[User]
Q1-1
Theory

                                                                                English (Official)

Hydrogen and galaxies (10 points)
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 ħ, defined as ħ = h/2π.

Part A - Introduction
Bohr model
We assume that the hydrogen atom consists of a non-relativistic electron, with mass me , orbiting a fixed
proton. Throughout this part, we assume its motion is on a circular orbit.

     A.1     Determine the electron's velocity v in a circular orbit of radius r.                      0.2pt

In the Bohr model, we assume the magnitude of the electron's angular momentum L is quantized, L = nħ
                                             2
where n > 0 is an integer. We define α = 4πεe ħc ≈ 7.27 × 10−3 .
                                               0

     A.2     Show that the radius of each orbit is given by rn = n2 r1 , where r1 is called the        0.5pt
             Bohr radius. Express r1 in terms of α, me , c and ħ and calculate its numerical
             value with 3 digits. Express v1 , the velocity on the orbit of radius r1 , in terms of
             α and c.

     A.3     Determine the electron's mechanical energy En on an orbit of radius rn in terms           0.5pt
             of e, ε0 , r1 and n. Determine E1 in the ground state in terms of α, me 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 r1 . This
                            →−
produces a magnetic field B 1 .

                                           →
                                           −
     A.4     Determine the magnitude B1 of B 1 at the position of the electron in terms of μ0 ,        0.5pt
             e, α, c and r1 .

                                                          −→
Second, the electron spin creates a magnetic moment M s . Its magnitude is roughly Ms = me ħ. The fine
                                                                                e
                                                                                −→
(F) structure is related to the energy difference ΔEF between an electron with a magnetic moment M s
            →
            −                                −→                   →
                                                                  −
parallel to B 1 and that of an electron with M s anti-parallel to B 1 . Similarly, the hyperfine (HF) structure is
related to the energy difference ΔEHF , due to the interaction between parallel and anti-parallel magnetic
moments of the electron and the proton. It is known to be approximately ΔEHF ≃ 3.72 m         me
                                                                                ΔEF where mp is
                                                                                p
the proton mass.

                                                                                Q1-2
Theory

                                                                                English (Official)

    A.5     Express ΔEF as a function of α and E1 .                                                 0.5pt
            Express the wavelength λHF 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 kpc = 3.09 × 1019 m
   • Solar mass : 1 M⊙ = 1.99 × 1030 kg
We consider a spherical galaxy centered around a fixed point O. At any point P, let ρ = ρ(P) be the
volumetric mass density and φ = φ(P) the associated gravitational potential (i.e. potential energy per
                                            −−→
                                            ‖OP ‖. The motion of a mass m located at P, due to the field
unit mass). Both ρ and φ depend only on r =    
φ, is restricted to a plane containing O.

    B.1     In the case of a circular orbit, determine the velocity vc of an object on a circular   0.2pt
                                                         dφ
            orbit passing through P in terms of r and       .
                                                         dr

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

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

    B.2     Deduce the mass Mb of the bulge of NGC 6946 from the red rotation curve in              0.5pt
            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

                                                                                Q1-3
Theory

                                                                                English (Official)

                                                               Cm
                                               ρm (r) =                                                     (1)
                                                            rm2 + r 2

where Cm > 0 and rm > 0 are constants.

    B.3      Show that the velocity profile vc,m (r), corresponding to the mass density in Eq.      1.8pt
                                               k2 ⋅arctan( rr )
             1, can be written vc,m (r) = k1 −     r
                                                        m
                                                          . Express k1 and k2 in terms of Cm , rm
             and G .
                        r  x2
             ( Hints:  2       dx = r − a arctan(r/a), and: arctan(x) ≃ x − x3 /3 for x ≪ 1. )
                       0 a + x2

             Simplify vc,m (r) when r ≪ rm and when r ≫ rm .
             Show that if r ≫ rm , the mass Mm (r) embedded in a sphere of radius r with the
             mass density given by Eq. 1 simplifies and depends only on Cm 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
                                           2
                               r     ⎡   z ⎤
is given by φG (r, z) = φ0 ln (l) (r) exp − (l) (r) , where z is the distance to the galactic plane (defined by z = 0
                               r0    ⎣ z0 ⎦
), and r < r0 is now the axial radius and φ0 > 0 a constant to be determined. r0 and z0 are constant values.

    C.1      Find the equation of motion on z for the vertical motion of a point mass m             0.5pt
             in such a potential, assuming r is constant. Show that, if r < r0 , the galactic
             plane is a stable equilibrium state by giving the angular frequency ω0 of small
             oscillations around it.

From here on, we set z = 0.

    C.2      Identify the regime, either r ≫ rm or r ≪ rm , in which the model of Eq. 1 recovers    0.6pt
             a potential of the form φG (r, 0) with a suitable definition of φ0 .
             Under this condition vc (r) no longer depends on r. Express it in terms of φ0 .

Therefore, outside the bulge the velocity modulus vc 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 u v (see Fig. 2).

                                                                                Q1-4
Theory

                                                                                English (Official)

                                    Fig. 2: Geometry of the measurement

Let l be the galactic longitude, measuring the angle between SC and the SE. The sun's velocity on its
                                                  →
                                                  −
circular orbit of radius R⊙ = 8.00 kpc is denoted v ⊙ . A galactic object in E orbits on another circle of radius
              →
              −
R at velocity v E . Using a Doppler effect on the previously studied 21 cm line, one can obtain the relative
radial velocity vrE/S of the emitter E with respect to the sun S : it is the projection of v⃗E − v⃗⊙ on the line of
sight.

     C.3     Determine vrE/S in terms of l, R, R⊙ and v⊙ . Then, express R in terms of R⊙ , v⊙ ,        0.7pt
             l and vrE/S .

Using a radio telescope, we make observations in the plane of our galaxy toward a longitude l = 30°. The
frequency band used contains the 21 cm line, whose frequency is f0 = 1.42 GHz. The results are reported
in Fig. 3.

     Fig. 3: Electromagnetic signal as a function of the frequency shift, measured in the radio
     frequency band at l = 30° using EU-HOU RadioAstronomy

                                                                                Q1-5
Theory

                                                                                English (Official)

    C.4      In our galaxy, v⊙ = 220 km ⋅ s−1 . Determine the values of the relative radial ve-                           0.6pt
             locity (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⊙ .

    C.5      On the top view of our galaxy (in the answer box), indicate the positions of the                             0.6pt
             sources observed in Fig. 3.
             What could be deduced from repeated measurements changing l?

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 vc,∞ as a function of the measured
total mass Mtot of each galaxy gives an interesting correlation called the Tully-Fischer relation, see Fig. 4
(right).

                          v c,1

                   NGC 3621                  NGC 5055

                                                          log 10 ( M tot / M )
                                                                  ̄

                    NGC 2903                 NGC 3521
                                                                                Spiral galaxies
                                                                                from different surveys

                                                                                Tully Fischer Relation

                   NGC 7321                  NGC 2841
                                                                                log 10( vc,1 / 1 km/s )

     Fig. 4. Left: Rotation curves for typical spiral galaxies - Right: log10 (Mtot ) as a function of
     log10 (vc,∞ ) on linear scales. Colored dots correspond to different galaxies and different sur-
     veys. 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      Assuming that the radius R of a galaxy doesn't depend on its mass, show that                                 0.4pt
                                                                               γ
             the model of Eq. 1 (part B) gives a relation of the form Mtot = ηvc,∞ where γ and
             η should be specified.
             Compare this expression to the Tully-Fischer relation by computing γT F .

In the extremely low acceleration regime, of the order of a0 = 10−10 m ⋅ s−2 , the MOdified Newtonian Dy-
                                                                                →
                                                                                −       a −→
namics (MOND) theory suggests that one can modify Newton's second law using F = mμ (l) (r) a where
                                                                                a0
     →
     −                                                                                x
a= ‖a ‖ is the modulus of the acceleration and the μ function is defined by μ(x) = 1 + x .

                                                                          Q1-6
Theory

                                                                                English (Official)

  D.2    Using data for NGC 6946 in Fig. 1, estimate, within Newton's theory, the mod-       0.2pt
         ulus of the acceleration am of a mass in the outer regions of NGC 6946.

  D.3    Let m be a mass on a circular orbit of radius r with velocity vc,∞ in the gravity   0.8pt
         field of a fixed mass M .
         Within the MOND theory, with a ≪ a0 , determine the Tully-Fischer exponent.
         Using data for NGC 6946 and/or Tully-Fischer law, calculate a0 to show that
         MOND operates in the correct regime.

  D.4    Considering relevant cases, determine vc (r) for all values of r in the MOND the-   0.9pt
         ory in the case of a gravitational field due to a homogeneously distributed mass
         M with radius Rb .

