mathrm B}\) \(1.380\,649\times10^{-23}\) \(\mathrm{kg\cdot m^2\cdot s^{-2}\cdot K^{-1}}\)
Avogadro constant \(N_{\mathrm A}\) \(6.022\,140\,76\times10^{23}\) \(\mathrm{mol^{-1}}\)
Molar gas constant \(R\) \(8.314\,462\,618\,153\,24\) \(\mathrm{kg\cdot m^2\cdot s^{-2}\cdot mol^{-1}\cdot K^{-1}}\)
Elementary charge \(e\) \(1.602\,176\,634\times10^{-19}\) \(\mathrm{A\cdot s}\)
Universal constant of gravitation \(G\) \(6.674\,30(15)\times10^{-11}\) \(\mathrm{m^3\cdot kg^{-1}\cdot s^{-2}}\)
Standard acceleration due to gravity \(g\) \(9.806\,65\) \(\mathrm{m\cdot s^{-2}}\)
Stefan Boltzmann constant \(\sigma\) \(5.670\,374\,419\times10^{-8}\) \(\mathrm{kg\cdot s^{-3}\cdot K^{-4}}\)
Vacuum permeability (magnetic constant) \(\mu_0\) \(1.256\,637\,061\,27(20)\times10^{-6}\) \(\mathrm{kg\cdot m\cdot A^{-2}\cdot s^{-2}}\)
Vacuum permittivity (electrical constant) \(\epsilon_0\) \(8.854\,187\,818\,8(14)\times10^{-12}\) \(\mathrm{A^2\cdot s^4\cdot kg^{-1}\cdot m^{-3}}\)
Rydberg constant \(R_\infty\) \(1.097\,373\,156\,815\,7(12)\times10^7\) \(\mathrm{m^{-1}}\)
Mass of the electron \(m_e\) \(9.109\,383\,713\,9(28)\times10^{-31}\) \(\mathrm{kg}\)
Mass of the proton \(m_p\) \(1.672\,621\,925\,95(52)\times10^{-27}\) \(\mathrm{kg}\)
Mass of the neutron \(m_n\) \(1.674\,927\,500\,56(85)\times10^{-27}\) \(\mathrm{kg}\)
Atomic mass constant \(m_u\) \(1.660\,539\,068\,92(52)\times10^{-27}\) \(\mathrm{kg}\)
Electronvolt \(\mathrm{eV}\) \(1.602\,176\,634\times10^{-19}\) \(\mathrm{kg\cdot m^2\cdot s^{-2}}\)

Official English text of the problem.

Answers

  • A.1 \[ \boxed{v=\sqrt{\frac{e^2}{4\pi\varepsilon_0m_er}}}. \]

  • A.2 \[ \boxed{r_n=n^2r_1,\qquad r_1=\frac{\hbar}{m_e\alpha c}=5.29\times10^{-11}\,\mathrm m,\qquad v_1=\alpha c}. \]

  • A.3 \[ \boxed{E_n=-\frac{e^2}{8\pi\varepsilon_0r_1n^2},\qquad E_1=-\frac12m_e\alpha^2c^2=-13.6\,\mathrm{eV}}. \]

  • A.4 \[ \boxed{B_1=\frac{\mu_0e\alpha c}{4\pi r_1^2}}. \]

  • A.5 \[ \boxed{\Delta E_{\mathrm F}=-4\alpha^2E_1}, \qquad \boxed{\lambda_{\mathrm{HF}} =\frac{\pi m_p\hbar}{3.72\,m_e^2\alpha^4c} \simeq0.21\,\mathrm m}. \]

  • B.1 \[ \boxed{v_c=\sqrt{r\frac{d\varphi}{dr}}}. \]

  • B.2 \[ \boxed{M_b\sim1\times10^9\,M_\odot}. \]

