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T1 Solutions/ Marking Scheme Dark Matter A Cluster of Galaxies Question A.1 Answer Marks Potential energy for a system of a spherical object with mass M (r ) r and a test particle with mass dm at a distance r is given by dU G 0.2 pts M (r ) dm r Thus for a sphere of radius R R U G R R M (r ) 4 r 16 dm G 4 r dr G r dr 0 r 3r 0.6 pts 16 G R5 15 Then using the total mass of the system M R3 0.2 pts we have U GM R Total Page 1 of 14 1.0 pts T1 Solutions/ Marking Scheme Question A.2 Answer Marks Using the Doppler Effect, fi f0 f (1 ) , 1 where v c and v c . Thus the i‐th galaxy moving away (radial) speed is Vri fi f0 c f0 0.2 pts Alternative without approximation: fi f0 1 f Vri c 1 fi All the galaxies in the galaxy cluster will be moving away together due to the cosmological expansion. Thus the average moving away speed of the N galaxies in the cluster is Vcr c N fi f0 c Nf i 1 N N fi i 1 f 1 0.3 pts Alternative without approximation: Vcr cf N 1 c N i 1 f i f N f0 1 i 1 f i N Total Page 2 of 14 0.5 pts T1 Solutions/ Marking Scheme Question A.3 Answer Marks The galaxy moving away speed Vi , in part A.2, is only one component of the three component of the galaxy velocity. Thus the average square speed of each galaxy with respect to the center of the cluster is N N (V i V c ) i 1 N N (V i 1 xi Vxc ) (Vyi Vyc ) (Vzi Vzc ) 0.5 pts Due to isotropic assumption N N (V i V c )2 i 1 N N (V i 1 ri Vcr ) And thus the root mean square of the galaxy speed with respect to the cluster center is vrms vrms N (Vri Vrc ) N i 1 N 2 (Vri 2VcrVri Vcr ) N i 1 N 2 Vri 3Vcr N i 1 2 1 N f 1 N f i i c 1 1 N i 1 f N f i 1 N 2 c 1 N f N fi fi f0 f0 fi fi f0 f N i 1 N i 1 N i 1 c N 2 N N fi fi f N i 1 i 1 Alternative without approximation: Page 3 of 14 0.7 pts T1 Solutions/ Marking Scheme vrms 2 1 N f N f 1 1 c N i 1 f i N i 1 f i 1 N 1 N 1 c N 2 f N i 1 f i f i f f N i 1 f i N f i 1 f i f 2 cf N N N N f f i 1 i i 1 i The mean kinetic energy of the galaxies with respect to the center of the cluster is K ave m N N (V i 1 i V c )2 m vrms 0.3 pts Total 1.5 pts Page 4 of 14 T1 Solutions/ Marking Scheme Question A.4 Answer Marks The time average of d / dt vanishes d dt 0 t Now 0.6 pts dpi d d dri pi ri ri pi dt dt i dt dt i i Fi ri mi vi vi Fi ri K i i i Where K is the total kinetic energy of the system. Since the gravitational force on i‐th particle comes from its interaction with other particles then Fi ri Fji ri Fji ri Fij ri Fji ri Fji rj i i , j i i j i j i j i j mi m j (ri rj ) mi m j Fji (ri rj ) G (ri rj ) G U tot | ri rj | | ri rj | | ri rj | i j i j i j Alternative proof: , ⋯ 0.9 pts Collecting terms and noting that ⋯ we have Page 5 of 14 ⋯ ⋯ … T1 Solutions/ Marking Scheme ⋯ ⋯ ⋯ Thus we have d U 2K dt d And by taking its time average we obtain U 2K dt K t and thus 0.2 pts t 1 U t Therefore 2 Total 1.7 pts Page 6 of 14 T1 Solutions/ Marking Scheme Question A.5 Answer Marks Using Virial theorem, and since the dark matter has the same root mean square speed as the galaxy, then we have K t U t 0.3 pts M GM vrms 25 R From which we have M 0.1 pts Rv rms 3G And the dark matter mass is then M dm 0.1 pts Rv rms Nm g 3G Total 0.5 pts Page 7 of 14 T1 Solutions/ Marking Scheme B Dark Matter in a Galaxy Question B.1 Answer Marks Answer B.1: The gravitational attraction for a particle at a distance r from the center of the sphere comes only from particles inside a spherical volume of radius r. For particle inside the sphere with mass ms , assuming the particle is orbiting the center of mass in a circular orbit, we have G 0.3 pts m' (r )m s m s v02 r