Solution:
Q1-1
Theory

                                                                                English (Official)

Hydrogen and galaxies (10 points)
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 ħ, defined as ħ = h/2π.

Part A - Introduction
Bohr model
We assume that the hydrogen atom consists of a non-relativistic electron, with mass me , orbiting a fixed
proton. Throughout this part, we assume its motion is on a circular orbit.

     A.1     Determine the electron's velocity v in a circular orbit of radius r.                                  0.2pt

SOLUTION:
Newton's second law on the electron in the electrical field of the proton for a circular orbit and projected
   →
   −          2        2                          e2
on ur : −me vr = − 4πεe r 2 hence v = 
                        0                      4πε0 me r

Marking Scheme

                      A.1.1 : Using Newton's second law                                         0.1
                      A.1.2 : Expression of the velocity                                        0.1

In the Bohr model, we assume the magnitude of the electron's angular momentum L is quantized, L = nħ
                                             2
where n > 0 is an integer. We define α = 4πεe ħc ≈ 7.27 × 10−3 .
                                                      0

     A.2     Show that the radius of each orbit is given by rn = n2 r1 , where r1 is called the                    0.5pt
             Bohr radius. Express r1 in terms of α, me , c and ħ and calculate its numerical
             value with 3 digits. Express v1 , the velocity on the orbit of radius r1 , in terms of
             α and c.

SOLUTION:
If the norm L of the angular momentum is quantified, for a circular orbit of radius rn it is L = me rn vn = nħ.
In the previous question, we have already obtained a relation between r and v that can be used for rn
and vn and gives vn =  4πε em r =  mαħcr . Then using the quantifed expression we get rn = mnħv = m      me rn
                             2                                                                       nħ
                                 0   e n       e n
                                                                                 αħce n     e
             2
thus rn = αm
           ħn
              c
                and then r1 = αmħ c . For the numerical value we previously compute α = 7.27 × 10−3 and
             e                             e

                                                                     2      2                         2
then r1 = 5.31 × 10   −11
                            m . For the velocity, we get me v12 = 4πεe r = e4πε
                                                                             me v 1
                                                                                ħ
                                                                                and then v1 = 4πε
                                                                                e
                                                                                ħ
                                                                                = αc .
                                                                     0 1        0                     0

Marker Scheme

                                                                                Q1-2
Theory

                                                                                English (Official)

                       A.2.1 : Expression of rn                                                   0.1
                       A.2.2 : Expression of r1                                                   0.1
                       A.2.3 : Numerical value for r1                                             0.1
                       A.2.4 : Expression of v1                                                   0.2

     A.3         Determine the electron's mechanical energy En on an orbit of radius rn in terms                0.5pt
                 of e, ε0 , r1 and n. Determine E1 in the ground state in terms of α, me and c.
                 Compute its numerical value in eV.

SOLUTION:
                                                   2      2                        2
The mechanical energy is En = 12 me vn2 − 4πεe r = − 8πεe r , hence En = − 8πεen2 r        then for the ground state
                                                   0 n    0 n                  0       1

           2
E1 = − 8πεe r . Using the expression of α , we get the beautiful formula E1 = − 12 α2 me c2 . The numerical
           0 1

value is E1 = −2.17 × 10−18 J which corresponds to E1 = −13.6 eV .

Marker Scheme

                       A.3.1 : Expression for En                                                  0.2
                       A.3.2 : Expression for E1 with α                                           0.2
                       A.3.3 : Numerical value for E1                                             0.1

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 r1 . This
                            →−
produces a magnetic field B 1 .

                                               →
                                               −
     A.4         Determine the magnitude B1 of B 1 at the position of the electron in terms of μ0 ,             0.5pt
                 e, α, c and r1 .

SOLUTION:
The period of the motion is : T = 2πr
                                   v
                                      1
                                        .
                                         1

The current i corresponding to the orbit of the proton is i = Te hence i = 2πr
                                                                           ev1   eαc
                                                                               = 2πr .
                                                                                1      1

The magnetic field created by a loop with current i and radius R is : B = μ2R0 i , which here gives B1 = μ4πr
                                                                                0 eαc
                                                                                2
.
                                                                                1

Marker Scheme

                                                                                Q1-3
Theory

                                                                                English (Official)

                          A.4.1 : Expression for the period                               0.1
                          A.4.2 : Expression for the current                              0.2
                          A.4.3 : General expression for B                                0.1
                          A.4.4 : Inject i into B                                         0.1

                                                          −→
Second, the electron spin creates a magnetic moment M s . Its magnitude is roughly Ms = me ħ. The fine
                                                                                e
                                                                                −→
(F) structure is related to the energy difference ΔEF between an electron with a magnetic moment M s
            →
            −                                −→                   →
                                                                  −
parallel to B 1 and that of an electron with M s anti-parallel to B 1 . Similarly, the hyperfine (HF) structure is
related to the energy difference ΔEHF , due to the interaction between parallel and anti-parallel magnetic
moments of the electron and the proton. It is known to be approximately ΔEHF ≃ 3.72 m         me
                                                                                ΔEF where mp is
                                                                                p
the proton mass.

     A.5          Express ΔEF as a function of α and E1 .                                                0.5pt
                  Express the wavelength λHF of a photon emitted during a transition between
                  the two states of the hyperfine structure and give its numerical value with two
                  digits.

SOLUTION:
                                                                                −→
The potential energy corresponding to the interaction between the spin magnetic moment M s and the
                               −→ − →
nuclear magnetic field : Ep = −M s ⋅ B 1
                                                                                →
                                                                                −
The difference ΔEF between the energy of two electrons with a spin parallel and antiparallel to B 1
is then ΔEF = 2Ms B1 . Using previous expressions one finds: ΔEF = 2 me ħB1 = 2 me ħ μ4πr 2 which writes
                                                                                0 eαc
                                                                                e       e     1

ΔEF = −4α E1 hence ΔEHF = −3.72 m
              2                 me
                                    4α2 E1 .
                                  p

The wavelength of the photon corresponding to this transition is then λhc = ΔEHF = −3.72. m
                                                                                me
                                                                                4α2 E1 hence
                                                                              HF                     p

λHF = − 3.72. mhc
                e 4α2 E   whose value is λHF = 21 cm .
               mp     1

Marker Sheme

                          A.5.1 : Expression for the potential energy                     0.1
                          A.5.2 : Expression for ΔEF                                      0.1
                          A.5.3 : Expression for ΔEHF in term of α                        0.1
                          A.5.4 : Expression for λHF                                      0.1
                          A.5.5 : Numerical value for λHF                                 0.1

Part B - Rotation curves of galaxies
Data
   • Kiloparsec: 1 kpc = 3.09 × 1019 m
   • Solar mass : 1 M⊙ = 1.99 × 1030 kg

                                                                                Q1-4
Theory

                                                                                English (Official)

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

     B.1     In the case of a circular orbit, determine the velocity vc of an object on a circular       0.2pt
                                                          dφ
             orbit passing through P in terms of r and       .
                                                          dr

SOLUTION:
                                      →
                                      −    →
                                           −                −
                                                            →
The force created by the potential is F = −∇(mφ(r)) = −m dφ u . Newton's second law for a circular orbit
                                                         dr r
               2
then gives m vrc = m dφ
                     dr
                        hence vc = r dφ
                                      dr
                                         .

SOLUTION:

                    B.1.1 : Using Newton's second law                                      0.1
                    B.1.2 : Expression for the velocity.                                   0.1

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

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

     B.2     Deduce the mass Mb of the bulge of NGC 6946 from the red rotation curve in                  0.5pt
             Fig. 1(B), in solar mass units.

SOLUTION:
Either by Gauss's theorem 4πr 2 g(r) = −4πG Mint (r) , then one gets g(r) = G Mint (r)/r 2 g(r) = −G Mint (r)/r 2 .
                                                                                :::::::::::::::::
or one knows the law g(r) = G M /r g(r)
                                     2
                                           = −G M /r and intuits that one can use the interior mass
                                       ::::::::::::::
                                                       2

                                                                                Q1-5
Theory

                                                                                English (Official)

g(r) = G Mint (r)/r 2 g(r) = −G Mint (r)/r 2
                      :::::::::::::::::

If there is almost no more mass after the bulge radius rb

then if r > rb , Mint (r) = Mb and ⃗
                                   g(r > rb ) = − GrM2 b u⃗r . But g⃗ = − dφ u⃗ .
                                                                          dr r

This gives vc (r > rb ) =  G M
                              r
                                b
                                  .
One can then deduce that if the velocity is given only by the bulge, at a given distance R we
must have Mb = vc2 R/G .              On the red curve we can read vc = 20 km ⋅ s−1 at R = 10 kpc hence
        2        8    20
Mb = vGc R ≃ 4.10  ×3.10
               6,7.10−11
                         ≃ 1, 8.1039 kg so that Mb ≃ 9.108 M⊙ .

