341 Phys Final Exam: Solutions 1. Consider a galaxy with a flat rotation curve with rotation speed vg . There is a dwarf galaxy orbiting the big galaxy in a circular orbit with radius Rg . The dwarf galaxy has a flat rotation curve with rotation velocity vd . (a) The tidal force from the big galaxy strips stars out of the "front" and "back" of the dwarf galaxy. We can estimate the tidal radius of the dwarf galaxy to be the distance from the center of the dwarf galaxy where the tidal force pulling "out" on a star equals the gravity from the dwarf galaxy pulling back "in." Derive an expression for the tidal radius in terms of Rg , vg , and vd . We say that the tidal radius is the radius rT at which the tidal force from the big galaxy equals the gravity from the dwarf galaxy: 2GMg (Rg )rT GMd (rT ) = 2 3 Rg rT If the dwarf galaxy has a flat rotation curve, the mass inside rT is Md (rT ) = 2 rT vd /G. As for Mg (Rg ), we know from Newton that for a spherical mass distribution the only mass that matters is the mass inside the orbit. Assuming the dark matter halo of the big galaxy extends at least as far as the dwarf galaxy's orbit, 2 we have Mg (Rg ) = Rg vg /G. (You may wonder whether we should use the mass within Rg - rT rather than Rg itself. That leads to a relatively small correction, which we will ignore.) Putting these masses in the equation above, we get
2 2vg rT v2 = d 2 Rg rT
Solving for rT gives
RG vd rT = 2 vg
(b) The Large Magellanic Cloud is a dwarf galaxy orbiting about 50 kpc. from the center of the Milky Way. It rotates at about 50 km s-1 . Recall that the Milky Way's rotation speed is about 220 km s-1 . Compute the tidal radius rT of the LMC (in kpc). Plugging in numbers, Rg = 50 kpc, vg = 220 km s-1 , and vd = 50 km s-1 , we have 50 kpc 50 km s-1 rT = = 8.0 kpc 2 220 km s-1 1
(c) Roughly speaking, only matter within the tidal radius is gravitationally bound to the dwarf galaxy. What is the total mass (in M ) bound to the LMC? Treating the dwarf galaxy as an isothermal sphere with rotation speed vd = 50 km s-1 , the mass within rT is
2 rT v d Md = = 9.3 1042 g = 4.7 109 M G
(d) Suppose the LMC moves closer to the Milky Way, into an orbit 40 kpc in radius. (Assume the LMC still rotates at the same speed.) What is the total mass (in M ) bound to the LMC in its new orbit? In its new orbit Rg = 40 kpc, the new tidal radius is rT = and the mass is 40 kpc 50 km s-1 = 6.4 kpc 2 220 km s-1
2 rT vd = 3.8 109 M G (e) The mass in part (d) is smaller than that in part (c). Where did the other mass go?
Md =
The mass that was tidally stripped from the dwarf galaxy went into the Milky Way's halo. The stars and gas stripped from the Large Magellanic Cloud actually form the "Magellanic stream" that is observed around the Milky Way. This process of small galaxies falling into big galaxies and being tidally disrupted is thought to play an important role in how galaxies form and grow. 2. According to general relativity, moving masses can create ripples in spacetime known as gravitational waves. In a binary star system, the gravitational radiation carries away energy; specifically, if the stars each have mass M and their orbit has semimajor axis a, the power emitted in gravitational radiation is 2c5 Pgr = 5G 2GM c2 a
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(a) Explain conceptually what happens to the stars' orbits. When the system loses energy, the two stars move closer together. As they lose more energy, they spiral together and eventually merge. (b) Using conservation of energy, derive a differential equation for the semimajor axis as a function of time. Solve the equation to find how long it takes to go from some initial semimajor axis a = a0 to a = 0. (Work with symbols, not numbers.) 2
From our work on binary star systems (e.g., HW#2), we know the total energy of the orbit is GM 2 Eorb = - 2a Since the gravitational radiation carries away energy, the orbital energy must decrease. Specifically, we must have dEorb = -Pgr dt GM 2 da 2c5 2GM = - 2a2 dt 5G c2 a da 128G3 M 3 = - dt 5c5 a3
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This is our differential equation for the semimajor axis. We can solve by it moving the a3 to the left-hand side and the dt to the right-hand side, and integrating from some initial time t = 0 and a = a0 to the final time t = tf and a = 0:
