SPA7010U/SPA7010P: THE GALAXY. Solutions for Coursework 1. Questions distributed on: 25 January 2018.
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1 SPA7U/SPA7P: THE GALAXY Solutions fo Cousewok Questions distibuted on: 25 Jnuy 28. Solution. Assessed question] We e told tht this is fint glxy, so essentilly we hve to ty to clssify it bsed on its spectl chcteistics lone. The glxy hs stong continuum with bsoption lines nd some emission lines supeimposed. The emission lines e chcteistic of es of st fomtion. This elly ules out n ellipticl glxy, since ellipticls typiclly hve no ctive st fomtion. The combintion of stong continuum with both bsoption nd emission lines suggests tht this is most likely to be spil glxy. b Vey stong emission lines on stell continuum is typicl chcteistic of n iegul glxy. So bsed on the spectl chcteistics lone, this is pobbly n iegul glxy. It could be nothe type of glxy, such s spil, expeiencing stong bust of st fomtion becuse of n intection with nothe glxy, but we e not consideing mege pocesses in this couse, nd in the bsence of othe infomtion, the simplest exption is tht is n iegul glxy. The vey stong emission lines indicte stong st fomtion. The st fomtion te is stonge in this glxy comped to the glxy in pt. c When two bodies intect, the intection is collisionl if the intections between the individul pticles in the bodies ffect the motions substntilly. The intections e collisionless if the intections between the individul pticles do not ffect the motions. The intections between two gs clouds e collisionl. d The meging of the two smll systems of sts is collisionl becuse the intections between individul sts e impotnt in this cse. This is diffeent fom the cse of two meging glxies: the st-st encountes e not impotnt in the mege of two glxies, so this would be collisionless pocess. e Some of the totl mechnicl enegy potentil + kinetic is conveted into het, so the mechnicl enegy is not conseved nd the collpse is dissiptive. continued...]
2 f Integting ove the sufce bightness gives the luminosity of glxy to be L I R 2 Becuse I is constnt fo ll glxies of this type, L R 2 fo ll The viil theoem implies M/R v 2, whee M is the mss of glxy nd v is typicl velocity of sts in glxy Eliminting R gives L M 2 v 4 Since the mss-to-light tio M/L = constnt, M L Substituting fo M in L M 2 v 4 gives L L 2 v 4 so finlly: L v 4 This is the sme s the Tully-Fishe eltion fo spil glxies, o the Fbe-Jckson eltion fo ellipticl glxies. End of ssessed question
3 Solution.2 To find M, conside thin spheicl shell of dius nd thickness d concentic with the glxy. The mss in the shell will be dm = 4π 2 d ρ this is the eqution of continuity of mss. Integting fom the cente of the glxy dius = to dil distnce, M dm = 4π 2 d ρ = 4π 2 d q 4π q 3 + q+ M tot. M = q M tot nd using the stndd integl povided. q + q+ d = M tot q + q ] = This gives, M = M tot q + q q + q = M tot q + q, the equied esult fo ll q. b We need to clculte the potentil Φ fo q = the Jffe model. The simplest wy to do this is to use dφ = GM. d 2 Fom the nswe to the fist pt of the question, putting q =, the mss M inteio to dius is M = M tot fo q =. + Theefoe, dφ d = G 2 M tot + = GM tot +. Integting fom dius to infinity emembeing tht Φ = fom the definition of gvittionl potentil, Φ dφ = GM tot + d.
4 This cn be solved using ptil fctions: Φ = G M tot d + Φ = G M ] tot + = ] = G M tot = G M tot + + /, = G M tot + which gives Φ = G M tot + fo the potentil t dius in the Jffe q = model. c If q, the density pofile gives ρ = fo >. Howeve, ρ = M tot 4π lim q, q q 3 + q+. So q implies tht ll the mss M tot is concentted t the cente: it coesponds to point mss. Solution.3 The density depends on dius nd does not depend on ngle, so the pofile is spheiclly symmetic. Theefoe the eqution of continuity of mss dm/d = 4π 2 ρ pplies. Substituting fo the pofile ρ, dm d = 4π 2 k + 2. Integting fom the cente to dius, M dm = 4π 2 k + 2 d M = 4πk + 2 d.
5 This integl cn be solved using + + d = d = = + d d 2 = d + c. M = 4πk + + = 4πk + + So + M = 4πk + ] + + which is the mss inside dius tht the question sks fo., b c As, M. So the model is not physiclly elistic t lge dii. To clculte the potentil, we mke use of the spheicl symmety once moe. Fo spheicl symmety, GM 2 = dφ d. Substituting fo M nd integting fom infinity to dius, Φ dφ = 4πGk 2 becuse the potentil is t. Theefoe, + Φ = 4πGk The integls cn be solved using ptil fctions: + d = d + = + + c = + d + c d
6 Then, using integtion by pts, + d = d = c using the integl bove. + Using these integls, we get fo the potentil Φ = 4πGk + + ] + + = 4πGk ] + = 4πGk + So the potentil t distnce fom the cente is Φ = 4πGk +. d The centl density is infinite. This ppes not to be physiclly elistic. Howeve, one of the people who fist used this pofile to descibe glxies Clos Fenk hs gued tht the pofile might be elistic in the coes of glxies fte ll. He ws thinking bout the possibility tht most mssive glxies hve blck holes nd theefoe singulity t thei coes.
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