Marker Scheme

                       B.2.1 : g(r) = G Mint (r)/r 2 g(r) = −G Mint (r)/r 2:via Gauss'
                                                     :::::::::::::::::
                                                                                0.1
                       Theorem or another method resulting in an equivalent
                       result.
                       B.2.2 : Expression for ⃗
                                             g(r > rb )                                    0.1
                       B.2.3 : Expression for Mb                                           0.1
                       B.2.4 : Taking the right value of vc in the figure                  0.1
                       B.2.5 : Numerical value for Mb with a tolerance of ±25%             0.1

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

                                                                      Cm
                                                       ρm (r) =      2
                                                                                (1)
                                                                    rm + r 2

where Cm > 0 and rm > 0 are constants.

     B.3       Show that the velocity profile vc,m (r), corresponding to the mass density in Eq.         1.8pt
                                                       k2 ⋅arctan( rr )
               1, can be written vc,m (r) = k1 −     r
                                                          m
                                                            . Express k1 and k2 in terms of Cm , rm
               and G .
                          r  x2
               ( Hints:  2       dx = r − a arctan(r/a), and: arctan(x) ≃ x − x3 /3 for x ≪ 1. )
                         0 a + x2

               Simplify vc,m (r) when r ≪ rm and when r ≫ rm .
               Show that if r ≫ rm , the mass Mm (r) embedded in a sphere of radius r with the
               mass density given by Eq. 1 simplifies and depends only on Cm and r.
               Estimate the mass of the galaxy NGC 6946 actually present in the picture in Fig.
               1(A).

SOLUTION:
                                                                                →
                                                                                −     −→
On the one hand, writing Gauss' theorem on a sphere of radius r gives ∫ g(r) ⋅ dS = 4πr 2 g(r) = −4πG Mint
and thus g(r) = G Mint (r)/r 2 g(r) = −G Mint (r)/r 2 . As long as this final formula is given it doesn't matter the
                               :::::::::::::::::
method.
                                       r
But, on the other hand Mint = ∫0 4πx2 ρ(x)dx = 4πCm r − rm arctan (l) rr (r) hence
                                                                                m

                                                                                Q1-6
Theory

                                                                                English (Official)

                                                         4πCm G r−rm arctan(l) rr (r)
                                            gm (r) = −              r2
                                                                               m
                                                                                (2)

            v2                 v2
But as −m c,m
          r
              = −mgm (r) −m c,m = mgm (r) we finally get vc,m = √rgm (r) vc,m
                        ::::::::::::::::
                             r
                                                                              = √−rgm (r):wich writes
                                                                         ::::::::::::::

                                                
                                                                          r
                                                 4πCm G r − rm arctan (l) rm (r)
                                         vc,m = ⎷                                                                  (3)
                                                               r

One can then read k1 = 4πCm G and k2 = 4πCm G rm

Two regime could be considered:

                                                                                2
   • if r ≪ rm , a third order Taylor expansion of arctan gives vc,m ≃  4πC3rm2G r ,
                                                                                m

   • and if r ≫ rm then arctan (l) rr (r) ≃ π/2 and vc,m ≃ √4πCm G .
                                     m

The function vc,m (r) is vanishing when r → 0 and is asymptotically constant with value √4πCm G when
r → +∞ : this corresponds to the observational curve for the galaxy considered (black circles on the right
part of figure 1(B). A natural interpretation for rm is the typical radius beyond which the circular velocity
is constant. On this picture one can read vc ≃ 160 km ⋅ s−1 for the constant value of vc,m after rm , then one
                    2          5 2
can deduce Cm = 4πG
                 vc     (1,6.10 )
                    ≃ 4π×6.67.10  −11 ≃ 3.10
                                             19
                                                kg ⋅ m−1 . The mass embedded in a sphere of radius r is given by
        r
Mint = ∫0 4πx2 ρm (x)dx = 4πCm r − rm arctan (l) rr (r) which reduces to Mint ≃ 4πCm r if r ≫ rm . In the picture we
                                                  m

have a radius R = 9 kpc = 2.27 × 1020 m of the galaxy, then a mass Min the figure ≃ 4πCm R ≃ 1041 kg ≃ 1011 M⊙ .
This mass corresponds to more than ten times the value of the mass actually visible in this picture : this
is the dark matter concept.
Marker Scheme

                    B.3.1 : g(r) = G Mint (r)/r 2 via Gauss' Theorem or another               0.2
                    method resulting in an equivalent result.
                    B.3.2 : Interior mass                                                     0.3
                    B.3.3. : Expression for g(r)                                              0.1
                    B.3.4 : Using Newton's second law                                         0.1
                    B.3.5 : Expression for k1                                                 0.1
                    B.3.6 : Expression for k2                                                 0.1
                    B.3.7 : Simplification for vc in the case r ≪ rm                          0.2
                    B.3.8 : Simplification for vc in the case r ≫ rm                          0.2
                    B.3.9 : Value of Cm                                                       0.2
                    B.3.10 : Expression for Mm in the case r ≫ rm                             0.2
                    B.3.11 : Mass in the figure (good if nearest power of ten)                0.1

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

                                                                                Q1-7
Theory

                                                                                English (Official)

                                              2
                               r     ⎡   z ⎤
is given by φG (r, z) = φ0 ln (l) (r) exp − (l) (r) , where z is the distance to the galactic plane (defined by z = 0
                               r0    ⎣ z0 ⎦
), and r < r0 is now the axial radius and φ0 > 0 a constant to be determined. r0 and z0 are constant values.

     C.1      Find the equation of motion on z for the vertical motion of a point mass m                                0.5pt
              in such a potential, assuming r is constant. Show that, if r < r0 , the galactic
              plane is a stable equilibrium state by giving the angular frequency ω0 of small
              oscillations around it.

SOLUTION:
                                                        → −
                                                        −   →     →
                                                                  −                →
                                                                                −
The equation of motion is given by Newton's second law ma = F = −m∇φ, projected on uz , it gives
                                                                                2
mz̈ = −m ∂φ
         ∂z
            . Using the given potential we have z̈ = 2z φ ln (l) rr (r) exp − (l) zz (r)  . Near the galactic plane (
                                                     z2 0        0        0                   0

z = 0 ) the exponential is equal to 1 and can be simplified to give z̈ ≃ 2z φ ln (l) rr (r) . If r < r0 the ln is negative
                                                                         z2 0         0               0

                                                                     0 
                                                                       ln (l) rr (r)| . This proves that z is oscillating
and the equation of motion is of the form z̈ ≃ −ω02 z with ω0 =  2φ
                                                                   z2 |       0           0

around z = 0 and that the motion is stable.
Marker Scheme

                     C.1.1 : Newton's second law, or equivalent method                                    0.1
                     C.1.2 : Projection on the z axis                                                     0.1
                     C.1.3 : Equation of motion                                                           0.1
                     C.1.4 : Equation near the galactic plane                                             0.1
                     C.1.5 : Expression for ω0                                                            0.1

From here on, we set z = 0.

     C.2      Identify the regime, either r ≫ rm or r ≪ rm , in which the model of Eq. 1 recovers                       0.6pt
              a potential of the form φG (r, 0) with a suitable definition of φ0 .
              Under this condition vc (r) no longer depends on r. Express it in terms of φ0 .

SOLUTION:
Using the density given by equation (1) in part B, we have obtained

                                                     4πCm G r − rm arctan (l) rr (r)
                                        gm (r) = −                                m
                                                                                (4)
                                                                  r2
                                                                     4πCm G
Hence, considering r ≫ rm , one can simplify this relation to gm (r) ≃ −      . The gravitational potential
                                                                         r
can be obtained by integration, we then have : φ(r) = +4πCm G ln(r) + cst . The constant can be found by
correctly choosing the origin of the potential. This potential corresponds to: φG (r, z = 0) = φ0 ln (l) rr (r) with
                                                                                0
                                                                                2
φ0 = +4πCm G . In that case, the equation of motion in the galactic plane gives −m vrc = −mgm (r) which
writes vc = √rgm (r) = √4πCm G , so that vc = √φ0 .

                                                                                Q1-8
Theory

                                                                                English (Official)

Marker Scheme

                    C2.1 : Condition for simplication r ≫ rm                               0.1
                    C2.2 : Expression for φ(r)                                             0.2
                    C2.3 : Identification of φ0                                            0.1
                    C2.4 : Newton's second law                                             0.1
                    C2.5 : Expression for vc                                               0.1

Therefore, outside the bulge the velocity modulus vc 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 u v (see Fig. 2).