  • B.3 \[ \boxed{k_1=4\pi GC_m,\qquad k_2=4\pi GC_mr_m}, \] \[ \boxed{v_{c,m}(r\ll r_m)\simeq \sqrt{\frac{4\pi GC_m}{3}}\frac r{r_m}}, \qquad \boxed{v_{c,m}(r\gg r_m)\simeq\sqrt{4\pi GC_m}}, \] \[ \boxed{M_m(r\gg r_m)\simeq4\pi C_mr,\qquad M_{\text{in the figure}}\simeq6.1\times10^{10}\,M_\odot}. \]

  • C.1 \[ \boxed{\ddot z= \frac{2\varphi_0}{z_0^2}\ln\!\left(\frac r{r_0}\right) z\,e^{-z^2/z_0^2}}, \qquad \boxed{\omega_0= \sqrt{\frac{2\varphi_0}{z_0^2}\ln\!\left(\frac{r_0}r\right)}}. \]

  • C.2 \[ \boxed{r\gg r_m,\qquad \varphi_0=4\pi GC_m,\qquad v_c(r)=\sqrt{\varphi_0}}. \]

  • C.3 \[ \boxed{v_{rE/S}=v_\odot\left(\frac{R_\odot}{R}-1\right)\sin\ell}, \qquad \boxed{R=\frac{R_\odot v_\odot\sin\ell} {v_\odot\sin\ell+v_{rE/S}}}. \]

  • C.4 Numbering the peaks by increasing frequency shift: \[ \boxed{(v_{r,1},v_{r,2},v_{r,3}) =(6.33,\ 31.7,\ 54.9)\,\mathrm{km\,s^{-1}}}, \] \[ \boxed{(R_1,R_2,R_3)=(0.95,\ 0.78,\ 0.67)\,R_\odot}. \]

  • C.5 On the \(\ell=30^\circ\) sightline, mark both intersections with each of the three circles of radii \(0.95R_\odot\), \(0.78R_\odot\), and \(0.67R_\odot\). The near and far distances from the Sun are \[ \boxed{\frac{(d_{1,-},d_{2,-},d_{3,-})}{R_\odot} \simeq(0.0635,\ 0.272,\ 0.424)}, \] \[ \boxed{\frac{(d_{1,+},d_{2,+},d_{3,+})}{R_\odot} \simeq(1.67,\ 1.46,\ 1.31)}. \] Repeated longitude scans reveal the hydrogen distribution and spiral structure, subject to this near–far ambiguity.

  • D.1 \[ \boxed{\eta=\frac RG,\qquad \gamma=2,\qquad \gamma_{TF}\approx3.7\ \text{(graphically close to }4\text{)}}. \]

  • D.2 \[ \boxed{a_m\simeq9.4\times10^{-11}\,\mathrm{m\,s^{-2}}}. \]

  • D.3 \[ \boxed{v_{c,\infty}^4=GMa_0,\qquad \gamma_{\mathrm{MOND}}=4,\qquad a_0\sim2\times10^{-10}\,\mathrm{m\,s^{-2}}}. \] This is the relevant acceleration scale; the plotted NGC 6946 data alone do not establish the stronger inequality \(a\ll a_0\).

  • D.4 \[ \boxed{ v_c(r)= \begin{cases} \displaystyle \left[ \frac{GMr^2}{2R_b^3} \left(1+\sqrt{1+\frac{4a_0R_b^3}{GMr}}\right) \right]^{1/2},&0<r\le R_b,\\[1.2em] \displaystyle \left[ \frac{GM}{2r} \left(1+\sqrt{1+\frac{4a_0r^2}{GM}}\right) \right]^{1/2},&r\ge R_b . \end{cases}} \] The continuous limit is \(v_c(0)=0\).

Solution

For the atomic numerical values, I evaluate \(\alpha\) from its defining expression and the supplied constants: \[ \alpha=\frac{e^2}{4\pi\varepsilon_0\hbar c} =7.29735\times10^{-3}. \] This differs slightly from the approximation printed in the statement.

