Astro 7B Midter 1 Practice Worksheet For all the questions below, ake sure you can derive all the relevant questions that s not on the forula sheet by heart (i.e. without referring to your lecture notes). 1 Tides 1. The oon pulls a tide on the Earth, and the axiu height of the tidal bulge on Earth is h. The oon is distance a away fro the Earth, and the Earth radius is R. The oon has ass and the Earth has ass M. Express h in ters of given variables. The axiu h is raised when the tidal acceleration is equal to the gravitational acceleration of the bulge. The tidal acceleration is the differential gravitational acceleration due to at the centre of M and at the tip of the bulge: a tide = (a (R + h)) 2 ( a 2 ) (a R) 2 + a 2 = a 2 (1 (R/a)) 2 + a 2 ( a 2 1 + 2R ) + a a 2 a 2 R a where the second equality is taking h R approxiation, fourth equality is taking R a approxiation and binoial expansion (recall: this is taking up to 1st order of Taylor expansion), and we ignore factor 2 in the last equality. The gravitational acceleration due to the bulge is a grav = bulge R 2 GρR2 h R 2 = Gρh (1) h R 3 (2) where the second equality is assuing M to be of unifor density ρ and taking the bulge to be approxiatly a cylinder of radius R and height h and ρ M/R 3 in the last equality. Equate a tide with a grav to get h = M ( ) R 3 R (3) a 1
2. Which is faster? Earth s spin or oon s orbital frequency? In this set-up, draw the tidal bulge relative to oon s orbit, and discuss in which direction angular oentu transfer occurs. Coent on the behaviour of the following properties: Earth s spin, oon s orbital distance, oon s orbital frequency. Earth s spin period is about a day while the oon s orbital period is about a onth so Earth s spin is faster. Therefore, you should draw the tidal bulge to be leading with respect to the line connecting the centres of the Earth and the oon. In this scenario, the oon is pulling the tidal bulge back so the angular oentu is transferred fro the spin to the orbit. Therefore, Earth s spin slows down while the oon s orbital distance increases. By Kepler s third law, P a 3/2 and P ω 1 so ω a 3/2. Consequently, the oon s orbital frequency decreases. 2 Death or Spaghettification? 1. A star of ass and radius R floats toward a black hole of ass M. The current orbital distance is a. What is the tidal force on this star? Don t just look it up but derive it. Recall that tidal is just a another word for differential. The derivation is exactly the sae as the derivation of a tide in tidal bulge calculation except you re taking the differential force at different positions. Since you re interested in the force that will rip a star of ass apart, consider splitting it in half. Take the differential gravitation force at the tip of these halved pieces of the star: F tide = (/2) (a R) 2 ( (/2) (a + R) 2 ) (4) I ll leave the ath to you but in the end, you should get (within a few factors) F tide = R a 2 a (5) 2. What kind of force binds the star together? Write down the binding energy of the star. You can derive it by integration but an order of agnitude estiate is fine. Gravity is the ain binding force of stars, and it is F grav = 3G(/2)2 5R 2 (6) I won t show it here but if you take the integral to calculate the work it takes to build up the sphere in shells, you ll see where the factor 3/5 coes fro. In this order of agnitude estiate, however, it really doesn t atter. If you re curious, I suggest you to look at Caroll & Ostlie or ask GSIs. 2
3. At what radius does the star spaghettify? In other words, at what radius does the star tidally disrupt? Equation the two forces. You ll find that it is exactly the sae thing as the tidal bulge claculation except h is R. So you could ve just started fro the tidal bulge calculation and with a little thinking, conclude that the star will be ripped apart when the tidal bulge is on the order the star s radius because that really is the axiu possible bulge height for the star to be reained as one single object. In any case, you should find a = ( ) M 1/3 R (7) Rearrange this to find R 3 = M a 3 (8) so the tidal disruption is the radius up to which if you sear out the ass of the disruptor unifortly, the resultant density is equal to the density of the star that is being disrupted. 4. Given a star of ass, what is the critical ass of the black hole that causes the star to plunge towards the black hole and die instead of being spaghettified? Assue the black hole to be non-spinning. If you need a forula that is not given on the forula sheet, you should derive it. Once your e inside R ISCO, you ll be spiralling towards the black hole. So equate tidal disruption radius to R ISCO, and recall that for non-spinning black holes, R ISCO = 6R grav and R grav = /c 2 : ( ) M 1/3 R = 6 c 2 ( ) M 1/3 = 6 c 2 R M 2/3 = 6G1/3 c 2 R ( c 2 ) 3/2 R M = 6G 1/3 (9) 3 Lagrange Points 1. Consider a three-body syste: 2 stars of ass M and radius R and with M and radius r and a test particle. The separation between the two stars is a. How any Lagrange points are there? In other words, how any equilibriu points are there (positions at which the particle will stay stationary unless perturbed)? Are any of the points stable equilibriu? 3