                                    Fig. 2: Geometry of the measurement

Let l be the galactic longitude, measuring the angle between SC and the SE. The sun's velocity on its
                                                  →
                                                  −
circular orbit of radius R⊙ = 8.00 kpc is denoted v ⊙ . A galactic object in E orbits on another circle of radius
              →
              −
R at velocity v E . Using a Doppler effect on the previously studied 21 cm line, one can obtain the relative
radial velocity vrE/S of the emitter E with respect to the sun S : it is the projection of v⃗E − v⃗⊙ on the line of
sight.

     C.3     Determine vrE/S in terms of l, R, R⊙ and v⊙ . Then, express R in terms of R⊙ , v⊙ ,         0.7pt
             l and vrE/S .

                                                                                Q1-9
Theory

                                                                                English (Official)

SOLUTION:
             −→                        −→
We have Ss = v⊙ sin(α)u v and Ee = vE cos(β)u v . In the right triangle SsSv the sum of angles gives
 
 −
 →   −−→                                                                           
                                                                                −
                                                                                →   −−→
(l)Ss, SSv (r) = π2 − α, but, as v⊙⃗ is perpendicular to the radius C S, we also have (l)Ss, SSv (r) = π2 − l : then α = l.

On the other side, we have C T = R⊙ sin(l) = R sin( π2 − β), which gives cos(β) = RR⊙ sin(l) . Merging all of
                                                                    −→ − →
these results and taking into account that vE = v⊙ and that vrE/S
                                                              ⃗   = Ee − Ss we have vrE/S = v⊙ (l) RR⊙ − 1(r) sin(l)
                      R⊙
and finally R =                 .
                  1 + v vsin(l)
                         rE/S
                       ⊙

Marker Scheme

                                           −
                                           →
                     C3.1 : Expression for Ss                                              0.1
                                           −→
                     C3.2 : Expression for Ee                                              0.1
                     C3.3 : α = l                                                          0.1
                     C3.4 : Expression for cos(β)                                          0.1
                     C3.5 : Expression for vr,E/S                                          0.2
                     C3.6 : Expression for R                                               0.1

Using a radio telescope, we make observations in the plane of our galaxy toward a longitude l = 30°. The
frequency band used contains the 21 cm line, whose frequency is f0 = 1.42 GHz. The results are reported
in Fig. 3.

     Fig. 3: Electromagnetic signal as a function of the frequency shift, measured in the radio
     frequency band at l = 30° using EU-HOU RadioAstronomy

     C.4     In our galaxy, v⊙ = 220 km ⋅ s−1 . Determine the values of the relative radial ve-          0.6pt
             locity (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⊙ .

SOLUTION:

                                                                          Q1-10
Theory

                                                                                English (Official)

In Fig. 3 one can measure the 3 frequency shifts ( f − f0 ) corresponding to each peak : Δf1 = 0.03 MHz ,
Δf2 = 0.15 MHz and f3 = 0.26 MHz. One can then compute the relative Doppler velocity using vr,i = cΔfi /f0
, with f0 = 1 420 MHz one gets

   • vr,1 = 6.33 km ⋅ s−1

   • vr,2 = 31.7 km ⋅ s−1

   • vr,3 = 54.9 km ⋅ s−1

As peaks are placed on grid points, the tolerance in the value is due to fact that candidates could use
c = 3.00 × 108 m/s in the place of the 9 digits given in the formulary.
The corresponding distances from the galactic center are then obtained using the relation Ri =        R⊙
                                                                                v         ,
                                                                                1+ v r,i
                                                                                sin l
                                                                                ⊙
with l = 30° we obtain :
   • R1 = 0.95R⊙

   • R2 = 0.78R⊙

   • R3 = 0.67R⊙

Marker Scheme

                    C4.1 : Doppler formula for vr                                  0.1
                    C4.2 : Getting the 3 numerical values for Δf                   0.2
                    C4.3 : Numerical values of the 3 velocities (± 0.01 km ⋅ s )
                                                                              −1
                                                                                0.2
                    C4.4 : Numerical values of the 3 distances (± 0.01 R⊙ )        0.1

    C.5      On the top view of our galaxy (in the answer box), indicate the positions of the    0.6pt
             sources observed in Fig. 3.
             What could be deduced from repeated measurements changing l?

SOLUTION:
As indicated on the figure below, the right line of sight could be obtained geometrically (i.e. without
protractor) : using the 15° grid graduation one can going back from 30° from the perpendicular line to
CS, we then obtain a radius which perpendicular to the line of sight, in other words as sin(30°)=0.5 the
line of sight is passing by S and is tangenting the circle of radius CS/2.
Drawing the circles or radius , and , the line of sight with from we get 2 possible intersection for each
peak : a near one and a far one. We plot only the nearest for each source on the answer figure.

                                                                        Q1-11
Theory

                                                                                English (Official)

                                                                 .
                                                                 ..
                                       R
                                         .  3
                                                            `=30°

                                                                                30°
                                                       R2
                                                      R1

The far intersections for each source is much further away and hence is likely less intense. Astronomers
could also use the variation in the radio signal when they slowly vary the longitude to determine the
right position of the actual source. A continuous variation of l in the interval [0, 2π] makes hydrogen
sources appear in the galaxy, as the galaxy is essentially composed of hydrogen, one can trace its mass
distribution : i.e. the spiral structure.
Marker Scheme

                   C5.1 : Getting the right line of sight                             0.1
                   C5.2 : Drawing for the 3 circles                                   0.2
                   C5.3 : Drawing for the 3 points                                    0.2
                   C5.4 : Deduction                                                   0.1

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 vc,∞ as a function of the measured
total mass Mtot of each galaxy gives an interesting correlation called the Tully-Fischer relation, see Fig. 4
(right).

                                                                                Q1-12
Theory

                                                                                English (Official)

                          v c,1

                    NGC 3621                  NGC 5055

                                                           log 10 ( M tot / M )
                                                                   ̄

                    NGC 2903                  NGC 3521
                                                                                Spiral galaxies
                                                                                from different surveys

                                                                                Tully Fischer Relation

                    NGC 7321                  NGC 2841
                                                                                log 10( vc,1 / 1 km/s )

     Fig. 4. Left: Rotation curves for typical spiral galaxies - Right: log10 (Mtot ) as a function of
     log10 (vc,∞ ) on linear scales. Colored dots correspond to different galaxies and different sur-
     veys. 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     Assuming that the radius R of a galaxy doesn't depend on its mass, show that                                  0.4pt
                                                                               γ
             the model of Eq. 1 (part B) gives a relation of the form Mtot = ηvc,∞ where γ and
             η should be specified.
             Compare this expression to the Tully-Fischer relation by computing γT F .

SOLUTION:
We have obtained vc,∞    2
                           = 4πCm G and for a galaxy of radius R , we have Mtot ≃ 4πCm R. This gives Cm = M tot
                                                                                4πR
                                                                       γ
and vc,∞
     2        M
         = 4π 4πR
                tot
                    G . This relation is of the expected form Mtot = ηvc,∞ with γ = 2 and η = R/G . Analysing
the data we get the power law exponent of the Tully-Fisher relation as γT F ≃ 2.6−1.8
                                                                               12−9
                                                                                = 3.75 : the dark
matter model from part B is not able to reproduce this law.
Marker Scheme

                   D1.1 : Recall for vc,∞                                                          0.1
                   D1.2 : Expression for η                                                         0.1
                   D1.3 : Expression for γ                                                         0.1
                   D1.4 : Numerical value for γT F (correct if it is between 3.5                   0.1
                   and 4)

In the extremely low acceleration regime, of the order of a0 = 10−10 m ⋅ s−2 , the MOdified Newtonian Dy-
                                                                                →
                                                                                −       a −→
namics (MOND) theory suggests that one can modify Newton's second law using F = mμ (l) (r) a where
                                                                                a0
     →
     −                                                                                x
a= ‖a ‖ is the modulus of the acceleration and the μ function is defined by μ(x) = 1 + x .

                                                                                Q1-13
Theory

                                                                                English (Official)

     D.2        Using data for NGC 6946 in Fig. 1, estimate, within Newton's theory, the mod-          0.2pt
                ulus of the acceleration am of a mass in the outer regions of NGC 6946.