A.1

The attractive Coulomb force supplies the centripetal force: \[ \frac{m_ev^2}{r}=\frac{e^2}{4\pi\varepsilon_0r^2}. \] Therefore, \[ \boxed{v=\sqrt{\frac{e^2}{4\pi\varepsilon_0m_er}}}. \]

A.2

Bohr quantization gives \[ m_ev_nr_n=n\hbar. \] Squaring this and using the force balance from A.1, \[ n^2\hbar^2 =m_e^2v_n^2r_n^2 =\frac{m_ee^2r_n}{4\pi\varepsilon_0}. \] Consequently, \[ r_n=\frac{4\pi\varepsilon_0n^2\hbar^2}{m_ee^2} =n^2\frac{\hbar}{m_e\alpha c}. \] Thus \[ \boxed{r_1=\frac{\hbar}{m_e\alpha c} =5.29\times10^{-11}\,\mathrm m}. \]

For the ground orbit, \[ v_1=\frac{\hbar}{m_er_1}=\boxed{\alpha c}. \] More generally, \(v_n=\alpha c/n\). Since \(\alpha\ll1\), the non-relativistic approximation is consistent.

A.3

Taking the electrostatic potential energy to vanish at infinity, \[ U(r)=-\frac{e^2}{4\pi\varepsilon_0r}. \] The force balance gives \[ K=\frac12m_ev^2=\frac{e^2}{8\pi\varepsilon_0r}. \] Hence the total mechanical energy is \[ E=K+U=-\frac{e^2}{8\pi\varepsilon_0r}. \] At \(r_n=n^2r_1\), \[ \boxed{E_n=-\frac{e^2}{8\pi\varepsilon_0r_1n^2}}. \] In particular, \[ E_1=-\frac{\alpha\hbar c}{2r_1} =\boxed{-\frac12m_e\alpha^2c^2}. \] Numerically, \[ E_1=-2.17987\times10^{-18}\,\mathrm J =\boxed{-13.6\,\mathrm{eV}}. \]

A.4

In the prescribed electron-frame picture, the proton circulates with speed \(v_1=\alpha c\). Its orbital period and equivalent current are \[ T=\frac{2\pi r_1}{v_1}, \qquad I=\frac eT=\frac{ev_1}{2\pi r_1}. \] The magnetic field at the centre of a circular current loop is \[ B_1=\frac{\mu_0I}{2r_1}. \] Therefore, \[ \boxed{B_1=\frac{\mu_0e\alpha c}{4\pi r_1^2}}. \]

A.5

A magnetic dipole in a field has energy \[ U=-\vec{\mathcal M}_s\cdot\vec B_1. \] The parallel and anti-parallel states consequently have energies \(-\mathcal M_sB_1\) and \(+\mathcal M_sB_1\). Their positive energy separation is \[ \Delta E_{\mathrm F}=2\mathcal M_sB_1. \] Using the magnetic moment specified in the problem, \[ \Delta E_{\mathrm F} =2\frac{e\hbar}{m_e}\frac{\mu_0e\alpha c}{4\pi r_1^2}. \] With \(\mu_0\varepsilon_0c^2=1\) and \(e^2/(4\pi\varepsilon_0)=\alpha\hbar c\), \[ \Delta E_{\mathrm F} =\frac{2\alpha^2\hbar^2}{m_er_1^2} =2m_e\alpha^4c^2. \] Since \(E_1=-m_e\alpha^2c^2/2\), \[ \boxed{\Delta E_{\mathrm F}=-4\alpha^2E_1}. \]

The hyperfine photon satisfies \[ \frac{2\pi\hbar c}{\lambda_{\mathrm{HF}}} =\Delta E_{\mathrm{HF}} \simeq3.72\frac{m_e}{m_p}\Delta E_{\mathrm F}. \] Thus \[ \boxed{\lambda_{\mathrm{HF}} =\frac{\pi m_p\hbar}{3.72\,m_e^2\alpha^4c}}. \] The numerical intermediate values are \[ \Delta E_{\mathrm F}\simeq2.898\times10^{-3}\,\mathrm{eV}, \qquad \lambda_{\mathrm{HF}}\simeq0.21116\,\mathrm m. \] To two significant digits, \[ \boxed{\lambda_{\mathrm{HF}}=0.21\,\mathrm m=21\,\mathrm{cm}}. \]