0
-
a0
a3 da = a4 0 4 tf
128G3 M 3 dt 5c5 0 128G3 M 3 = tf 5c5 5c5 a4 0 = 512G3 M 3
tf
(1)
(c) Recently a binary pulsar system called J0737-3039 was discovered. The system is very close to edge-on. Each pulsar moves in a nearly circular orbit with a period of P = 0.102 day, at a speed of about 310 km s-1 . What are the masses of the pulsars? What is the semimajor axis of the orbit? When will the two stars merge? The period is P = 0.102 day = 8813 s. Since the stars have equal velocities they must have equal masses. From Kepler III we found an expression for the total mass, 3 P v1 + v2 m1 + m2 = 2G sin i We know the orbit is close to edge-on so we can take sin i 1. Then filling in numbers: m1 + m2 = 8813 s 2 6.67 10-8 cm3 g-1 s-2 = 5.01 1033 g = 2.52 M 3 (310 + 310) 105 cm s-1
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Therefore each star must have M = 1.26 M . The current semimajor axis is a0 = P 8813 s (v1 + v2 ) = (620 105 cm s-1 ) = 8.7 1010 cm = 0.0058 AU 2 2
From eq. (1) the time it will take for gravitational radiation to make the orbit decay so the stars merge is tf = 5 (3.0 1010 cm s-1 )5 (8.7 1010 cm)4 512 (6.67 10-8 cm3 g-1 s-2 )3 (1.26 1.99 1033 g)3 = 2.9 1015 s
= 9.2 107 yr In other words, this binary pulsar system will only be around for another 92 million years or so! 3. Because the expansion of the universe is accelerating, the universe in the future will be a lonely place. (a) What is the scale factor as a function of time for a flat universe with M = 0 and = 1? Express your answer in terms of the time today (t0 ), the scale factor today (R0 ), and the Hubble constant (H0 ). In class, we wrote the Friedmann equation as dR = H0 (1 - M - ) + M R-1 + R2 dt
1/2
where H0 is the Hubble constant, while M and are the density parameters associated with matter and the cosmological constant. For M = 0 and = 1 the equation becomes dR = H0 R dt We can solve this by rearranging and integrating: dR = H0 dt R ln R = H0 t + constant R(t) = R0 eH0 (t-t0 ) (2)
I have chosen the constant of integration such that the scale factor today is R0 .
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(b) Since each galaxy is accelerating away from us but the speed of light is finite, there is some time in the future when light emitted by a galaxy will not be able to reach us. (It would take an infinite amount of time for the light to travel to us.) We will lose contact with that galaxy and never know what happens to it after that time. How long in the future will we lose contact with a galaxy that is presently 10 Mpc away? Hint. Recall that the spacetime interval of a light ray is always zero. Use that to find out how long it would take light emitted from a given galaxy at a given time to reach us. You do not need to know the present age of the universe to answer the question. You may take the Hubble constant to be H0 = 70 km s-1 Mpc-1 . From the Robertson-Walker metric, the spacetime interval for light moving on a radial line in a flat universe is 0 = ds2 = c2 dt2 - R(t)2 dr2 Suppose light is emitted from radial coordinate rem at time tem and is observed by us at r = 0 at time tobs . From the spacetime interval we can deduce the relation between the quantites to be
rem tobs
dr =
0 tem
c dt R(t)
rem
c H0 t0 tobs -H0 t = e e dt R0 tem c = eH0 t0 e-H0 tem - e-H0 tobs H0 R0
I used R(t) from eq. (2). We are interested in situations such that the light never reaches us, meaning tobs . Then the equation becomes rem = We can solve for the time:
-1 tem - t0 = H0 ln
c e-H0 (tem -t0 ) H0 R0
c H0 R0 rem
-1 = H0 ln
c H0 Dem
where Dem = R0 rem is the present distance to the galaxy. Consider a galaxy that is now Dem = 10 Mpc away. We use for the Hubble constant H0 = 70 km s-1 Mpc-1 = 2.27 10-18 s-1 = 7.15 10-11 yr-1
-1 H0 = 1.40 1010 yr
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Then we have: tem - t0 = 1.40 1010 yr ln 3 105 km s-1 = 8.5 1010 yr 70 km s-1 Mpc-1 10 Mpc
In other words, after another 85 billion years the light emitted by a galaxy that is presently 10 Mpc away will no longer be able to reach us.
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