SOLUTION:
Considering that outer orbits are circular, the corresponding acceleration for a test mass m is radial and
given in newtonian theory by am ≃ vc2 /R . In the case of NGC 6946, the value of the velocity is roughly
constant and equal to vc = 160 km ⋅ s−1 as far R > 5 kpc. For this smallest distance from the center, the
                                  5 2
acceleration is am = (1.6.10 )
                      5.3.1019
                               ≃ 1.5 × 10−10 m ⋅ s−2 , this value is the maximal acceleration to which as star is
submitted in the outer regions of this galaxy. It corresponds to the MOND regime.
Marker Scheme

                            D2.1 : Expression for am                                     0.1
                            D2.2 : Numerical value for am (good nearest power of ten)    0.1

     D.3        Let m be a mass on a circular orbit of radius r with velocity vc,∞ in the gravity      0.8pt
                field of a fixed mass M .
                Within the MOND theory, with a ≪ a0 , determine the Tully-Fischer exponent.
                Using data for NGC 6946 and/or Tully-Fischer law, calculate a0 to show that
                MOND operates in the correct regime.

SOLUTION:
                                                                 →−         →
                                                                            −
If x = a/a0 ≪ 1, then μ(x ≪ 1) ≃ x and MOND theory gives F = m aa a. Considering a gravitational inter-
                                                                          0
action between M and m we then have for the radial component of the modified Newton's second Law
              2
GM r2
      m = m aa . The radial acceleration on a circular orbit of radius r is always given by a = vc,∞
                                                                                2
                                                                                /r, the mod-
            0
                                            v4
ified second law writes now G M
                              r2
                                 = r 2c,∞
                                       a
                                          which gives vc,∞ = (a0 G M )1/4 , and thus M = a 1G vc,∞
                                                                                4
                                                                                . Considering
                                                 0                                         0

the notation from D.1, this is a power law relation with γMOND = 4 in accordance with the Tully-Fischer
relation.
                                                                             v
For the NGC 6946 galaxy, we read vc,∞ = 160 km ⋅ s−1 thus log10 (l) 1 km⋅s
                                                                     c,∞
                                                                        −1 (r) = 2.2 and one can read the corre-

sponding total mass by the Tully-Fischer relation as log(Mtot /M⊙ ) = 10.5 thus Mtot = 2.1040,5 kg. One can
obtain similar numbers using experimental data on the curve of Fig. 4. Introducing these values in the
                 v4
relation a0 = G M
                c,∞
                            it gives a0 = 1.5 × 10−10 m ⋅ s−2 as expected.
                      tot

Marker Scheme

                                                                                Q1-14
Theory

                                                                                English (Official)

                           D3.1 : Considering the hypothesis a ≪ a0                                        0.1
                           D3.2 : Newton's second law                                                      0.1
                           D3.3 : Expression for vc,∞                                                      0.1
                           D3.4 : Numerical value for γM ON D                                              0.1
                           D3.5 : Numerical value for log10 (vc,∞ /1 km/s)                                 0.1
                           D3.6 : Numerical value for log10 (M )                                           0.1
                           D3.7 : Expression for a0                                                        0.1
                           D3.8 : Numerical value for a0 (good if nearest power of ten)                    0.1

     D.4      Considering relevant cases, determine vc (r) for all values of r in the MOND the-                          0.9pt
              ory in the case of a gravitational field due to a homogeneously distributed mass
              M with radius Rb .

SOLUTION:
Taking the full formula for μ, the modified second law with circular velocity vc at radius r writes now
                v2
                 f
                       vf2
G(r)m = −m     a0 r
                 v2    r
                                 where G(r) is the gravitational field of the homogeneous ball of mass M and with
                   f
              1+ a r
                   0
radius Rb . This field can be deduced from Gauss' theorem it is

                                                       ⎧ −G M /r 2             if    r > Rb
                                                G(r) = ⎨                                                                         (5)
                                                       ⎩ −G M r/Rb
                                                                  3
                                                                               if    r ≤ Rb

Outside the ball : r > Rb . After a small reorganisation, vc appears to be solution of the biquadratic equa-
tion vc4 − GrM vc2 − a0 G M = 0 . The positive root of this equation is

                                              
                                              GM ⎛         4a r 2 ⎞
                                     vc (r) =      1 + 1 + 0             which is valid only if r > Rb                         (6)
                                              ⎷ 2r ⎝         GM ⎠

                                                                   v4
When r → ∞, vc is asymptotically constant and M → ac,∞
                                                     G
                                                       which is the Tully-Fisher relation. Inside the ball
                                                                       0

: r ≤ Rb . With a similar reorganisation, vc appears now to be solution of another biquadratic equation
                                 3                   3
which is vc4 − GrM (l) Rr (r) vc2 − a0 G M (l) Rr (r) = 0 . The positive solution is now
                             b                   b

                                
                                GM r 3 ⎡        4a r 2 R 3 ⎤
                       vc (r) =     (l) (r) 1 + 1 + 0 (l) b (r)                           which is valid only if r ≤ Rb                (7)
                                ⎷ 2r Rb ⎣         GM r ⎦

When r → 0, we recover vc → 0 as in the experimental data.

Marker Scheme

                                                           Q1-15
Theory

                                                                  English (Official)

         D4.1 : Modified second law                         0.1
         D4.2 : Gravitational field in the case r > Rb      0.1
         D4.3 : Gravitational field in the case r < Rb      0.1
         D4.4 : Bi-quadratic equation in the case r > Rb    0.1
         D4.5 : Expression for vc in the case r > Rb        0.1
         D4.6 : Behaviour in the limit r → ∞                0.1
         D4.7 : Bi-quadractic equation for r < Rb           0.1
         D4.8 : Expression for vc when r < Rb               0.1
         D4.9 : Behaviour when r → 0                        0.1

Theory


[User]
We are given the integral
                                             Z ∞
                                                          2        √
                                                     e−x dx =           π.
                                              −∞

    For positive a, find the value of the integral
                                              Z ∞

                                                     e−ax +bx+c dx
                                               −∞

    and verify that your answer makes dimensional sense.

       Remark
       Consider the value of the definite integral
                                               Z x
                                                           ′2
                                                      e−x dx′ .
                                                 −∞

       You can try all day to compute the value of this integral, using all the integration tricks

       you know, but nothing will work. The function e−x simply doesn’t have an antiderivative
       in terms of the functions you already know, i.e. in terms of polynomials, exponents and
       logarithms, and trigonometric functions (for more discussion, see here).

        If you ask a computer algebra system like Mathematica, it’ll spit out something involv-

        ing erf(x), which is defined by being an antiderivative of e−x . But is this really an

       “analytic” solution? Isn’t that just saying “the integral of e−x is equal to the integral

        of e−x ”? Well, like many things in math, it depends on what the meaning of the word “is” is.

       The fact is, the set of functions we regard as “elementary” is arbitrary; we just choose a set
       that’s big enough to solve most of the problems we want, and small enough to attain fluency
       with. (Back in the days before calculators, it just meant all the functions whose values were
       tabulated in the references on hand.) If you’re uncomfortable with erf(x), note that a similar
       thing would happen if a little kid asked you what the ratio of the opposite to adjacent sides
       of a right triangle is. You’d say tan(x), but they could say it’s tautological, because the only
       way to define tan(x) at their level is as the ratio of opposite to adjacent sides. Similarly,
       1/x has no elementary antiderivative – unless you count log(x) as elementary, but ultimately
       log(x) is simply defined to be such an antiderivative. It’s all tautology, but it’s still useful.

Solution:
We can simply factor out the dependence on c, and get rid of the a in the exponent by
              √
    taking u = a x, to get                Z ∞
                                       ec         2    √
                                       √       e−u +(b/ a)u du.
                                         a −∞
    To get rid of the pesky linear term in the exponent, we note that
                                                      2
                                                                         b2

                                           b                        b

                                −v = − u − √               = −u2 + √ u −
                                          2 a                        a   4a

    so that we can “complete the square” in the exponent by working in terms of v, for

                                  ec+b /4a ∞ −v2
                                          Z                         r
                                                           c+b2 /4a   π
                                    √          e    dv = e              .
                                      a     −∞                        a

    To check this makes sense, let’s again suppose that [x] = m, which implies [a] = m−2 , [b] = m−1 ,
    and [c] = 1. The overall integral must have dimensions of m, which it does, and the argument of
    the exponent must be dimensionless, which it is.

       Remark
       Consider the value of the definite integral
                                               Z x
                                                           ′2
                                                      e−x dx′ .
                                                 −∞

       You can try all day to compute the value of this integral, using all the integration tricks

       you know, but nothing will work. The function e−x simply doesn’t have an antiderivative
       in terms of the functions you already know, i.e. in terms of polynomials, exponents and
       logarithms, and trigonometric functions (for more discussion, see here).