r2 with m' ( r ) is the total mass inside a sphere of radius r m' ( r ) r m s n 0.2 pts Thus we have 1/ 4Gnms v (r ) r While for particle outside the sphere, we have 0.2 pts 1/ 4Gnms R v (r ) 3r Page 8 of 14 T1 Solutions/ Marking Scheme The sketch is given below 0.1 pts Sketch of the rotation velocity vs distance from the center of galaxy Total 0.8 pts Question B.2 Answer Marks The total mass can be inferred from G m' ( R g ) m s Rg ms v02 Rg 0.5 pts Thus m R m' ( R g ) v02 R g G Total 0.5 pts Page 9 of 14 T1 Solutions/ Marking Scheme Question B.3 Answer Marks Base on the previous answer in B.1, if the mass of the galaxy comes only from the visible stars, then the galaxy rotation curve should fall proportional to r on the outside at a distance r > Rg But in the figure of problem b) the curve remain constant after r > Rg , we can infer from G m' (r )m s m s v02 r r2 0.3 pts to make v (r ) constant, then m'(r) should be proportional to r for r R g , i.e. for r > Rg , m' (r ) Ar with A is a constant. While for r R g , to obtain a linear plot proportional to r , then m ' ( r ) should be proportional to r , i.e. m' (r ) Br 0.3 pts Thus for r R g we have r m' (r ) t (r )4r ' dr ' Br 0.2 pts dm' ( r ) t ( r ) 4r dr 3Br dr Thus total mass density t (r ) 3B 4 v02 mR 3B B r ' dr ' BR or g 0 4 Rg GRg Rg mR Thus the dark matter mass density (r ) Page 10 of 14 3v 02 4GR g nm s 0.2 pts T1 Solutions/ Marking Scheme While for r R g we have Rg r Rg m' (r ) (r ' )4r ' dr ' (r ' )4r ' dr ' Ar r m' (r ) m R (r ' )4r ' dr ' Ar Rg 0.2 pts r (r ' )4r ' R dr ' Ar M (r )4r A , or (r ) A . 4r Now to find the constant A. r R A 4r ' dr ' A(r R g ) Ar m R 4r ' Thus AR g m R and A v02 G We can also find A from the following G 0.3 pts m' ( r ) m s Arm s m s v02 v02 A , thus G r G r2 r2 Thus the dark matter mass density (which is also the total mass density since n for r R g (r ) v02 for r R g 4Gr Total 1.5 pts Page 11 of 14 T1 Solutions/ Marking Scheme C Interstellar Gas and Dark Matter Question C.1 Answer Marks Consider a very small volume of a disk with area A and thickness r, see Fig.1 0.3 pts Figure 1. Hydrostatic equilibrium In hydrostatic equilibrium we have ( P(r ) P(r r )) A g (r ) Ar P Gm' (r ) r r2 dP Gm' ( r ) Gm' ( r ) n( r ) m p dr r r2 0.2 pts Total 0.5 pts Page 12 of 14 T1 Solutions/ Marking Scheme Question C.2 Answer Marks Using the ideal gas law P = n kT where n = N/V where n is the number density, we have dP dn( r ) dT GM ' (r ) Gm’(r) kT kn( r ) n(r )m p dr dr dr r 0.5 pts Thus we have m' ( r ) kT Gm p r dn(r ) r dT (r ) T (r ) dr n(r ) dr Total 0.5 pts Question C.3 Answer Marks If we have isothermal distribution, we have dT/dr = 0 and m' ( r ) kT0 Gm p r dn(r ) n(r ) dr 0.2 pts From information about interstellar gas number density, we have dn(r ) 3r n(r ) dr r (r ) Thus we have m' (r ) 0.2 pts kT0 r 3r Gm p ( r ) Page 13 of 14 T1 Solutions/ Marking Scheme Mass density of the interstellar gas is g (r ) m p r ( r )2 Thus m' (r ) g (r ' ) dm (r ' ) 4r ' dr ' r kT0 r 3r Gm p (r ) 0.3 pts m p kT r 3r dm (r ' ) 4r ' dr ' m' (r ) Gm p (r ) r ' ( r ' ) r m p kT0 3r 6r 4r ( ) r dm Gm p (r ) r ( r ) dm (r ) m p kT0 3r 6r 2 4Gm p (r ) r r ( r )2 0.3 pts Total 1.0 pts Page 14 of 14 ... Page 3 of? ?14 0.7 pts T1 Solutions/ Marking Scheme vrms 2 ? ?1 N f N f 1? ?? 1? ?? c N i ? ?1 f i N i ? ?1 f i 1 N 1 N 1 c ... N i ? ?1 N 2 (Vri 2VcrVri Vcr ) N i ? ?1 N 2 Vri 3Vcr N i ? ?1 2 ? ?1 N f ? ?1 N f i i c 1? ?? 1? ?? N i ? ?1 f N f i ? ?1 ... i ? ?1 N N fi i ? ?1 f 1? ?? 0.3 pts Alternative without approximation: Vcr cf N 1 c N i ? ?1 f i f N f0 1? ?? i ? ?1 f i N Total Page 2 of? ?14