B.1

The gravitational acceleration is \[ \vec g=-\frac{d\varphi}{dr}\hat r. \] Equating this to the inward acceleration of a circular orbit, \[ -\frac{v_c^2}{r}\hat r=-\frac{d\varphi}{dr}\hat r. \] Therefore, \[ \boxed{v_c=\sqrt{r\frac{d\varphi}{dr}}}. \]

B.2

Outside a spherical bulge, the shell theorem gives \[ \varphi(r)=-\frac{GM_b}{r}+\text{constant}. \] It follows that \[ v_c^2=\frac{GM_b}{r}, \qquad M_b=\frac{rv_c^2}{G}. \]

A convenient approximate reading of the red curve is \[ r\simeq10\,\mathrm{kpc}, \qquad v_c\simeq20\,\mathrm{km\,s^{-1}}. \] Hence \[ M_b\simeq \frac{(10\times3.09\times10^{19})(2.0\times10^4)^2} {6.67430\times10^{-11}} =1.85\times10^{39}\,\mathrm{kg}. \] In solar units this is approximately \(9.3\times10^8M_\odot\). Given the graphical precision, \[ \boxed{M_b\sim10^9M_\odot}. \]

B.3

The mass enclosed within radius \(r\) is \[ M_m(r)=4\pi\int_0^r\rho_m(x)x^2\,dx =4\pi C_m\int_0^r\frac{x^2}{r_m^2+x^2}\,dx. \] Applying the supplied integral, \[ \boxed{M_m(r)=4\pi C_m \left[r-r_m\arctan\!\left(\frac r{r_m}\right)\right]}. \]

By spherical symmetry, \[ v_{c,m}^2(r)=\frac{GM_m(r)}r =4\pi GC_m \left[1-\frac{r_m}{r}\arctan\!\left(\frac r{r_m}\right)\right]. \] Comparison with the requested form gives \[ \boxed{k_1=4\pi GC_m,\qquad k_2=4\pi GC_mr_m}. \]

For \(r\ll r_m\), set \(x=r/r_m\). Then \[ 1-\frac{\arctan x}{x} \simeq1-\frac{x-x^3/3}{x} =\frac{x^2}{3}. \] Therefore, \[ \boxed{v_{c,m}(r)\simeq \sqrt{\frac{4\pi GC_m}{3}}\frac r{r_m}}. \] The central rotation curve rises linearly, as expected for approximately uniform density.

For \(r\gg r_m\), \(\arctan(r/r_m)\to\pi/2\), while \(r_m/r\to0\). Thus \[ \boxed{v_{c,m}(r)\simeq\sqrt{4\pi GC_m} \equiv v_{c,\infty}}, \] and \[ \boxed{M_m(r)\simeq4\pi C_mr =\frac{v_{c,\infty}^2r}{G}}. \]

The pictured diameter is approximately \(18\,\mathrm{kpc}\), so take the enclosing radius as \(9\,\mathrm{kpc}\). The observed outer speed is approximately \(170\,\mathrm{km\,s^{-1}}\). Hence \[ M_{\text{in the figure}} \simeq \frac{(1.70\times10^5)^2(9\times3.09\times10^{19})} {6.67430\times10^{-11}} \simeq1.20\times10^{41}\,\mathrm{kg}. \] Therefore, \[ \boxed{M_{\text{in the figure}}\simeq6.1\times10^{10}M_\odot}. \]

C.1

For fixed \(r\), the vertical equation of motion is \[ m\ddot z=-m\frac{\partial\varphi_G}{\partial z}. \] Differentiating the prescribed potential gives \[ \boxed{\ddot z= \frac{2\varphi_0}{z_0^2}\ln\!\left(\frac r{r_0}\right) z\,e^{-z^2/z_0^2}}. \]