        If you ask a computer algebra system like Mathematica, it’ll spit out something involv-

        ing erf(x), which is defined by being an antiderivative of e−x . But is this really an

       “analytic” solution? Isn’t that just saying “the integral of e−x is equal to the integral

        of e−x ”? Well, like many things in math, it depends on what the meaning of the word “is” is.

       The fact is, the set of functions we regard as “elementary” is arbitrary; we just choose a set
       that’s big enough to solve most of the problems we want, and small enough to attain fluency
       with. (Back in the days before calculators, it just meant all the functions whose values were
       tabulated in the references on hand.) If you’re uncomfortable with erf(x), note that a similar
       thing would happen if a little kid asked you what the ratio of the opposite to adjacent sides
       of a right triangle is. You’d say tan(x), but they could say it’s tautological, because the only
       way to define tan(x) at their level is as the ratio of opposite to adjacent sides. Similarly,
       1/x has no elementary antiderivative – unless you count log(x) as elementary, but ultimately
       log(x) is simply defined to be such an antiderivative. It’s all tautology, but it’s still useful.

Kevin Zhou Physics Olympiad Handouts — educational use, attribute to Kevin Zhou (knzhou.github.io). LICENSE FLAG: must reconfirm permissions before public release.


[User]
\begin{wrapfigure}[4]{r}{3cm}%
\includegraphics[width=\linewidth]{2013-v2g-10-massspektromeeter_ipe_ing}%
\end{wrapfigure}
In a laboratory there was a quantity of a substance to be examined. The molar mass of the substance was measured to be $\mu_{1}$. The substance was ionized once (each atom lost one electron), then accelerated in an electric field with a potential difference $U$ and directed to a magnetic field of induction $B$ (see figure). The magnetic induction was perpendicular to the plane of the figure, the initial velocity of the ions was $y$-directional, the magnetic field was located in the region $y>0$ and the substance entered the magnetic field at the origin, $(0, 0, 0)$.\\
It was observed that a small amount of the substance fell by a distance $d$ further on the detector on the $x$-axis from where the rest of the substance fell on. From this it was assumed that a small amount of the isotope in the substance had a different molar mass. Find the molar mass $\mu_{2}$ of this isotope. The Avogadro number is $N_A$ and the charge of an electron is $-e$.

Solution:
Potentsiaalide vahes $U$ saab kiirendatud laetud osake kineetilise
energia
\[
\frac{mv^{2}}{2}=qU,
\]
siit avaldame osakese kiiruse:
\[
v=\sqrt{\frac{2qU}{m}}.
\]
Kuna magnetväli on kiirusega risti, hakkab laetud osake
magnetvälja jõudes liikuma seal mööda ringjoone kaart, kus ringjoone raadius
$R=\frac{mv}{qB}$. Selleks ajaks, kui osake jõuab detektorini,
on ta läbinud pool ringjoonest. Olgu $m_{2}$ raskema isotoobi
mass ja $m_{1}$ kergema isotoobi mass. Sel juhul
\[
2\left(\frac{m_{1}v_{1}}{qB}-\frac{m_{2}v_{2}}{qB}\right)=d,
\]
ehk
\[
m_{2}v_{2}=\frac{qBd}{2}+m_{1}v_{1 }.
\]
Arvestades, et $v_{1}=\sqrt{\frac{2eU}{m_{1}}}$ ja $v_{2}=\sqrt{\frac{2eU}{m_{2}}}$,
kus $e$ on elementaarlaeng, saame eelmise võrrandi ümber kirjutada kui
\[ \sqrt{m_{2}}=\sqrt{m_{1}}+\frac{Bd}{2}\sqrt{\frac{e}{2U}}. \]
Võttes arvesse, et $m_{2}=\frac{\mu_{2}}{N_{A}}$ ja $m_{1}=\frac{\mu_{1}}{N_{A}}$,
kus $N_{A}$ on Avogadro arv, saame:
\[ \mu_{2}=N_{A}\left(\sqrt{\frac{\mu_{1}}{N_{A}}}+\frac{Bd}{2}\sqrt{\frac{e}{2U}}\right)^{2}.\]

Mass-spektromeeter

\begin{wrapfigure}[4]{r}{3cm}%
\includegraphics[width=\linewidth]{2013-v2g-10-massspektromeeter_ipe}%
\end{wrapfigure}
Laboris oli uurimiseks hulk mingit atomaarset ainet, mille molaarmassiks mõõdeti
$\mu_{1}$. Ühekordselt ioniseeritud ainet (iga aatom oli kaotanud ühe
elektroni) kiirendati elektriväljas potentsiaalide vahega $U$ ja suunati magnetvälja
induktsiooniga $B$ (vaadake joonist). Magnetinduktsioon oli joonise tasandiga
risti,
ioonide algkiirus oli $y$-telje suunaline,
magnetväli asus piirkonnas $y>0$ ning aine sisenes magnetvälja punktis
$(0, 0, 0)$.

Täheldati, et väike kogus ainet langes $x$-teljel asuvale
detektorile kauguse
$d$ võrra kaugemal kohast, kuhu langes põhiosa ainest. Sellest järeldati,
et aine hulgas oli väike osa isotoopi erineva molaarmassiga. Leidke
selle isotoobi molaarmass $\mu_{2}$. Avogadro arv on $N_A$ ja elektroni laeng
on $-e$.

Osakeste kiirus magnetvälja sisenedes on avaldatav energia jäävusest. Magnetväljas algab ringliikumine, kusjuures detektorini jõutakse pärast poole ringjoone läbimist. Teisisõnu on detekteeritud $x$-koordinaat $2R$, kus $R$ on ringjoone raadius.


[User]
10. A binary star has a period (T ) of 2 earth years while distance L between its components having
    masses M1 and M2 is four astronomical units. If M1 = MS where MS is the mass of Sun, the
    mass of other component M2 is

     (a) 3MS
    (b) 7 MS
     (c) 15 MS
    (d) MS

    Note: The earth - sun distance is one astronomical unit.

                                    3

                             INPhO - 2009

                              A                            r

                                                               Figure 4:

Solution:
10.

     CS
    C /R
              5
              4
              3
              2
              1
              0                                      V/V
                  0.2    0.4   0.6    0.8   1              0

       E                HOMI BHABHA CENTRE FOR SCIENCE EDUCATION
                               Tata Institute of Fundamental Research
                           V. N. Purav Marg, Mankhurd, Mumbai, 400 088

INPhO 2009 Problem 10


[User]
3. MISSION TO MARS (7 points)

A crew of
astronauts is going to be sent to explore the
polar region of Mars and search for buried
water ice. Their spaceship will travel from
Earth to Mars along an elliptic transfer orbit
tangential to the orbits of both planets. Des-
pite its shortcomings, this orbit is commonly
used in space travel due to its relatively good
fuel economy. Future manned missions to
Mars are very likely based on this kind of
transfer. In this problem you will examine
some aspects of this orbit.
The mean orbital radius of Mars is Ra =
1.52AU. The mean orbital radius of Earth
is Rg = 1AU = 149600000km. Mars has a
mean radius of ra = 3397km and surface
gravity ga = 3.71m/s2
. Earth has a mean
radius of rg =6371km.
i) (1 point) Find the orbital period Ta of
Mars, i.e. find the length of Martian “year”
in Earth years.
ii) (1.5 points) How long (tt) does a one-way
trip to Mars take?
iii) (1.5 points) The spaceship is put into this
orbit by using a powerful rocket. It is more
efficient to burn fuel as a short burst when
the spaceship is still near Earth. How much
additional speed (∆v1) does the booster have
to be able to give to the spaceship to enter the
transfer orbit, starting from the north pole?
Neglect the air resistance.
iv) (1.5 points) Estimate the ∆v2 needed to
enter a circular orbit close to Mars.
v) (1.5 points) What is the minimal duration
of the trip to Mars and back?