Figure 1: Configuration of Lagrange points and the force balance. Draw out the configuration and note where these points are as well as the direction and the kind of forces that balance there. There are 5 Lagrange points and only the fourth and the fifth points are dynaically stable. 2. One of these points is called Hill radius or Roche lobe radius. Point out where it is and derive the expression for it. Note that M is critical in the derivation. It s the first Lagrange point. Understand that conceptually, Roche lobe radius is where the otion of the test particle start to be doinated by the gravitational pull of M rather than. The force balance, then, is between the tidal force fro M trying to pull atter away fro and the force that gravitationally binds the atter with. I going to call the Hill radius rh easured fro the position of. The tidal acceleration is then ( ) a2 (a rh )2 rh a2 a atide = ' (10) where I skipped soe algebra. Note that in your derivation, M is critical as this allows you to ake the approxiation rh a. The binding acceleration (really, the binding force per unit ass) is abind = 4 G 2. rh (11)
Equate the two to get r H = ( ) 1/3 a (12) M 3. Discuss what happens to the test particle as you perturb it fro the equilibriu points ever so slightly. For instance, if you deflect the particle fro the Hill radius upwards or downwards, what happens to the otion of the particle? L1,2,3 are saddle nodes so any perturbation will ake the particle run away fro the equilibriu point. You d think that if you deflect the directly vertically, they should oscillate about the equilibriu points but the coriolis force will deflect the away fro this directly vertical line aking the test particle run away. On the other hand, at L4,5, the Coriolis force will help the draw out soe sort of closed orbit around the equilibriu points as the perturbed test particle oves outwards. 4 Accretion Disk 1. In the evolution of binary star syste (be it red giant & ain sequence star, copact object & ain sequence star, etc.) an accretion disk can for. When does this happen? This happens when an object fills its Roche lobe. occurs through the Roche lobe radius. In other words, ass transfer 2. Do you understand why it s a disk and not just a straight free-fall? Explain. This is due to Coriolis force where the initial rotation is fro the spin of the stars. 3. Initially, disk aterial starts out on an eccentric orbit but in the end it circularizes. How? As the accreted aterial for a disk, to close the orbit, it ust hit itself to the strea of new accreted aterial. This causes energy dissipation via heat, etc. Recall that E = /2a. Loss of energy eans energy is ore negative; this eans the orbital sei-ajor axis a is decreasing. On the other hand, the angular oentu L = µ G(M + )a(1 e) is conserved. Since a is dropping, (1 e) ust be increasing to copensate. This eans e has to approach zero; in other words, the orbit becoes circular. 4. There s accretion throughout the accretion disk but there s also accretion onto the object. What s their relative budget to the total luinosity? Can you explain or derive how this is so? Equal budget so half and half. Since luinosity is E/ t, we only need to copare the change in energy or the work done to accrete. Consider the disk first. Let s say the radius of the object is R and the outer edge of the disk is a; there is no gap between the object s surface and the inner edge of the disk. The orbital 5
energy before accretion is /2a whereas the orbital energy after accretion is /2R. Consider R a, then E disk = 2R ( ) 2a (13) 2R Now consider the accretion onto the object. Once the object falls to the object, there should be zero kinetic energy. This is when the total energy is siply equal to the gravitational potential /R. The initial energy is the total orbital energy at R. The change in energy is then E obj = R = 2R You can see that E disk = E obj. + 2R 5. For an accretion disk around black hole, what is the relative budget between disk luinosity and object luinosity? You can t really define the surface of the black hole so there s only disk luinosity. 6. Take an annulus of the disk at soe arbitrary radius r and assue this annulus to be a blackbody. What is the teperature of the annulus? Which part of the disk is hotter? Inner part or the outer part? Sketch what the accretion disk spectra would look like. How would the flux scale with frequency? Again, don t just write it down. Derive it. Recall that accretion luinosity is Ṁ/2r but luinosity is also F Area where F = σt 4, the blackbody flux at the surface and area is π(2r) 2 πr 2. To account for total luinosity, you ll need to ultiply by 2 since the light is eitted fro both sides of the disk. Equate this two expressions: (14) Ṁ = σt 4 [π(2r) 2 πr 2 ] 2 2r = 6πσT 4 r 2 T 4 1 Ṁ = 12πσ r 3 T r 3/4 (15) Therefore, inner disk is hotter than the outer disk. Your disk spectra should siply be superposition of blackbody spectra of different teperature. Recall that fro Wien s peak law, T ν. Also, what you observe is specific flux F ν F/ν. Luinosity is 6πσT 4 r 2 but it is also F (4πd 2 ) where d is the distance between the disk and the observer. This eans F T 4 r 2 but T r 3/4 so F r 1. You know that r T 4/3 and T ν; this eans r ν 4/3 so F ν 4/3. Finally, F ν ν 1/3 which should be the result of the superposition of blackbody spectra you sketched earlier. Good luck on the idter! 6