At \(z=0\), the force vanishes. For small oscillations, \(|z|\ll z_0\), so \[ e^{-z^2/z_0^2}\simeq1. \] The linearized equation becomes \[ \ddot z+ \frac{2\varphi_0}{z_0^2}\ln\!\left(\frac{r_0}r\right)z=0. \] For \(r<r_0\), its coefficient is positive. Thus the plane is a stable equilibrium, with \[ \boxed{\omega_0= \sqrt{\frac{2\varphi_0}{z_0^2} \ln\!\left(\frac{r_0}r\right)}}. \]

C.2

For \(r\gg r_m\), B.3 gives \[ M_m(r)\simeq4\pi C_mr. \] The radial potential gradient is therefore \[ \frac{d\varphi}{dr}=\frac{GM_m(r)}{r^2} \simeq\frac{4\pi GC_m}{r}. \] Integrating and choosing the additive constant appropriately, \[ \varphi(r)\simeq4\pi GC_m\ln\!\left(\frac r{r_0}\right). \] Thus the required regime and coefficient are \[ \boxed{r\gg r_m,\qquad \varphi_0=4\pi GC_m}. \]

In the plane, \[ v_c^2=r\frac{\partial\varphi_G(r,0)}{\partial r} =r\frac{\varphi_0}{r}=\varphi_0, \] so \[ \boxed{v_c(r)=\sqrt{\varphi_0}}. \]

C.3

Let \(T\) be the foot of the perpendicular from \(C\) to the line of sight. The geometry gives \[ CT=R_\odot\sin\ell. \] The Sun's tangential velocity projects onto the sightline as \[ \vec v_\odot\cdot\hat u_v=v_\odot\sin\ell. \]

For the emitter, its tangential velocity is perpendicular to \(CE\). Its projection onto the sightline is \[ \vec v_E\cdot\hat u_v=v_E\frac{CT}{R} =v_E\frac{R_\odot}{R}\sin\ell. \] The flat rotation curve implies \(v_E=v_\odot\). Consequently, \[ v_{rE/S} =(\vec v_E-\vec v_\odot)\cdot\hat u_v =\boxed{v_\odot\left(\frac{R_\odot}{R}-1\right)\sin\ell}. \] Solving for \(R\), \[ \boxed{R=\frac{R_\odot v_\odot\sin\ell} {v_\odot\sin\ell+v_{rE/S}}}. \]

C.4

Following the intended convention, which is consistent with the answer sheet, use \[ v_r=+c\frac{f-f_0}{f_0}, \] with positive values denoting recession. The printed \(f-f_0\) axis therefore has the opposite sign from the usual frequency shift for positive recession.

The three peaks occur approximately at \[ \Delta f_1=0.030\,\mathrm{MHz},\qquad \Delta f_2=0.150\,\mathrm{MHz},\qquad \Delta f_3=0.260\,\mathrm{MHz}. \] Since \[ v_\odot\sin\ell =220\sin30^\circ =110\,\mathrm{km\,s^{-1}}, \] the radius formula becomes \[ \frac R{R_\odot}=\frac{110}{110+v_r}, \] where \(v_r\) is expressed in \(\mathrm{km\,s^{-1}}\).

Source, in increasing peak frequency \(\Delta f\) (\(\mathrm{MHz}\)) \(v_r\) (\(\mathrm{km\,s^{-1}}\)), 3 significant digits \(R/R_\odot\), 2 significant digits
1 \(0.030\) \(+6.33\) \(0.95\)
2 \(0.150\) \(+31.7\) \(0.78\)
3 \(0.260\) \(+54.9\) \(0.67\)

C.5

On the answer-sheet diagram, draw the observation ray from \(S\), making \(30^\circ\) with \(SC\) on the side indicated by Fig. 2. Draw, or identify, the circles centred at \(C\) with radii \[ 0.95R_\odot,\qquad0.78R_\odot,\qquad0.67R_\odot. \] Each circle intersects the observation ray twice.