Solution:
3. MISSION TO MARS

i) (1 point) We can find the orbital period
of Mars from Kepler’s third law Ra
3
/Rg
3
=
Ta
2
/Tg
2
, giving tt ≈1.87yr.
ii) (1.5 points) Again, we can use Kepler’s
third law to calculate half of the orbital
period.
tt =
Tg
2
(Ra +Rg)3/2
(2Rg)3/2
≈0.707yr.
iii) (1.5 points) Background. ∆v is import-
ant, because the sum of all ∆v determines
how much fuel is needed for a given mission.
The fuel needed is exponential of total ∆v and
is described by Tsiolkovsky rocket equation.
Kinetic energy per unit mass of such a
transfer orbit where it intersects the Earth’s
orbit is −
GMs
Rg+Ra
+
GMs
Rg
. Using the orbital
angular speed of Earth we can substitute
GMs =
4π2R3
g
Tg
. The speed at the beginning of
the transfer orbit becomes
vt0 =
s
2GMs
μ
1
Rg
−
1
Rg +Ra
¶
≈32.7km/s
. The speed in Earth’s inertial frame is
v0
t0
= vt0 −vg ≈2.94km/h. To achive that, we
first need to escape Earth’s gravity, so
∆v1 =
v
u
u
t2
Ã
v0
t0
2
2
+
GMg
rg
!
.
— 1 —
Using the surface gravity of Earth we can
substitute
GMg
rg
= ggrg so ∆v1 ≈11.2km/s.
iv) (1.5 points) We can calculate the speed
of the transfer orbit where it intersects the
orbit of Mars from Kepler’s second law vt1 =
vt0/1.52≈21.5km/s. The speed of the space-
craft relative to Mars is v0
t1
≈3.25km/s. The
speed of the spacecraft once near Mars sur-
face is
vtm =
v
u
u
t2
Ã
v0
t1
2
2
+raga
!
≈5.98km/s
Since the speed of low Mars orbit is vea =
p
raga ≈ 3.55km/s, we need to brake for
∆v2 ≈2.43km/s.
v) (1.5 points) The Earth–Sun–Mars angle
α at the launch of the mission needs to
be α = π−watt ≈ 0.77 for the spacecraft to
reach Mars. Likewise for the return trip
β = π−wgtt ≈ −1.301. If we go to the coro-
tating frame of referense with earth, we can
see that the minimal time between those two
angles is
2π−α+β
wa−wg
≈1.96yr. The minimal dura-
tion of the trip is therefore longer by tt, giving
2.67yr.

MISSION TO MARS


[User]
Question A2
A thin, uniform rod of length L and mass M = 0.258 kg is suspended from a point a distance R
away from its center of mass. When the end of the rod is displaced slightly and released it executes
simple harmonic oscillation. The period, T , of the oscillation is timed using an electronic timer.
The following data is recorded for the period as a function of R. What is the local value of g? Do
not assume it is the canonical value of 9.8 m/s2 . What is the length, L, of the rod? No estimation
of error in either value is required. The moment of inertia of a rod about its center of mass is
(1/12)M L2 .

                     R       T                        R       T
                    (m)     (s)                      (m)     (s)
                   0.050   3.842                    0.211   2.074
                   0.075   3.164                    0.302   1.905
                   0.102   2.747                    0.387   1.855
                   0.156   2.301                    0.451   1.853
                   0.198   2.115                    0.588   1.900

    You must show your work to obtain full credit. If you use graphical techniques then you must
plot the graph; if you use linear regression techniques then you must show all of the formulae and
associated workings used to obtain your result.
                   Copyright ©2011 American Association of Physics Teachers

2011 Semifinal Exam                             Part A                                                4

Solution:
Question A2
A thin, uniform rod of length L and mass M = 0.258 kg is suspended from a point a distance R
away from its center of mass. When the end of the rod is displaced slightly and released it executes
simple harmonic oscillation. The period, T , of the oscillation is timed using an electronic timer.
The following data is recorded for the period as a function of R. What is the local value of g? Do
not assume it is the canonical value of 9.8 m/s2 . What is the length, L, of the rod? No estimation

                   Copyright ©2011 American Association of Physics Teachers

2011 Semifinal Exam                                Part A                                       5

of error in either value is required. The moment of inertia of a rod about its center of mass is
(1/12)M L2 .

                     R       T                             R       T
                    (m)     (s)                           (m)     (s)
                   0.050   3.842                         0.211   2.074
                   0.075   3.164                         0.302   1.905
                   0.102   2.747                         0.387   1.855
                   0.156   2.301                         0.451   1.853
                   0.198   2.115                         0.588   1.900

    You must show your work to obtain full credit. If you use graphical techniques then you must
plot the graph; if you use linear regression techniques then you must show all of the formulae and
associated workings used to obtain your result.

                                             Solution
Using the parallel axis theorem, the period of such a physical pendulum is
                                       s             s
                                           I           R2 + L2 /12
                               T = 2π           = 2π               .
                                         mgR               gR

We can rearrange this into the linear form

                                                        T 2R                    L2
                     y = mx + b,     y = R2 ,      x=        ,   m = g,   b=−      .
                                                        4π 2                    12
Filling out a data table, we get
                                     R       T      T 2 R/4π 2     R2
                                   0.050   3.842     0.0187      0.0025
                                   0.075   3.164     0.0190      0.0056
                                   0.102   2.747     0.0195      0.0104
                                   0.156   2.301     0.0209      0.0243
                                   0.198   2.115     0.0224      0.0392
                                   0.211   2.074     0.0230      0.0445
                                   0.302   1.905     0.0278      0.0912
                                   0.387   1.855     0.0337      0.1498
                                   0.451   1.853     0.0392      0.2034
                                   0.588   1.900     0.0538      0.3457

The graph of T 2 R/4π 2 versus R2 is a line with slope g and intercept −L2 /12, which should be
plotted on graph paper.

                   Copyright ©2011 American Association of Physics Teachers

2011 Semifinal Exam                                Part A                                        6

Looking at the graph, we read off the results

                                     g = 9.79 m/s2 ,    L = 1.47 m.

Also note that most of the first few data points are not useful. To get full credit, only the five
useful data points had to be used.

USAPhO 2011 Question A2


[User]
A cylindrical space station of radius R can create artificial gravity by rotating with
    angular velocity ω about its axis.

     (a) For an observer rotating along with the spaceship on the rim, what gravitational acceleration
         g do they perceive?

     (b) The observer throws a ball parallel to the floor. For some launch speed v, the observer will
         see the ball perform a circular orbit along the spaceship, always parallel to the floor. Find v.

      (c) What does the motion of part (b) look like, in a frame that isn’t rotating with the ship?

Solution:
(a) The centrifugal acceleration is ω 2 R, and this is perceived as gravity, g = ω 2 R.

      (b) The Coriolis force points upward, so the acceleration upward is

                                                  a = 2ωv − ω 2 R.

          For the ball to perform a circular orbit, it needs to have a centripetal acceleration of a = ω 2 R.
          Combining these and solving gives
                                                       v = ωR.

      (c) This is an example of a case where working in an inertial frame is easiest. In an inertial frame,
          the ball just hovers in place (since there isn’t any gravity), while the space station’s floor
          rotates with speed v right under it.


[User] Inimene massiga $m_{1}=\SI{75}{kg}$ sõidab mootorrattaga, mille mass on $m_{2}=\SI{150}{kg}$. Mootorratturi ja mootorratta ühine masskese asub kõrgusel $h=\SI{0,6}{m}$ maapinnast ja horisontaalsihis kaugusel $l=\SI{0,5}{m}$ tagumise ratta teljest. Millise kiirendusega sõites kerkib esiratas maapinnast lahti? Kui suur peab olema hõõrdetegur tagaratta ja maapinna vahel, et niimoodi saaks sõita?

Solution:
Kuna mootorratas ei pöörle, siis mootorrattale mõjuvate jõumomentide summa mistahes punkti, näiteks masskeskme, suhtes peab olema null ehk $F_{h} h=N l$, kus $N$ on tagarattale mõjuv maapinna normaalreaktsioon ja $F_{h}-$ tagarattale mõjuv hõõrdejõud. Kuna $N$ kompenseerib mootorratta kaalu, $F_{h}$ aga annab kiirenduse $a$, siis $N=m g$ ja $F_{h}=m a$, kus $m=m_{1}+m_{2}$. Seega $a=\lg / h \approx\SI{8,2}{m/s^2}$. Kui
tagumine ratas on libisemise piiril, siis $\mu=F_{h} / N=l / h \approx \num{0,83}$.

Mootorratas

Inimene massiga $m_{1}=\SI{75}{kg}$ sõidab mootorrattaga, mille mass on $m_{2}=\SI{150}{kg}$. Mootorratturi ja mootorratta ühine masskese asub kõrgusel $h=\SI{0,6}{m}$ maapinnast ja horisontaalsihis kaugusel $l=\SI{0,5}{m}$ tagumise ratta teljest. Millise kiirendusega sõites kerkib esiratas maapinnast lahti? Kui suur peab olema hõõrdetegur tagaratta ja maapinna vahel, et niimoodi saaks sõita?


[User] Paigalseisust liikuma hakanud autol kulus teatud vahemaa läbimiseks $t = \SI{15}{s}$. Millise ajaga läbis auto viimase viiendiku sellest vahemaast? Auto liikumine lugeda ühtlaselt kiirenevaks.

Solution:
Olgu $s$ läbitud vahemaa, $a$ auto kiirendus ning $\tau$ aeg, millega auto läbis esimese $4/5$ teest. Kehtivad võrdused
\[
s=\frac{a t^{2}}{2}, \quad \frac{4}{5} s=\frac{a \tau^{2}}{2}.
\]
Siit $\tau = 2t/\sqrt 5\approx \SI{13,4}{s}$. Seega otsitav ajavahemik on $t-\tau\approx \SI{1,6}{s}$.

Auto

Paigalseisust liikuma hakanud autol kulus teatud vahemaa läbimiseks $t = \SI{15}{s}$. Millise ajaga läbis auto viimase viiendiku sellest vahemaast? Auto liikumine lugeda ühtlaselt kiirenevaks.

Lihtsam on leida, kui palju aega kulus autol esimese $4/5$ läbimiseks ning seejärel võtta $t$ ja leitud aja vahe.


[User]
AuPhO 2015, problem 14. Print out the relevant part of the answer booklet as well.
    This problem contains several instructive examples of higher-dimensional diffraction patterns.

    For a much harder multi-dimensional diffraction problem, beyond the scope of the Olympiad, see
    Physics Cup 2019, problem 5.

    5    Technological Applications
        Example 8

        Gratings split light into its components. If a grating can just resolve the two wavelengths λ
        and λ + ∆λ, its resolving power is R = λ/∆λ. Compute the resolving power of a diffraction
        grating with N slits at the nth order maximum.

       Solution
       Conventionally, we say that two diffraction peaks are distinguishable if the maximum of one
       falls outside the first minimum of the other. We know the nth order maximum for wavelength
       λ occurs when d sin θ = nλ, as here the path length difference between adjacent slits is nλ.
       Furthermore, the first minimum around this maximum occurs when there is a extra net path
       length difference of λ across the entire diffraction grating, i.e. when

                                            N d sin θ = N nλ + λ.

       Setting the value of sin θ equal to that for wavelength λ + ∆λ, we see that we can just resolve
       these two wavelengths if

                                n(λ + ∆λ)   (N n + 1)λ               λ
                                          =            ,     R=        = N n.
                                    d          Nd                   ∆λ
       Note that the resolving power is also the number of wavelengths by which the longest and
       shortest possible paths to the diffraction maximum differ (i.e. the paths through the very
       top and very bottom slits). The fact that a larger distance may be used to resolve smaller
       wavelength differences is another manifestation of the uncertainty principle.

Solution:
See the official solutions here.
    For a much harder multi-dimensional diffraction problem, beyond the scope of the Olympiad, see
    Physics Cup 2019, problem 5.

    5    Technological Applications

       Example 8

       Gratings split light into its components. If a grating can just resolve the two wavelengths λ
       and λ + ∆λ, its resolving power is R = λ/∆λ. Compute the resolving power of a diffraction
       grating with N slits at the nth order maximum.

       Solution
       Conventionally, we say that two diffraction peaks are distinguishable if the maximum of one
       falls outside the first minimum of the other. We know the nth order maximum for wavelength
       λ occurs when d sin θ = nλ, as here the path length difference between adjacent slits is nλ.
       Furthermore, the first minimum around this maximum occurs when there is a extra net path
       length difference of λ across the entire diffraction grating, i.e. when

                                           N d sin θ = N nλ + λ.

       Setting the value of sin θ equal to that for wavelength λ + ∆λ, we see that we can just resolve
       these two wavelengths if

                               n(λ + ∆λ)   (N n + 1)λ               λ
                                         =            ,      R=       = N n.
                                   d          Nd                   ∆λ
       Note that the resolving power is also the number of wavelengths by which the longest and
       shortest possible paths to the diffraction maximum differ (i.e. the paths through the very
       top and very bottom slits). The fact that a larger distance may be used to resolve smaller
       wavelength differences is another manifestation of the uncertainty principle.


[User]
IPhO 2012, problem 1A.

    5      Reading Graphs
    In some kinematics problems, you’ll have to infer what’s going on from a diagram. To make progress,
    you’ll have to print out the diagram to make measurements directly on it.

Solution:
01m   IPhO 2012, problem 1A.

   5    Reading Graphs
   In some kinematics problems, you’ll have to infer what’s going on from a diagram. To make progress,
   you’ll have to print out the diagram to make measurements directly on it.


[User]
Two diametrically opposite points on a ring of mass M and radius R are marked
    out. The ring is placed at rest on a frictionless floor. An ant of mass m starts at one point, then
    walks horizontally along the ring’s edge to the other. Through what total angle does the ring turn?

Solution:
Two diametrically opposite points on a ring of mass M and radius R are marked
    out. The ring is placed at rest on a frictionless floor. An ant of mass m starts at one point, then
    walks horizontally along the ring’s edge to the other. Through what total angle does the ring turn?

    Answer. πm/(M + m)


[User]
Question A2
A mixture of 32 P and 35 S (two beta emitters widely used in biochemical research) is placed next to
a detector and allowed to decay, resulting in the data below. The detector has equal sensitivity to
the beta particles emitted by each isotope, and both isotopes decay into stable daughters.
   You should analyze the data graphically. Error estimates are not required.

                        Day    Activity    Day    Activity    Day    Activity
                          0      64557      40      12441     200       1121
                          5      51714      60       6385     250        673
                         10      41444      80       3855     300        467
                         20      27020     100       2734
                         30      18003     150       1626

  a. Determine the half-life of each isotope. 35 S has a significantly longer half-life than 32 P.

  b. Determine the ratio of the number of 32 P atoms to the number of 35 S atoms in the original
     sample.

                   Copyright c 2009 American Association of Physics Teachers

2009 Semifinal Exam                                          Part A                                             4

Solution:
Question A2
A mixture of 32 P and 35 S (two beta emitters widely used in biochemical research) is placed next to
a detector and allowed to decay, resulting in the data below. The detector has equal sensitivity to
the beta particles emitted by each isotope, and both isotopes decay into stable daughters.
   You should analyze the data graphically. Error estimates are not required.

                        Day    Activity    Day     Activity     Day      Activity
                          0      64557      40       12441      200         1121
                          5      51714      60        6385      250          673
                         10      41444      80        3855      300          467
                         20      27020     100        2734
                         30      18003     150        1626

  a. Determine the half-life of each isotope. 35 S has a significantly longer half-life than 32 P.

  b. Determine the ratio of the number of 32 P atoms to the number of 35 S atoms in the original
     sample.

                                            Solution

                   Copyright ©2009 American Association of Physics Teachers

2009 Semifinal Exam                                          Part A                                                                                5

  a. The number of each isotope decays exponentially, N (t) = NS,0 2−t/tS + NP,0 2−t/tP where tS
     and tP are the half-lives, and NS,0 and NP,0 are the initial abundances. Because the detector is
     equally likely to detect any beta particle emitted in a decay, the total activity is proportional
     to the decay rate,
                                          dN     NS,0 −t/tS NP,0 −t/tP
                                 A(t) ∝ −     ∝       2      +       2    .
                                           dt     tS              tP
     We know that tS  tP , which implies that after a long time, decays of 35 S will dominate,
     since by then almost all the 32 P has decayed away. On the other hand, the decay rate of 35 S
     is suppressed by a factor of tS , which means that at early times, decays of 32 P dominate.
     Therefore, we can find the individual half-lives by fitting a single decaying exponential to the
     data, at both early and late times, which is equivalent to fitting a line to a plot of ln A(t)
     versus t. For the early times, we choose the first five data points; for the late times, we choose
     the last five data points. The results are shown below.

                                                                                8.00
                11.0
                                                                                7.75
                10.8                                                            7.50
                10.6                                                            7.25

      ln A(t)   10.4                                                  ln A(t)   7.00

                10.2                                                            6.75

                                                                                6.50
                10.0
                                                                                6.25
                 9.8
                       0   5    10       15        20   25     30                      100   125   150   175      200      225   250   275   300
                                     Time (days)                                                               Time (days)

     The slopes of these lines, in the units used in the plots, are −0.043 and −0.0088, and they
     should be equal to −(ln 2)/tP and −(ln 2)/tS , respectively. This gives half-lives tP = 16 days
     and tS = 78 days. Any answer within 10% of these values is reasonably good.

  b. Because the activity is dominated by 32 P at the beginning and 35 S at the end,

                                              A(300 days)   NS,0 2−(300 days)/tS /tS
                                                          =
                                               A(0 days)           NP,0 /tP

     which means that
                                         NP,0    A(0 days) tP −(300 days)/tS
                                              =                2             = 2.0.
                                         NS,0   A(300 days) tS
     Since this final result depends sensitively on your values of tP and tS , any answer within 25%
     is acceptable.

USAPhO 2009 Question A2