To specify the positions quantitatively, let \(d=SE\). The cosine rule gives \[ R^2=R_\odot^2+d^2-2R_\odot d\cos\ell. \] Thus \[ \boxed{d_\pm= R_\odot\cos\ell \pm\sqrt{R^2-R_\odot^2\sin^2\ell}}. \] At \(\ell=30^\circ\), the sightline's closest approach to the centre is \[ CT=0.50R_\odot, \qquad ST=R_\odot\cos30^\circ=0.866R_\odot. \] All three inferred radii exceed \(CT\) and are smaller than \(R_\odot\), so both intersections lie forward of the Sun.

Using the unrounded radii from C.4:

Source Near position \(d_-/R_\odot\) Far position \(d_+/R_\odot\)
1 \(0.0635\) \(1.67\)
2 \(0.272\) \(1.46\)
3 \(0.424\) \(1.31\)

Moving away from \(S\) along the ray, the candidate points occur in the order \[ 1_{\rm near},\quad2_{\rm near},\quad3_{\rm near}, \quad T,\quad 3_{\rm far},\quad2_{\rm far},\quad1_{\rm far}. \]

The spectrum determines the galactocentric radius of each source but cannot distinguish its near and far positions. Repeating the observations at different longitudes maps hydrogen-rich structures and reveals the galaxy's spiral arms; additional information is needed to resolve individual near–far ambiguities.

D.1

In the outer regime of B.3, \[ M_{\mathrm{tot}}\simeq4\pi C_mR, \qquad v_{c,\infty}^2=4\pi GC_m. \] Eliminating \(C_m\), \[ M_{\mathrm{tot}}\simeq\frac RGv_{c,\infty}^2. \] For a mass-independent radius, \[ \boxed{\eta=\frac RG,\qquad\gamma=2}. \]

For an empirical power law, taking logarithms gives \[ \log_{10}\!\left(\frac{M_{\mathrm{tot}}}{M_\odot}\right) =\gamma_{TF} \log_{10}\!\left(\frac{v_{c,\infty}}{1\,\mathrm{km\,s^{-1}}}\right) +\text{constant}. \] Thus \(\gamma_{TF}\) is the slope in Fig. 4.

Approximate readings of the green line are \[ (x,y)\simeq(1.6,8.2),\qquad(2.8,12.6). \] They give \[ \boxed{\gamma_{TF}\simeq \frac{12.6-8.2}{2.8-1.6}\simeq3.7}. \] The observed scaling is close to a fourth-power law and substantially steeper than the second-power prediction of the fixed-radius Newtonian model.

D.2

The magnitude of the acceleration in a circular orbit is \[ a_m=\frac{v_c^2}{r}. \] Using an outer observed point, \[ v_c\simeq170\,\mathrm{km\,s^{-1}}, \qquad r\simeq10\,\mathrm{kpc}, \] gives \[ a_m\simeq \frac{(1.70\times10^5)^2} {10\times3.09\times10^{19}} =\boxed{9.4\times10^{-11}\,\mathrm{m\,s^{-2}}}. \]

D.3

In the deep-MOND limit \(a\ll a_0\), \[ \mu\!\left(\frac a{a_0}\right) =\frac{a/a_0}{1+a/a_0} \simeq\frac a{a_0}. \] The modified equation of motion in the Newtonian gravitational field of \(M\) becomes \[ \frac{GMm}{r^2}\simeq m\frac{a^2}{a_0}. \] For a circular orbit, \(a=v_{c,\infty}^2/r\), so \[ \frac{GM}{r^2} =\frac{v_{c,\infty}^4}{a_0r^2}. \] The radius cancels: \[ \boxed{v_{c,\infty}^4=GMa_0}. \] Consequently, \[ M=\frac1{Ga_0}v_{c,\infty}^4, \qquad \boxed{\gamma_{\mathrm{MOND}}=4}. \] This also explains why the asymptotic circular speed is independent of radius.

For a graphical estimate of \(a_0\), Fig. 4 gives approximately \[ v_{c,\infty}\simeq100\,\mathrm{km\,s^{-1}}, \qquad M\simeq4\times10^9M_\odot. \] Therefore, \[ a_0\simeq \frac{(1.0\times10^5)^4} {(6.67430\times10^{-11})(4\times10^9)(1.99\times10^{30})} \simeq1.9\times10^{-10}\,\mathrm{m\,s^{-2}}. \] At the precision of the plot, \[ \boxed{a_0\sim2\times10^{-10}\,\mathrm{m\,s^{-2}}}. \]

This is of the stated order \(10^{-10}\,\mathrm{m\,s^{-2}}\). Comparing with D.2 gives \(a_m/a_0\sim0.5\): the displayed outer-galaxy data probe the MOND acceleration scale, but do not establish \(a_m\ll a_0\).

For an asymptotically flat speed of \(170\,\mathrm{km\,s^{-1}}\), the strict deep-MOND condition is \[ r\gg\frac{v_{c,\infty}^2}{a_0}\sim5\,\mathrm{kpc}. \] It becomes increasingly accurate farther out. Near the transition, the full interpolation function must be used, as in D.4.

D.4

The homogeneous sphere has density \[ \rho=\frac{3M}{4\pi R_b^3}. \] Its enclosed mass is \[ M(<r)= \begin{cases} M(r/R_b)^3,&r\le R_b,\\ M,&r\ge R_b. \end{cases} \] The gravitational force remains Newtonian in the modified-second-law formulation specified in the problem. Define its magnitude per unit test mass as \[ g_N(r)=\frac{GM(<r)}{r^2} = \begin{cases} \dfrac{GMr}{R_b^3},&r\le R_b,\\[0.5em] \dfrac{GM}{r^2},&r\ge R_b. \end{cases} \]

Using the full interpolation function, \[ g_N=\mu\!\left(\frac a{a_0}\right)a =\frac{a^2}{a+a_0}. \] Hence \[ a^2-g_Na-g_Na_0=0. \] The physical, non-negative root is \[ a=\frac{g_N+\sqrt{g_N^2+4a_0g_N}}2. \] Since \(v_c^2=ra\), \[ v_c(r)= \left[\frac r2 \left(g_N+\sqrt{g_N^2+4a_0g_N}\right)\right]^{1/2}. \]

Substitution gives the expression valid throughout both regions: \[ \boxed{ v_c(r)= \begin{cases} \displaystyle \left[ \frac{GMr^2}{2R_b^3} \left(1+\sqrt{1+\frac{4a_0R_b^3}{GMr}}\right) \right]^{1/2}, &0<r\le R_b,\\[1.2em] \displaystyle \left[ \frac{GM}{2r} \left(1+\sqrt{1+\frac{4a_0r^2}{GM}}\right) \right]^{1/2}, &r\ge R_b. \end{cases}} \] The two expressions agree at \(r=R_b\), and the limit at the centre is \(v_c(0)=0\).

The limiting cases clarify the shape of the rotation curve:

Region Acceleration regime Circular speed
Inside, \(r<R_b\) \(g_N\gg a_0\) \(\displaystyle v_c\simeq\sqrt{\frac{GM}{R_b^3}}\,r\)
Inside, \(r<R_b\) \(g_N\ll a_0\) \(\displaystyle v_c\simeq\left(\frac{GMa_0}{R_b^3}\right)^{1/4}r^{3/4}\)
Outside, \(r>R_b\) \(g_N\gg a_0\) \(\displaystyle v_c\simeq\sqrt{\frac{GM}{r}}\)
Outside, \(r>R_b\) \(g_N\ll a_0\) \(\displaystyle v_c\simeq(GMa_0)^{1/4}\)

Thus the curve starts at zero, behaves as \(r^{3/4}\) sufficiently close to the centre, and approaches a constant far outside. Newtonian intermediate regions occur only if the sphere's surface acceleration \(GM/R_b^2\) is sufficiently larger than \(a_0\).

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.