Astro 7B – Midterm 1 Practice Worksheet

For all the questions below, make sure you can derive all the relevant questions that’s not onthe formula sheet by heart (i.e. without referring to your lecture notes).

1 Tides

1. The moon pulls a tide on the Earth, and the maximum height of the tidal bulge on Earth ish. The moon is distance a away from the Earth, and the Earth radius is R. The moon hasmass m and the Earth has mass M. Express h in terms of given variables.

The maximum 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 m at the centre of M and at the tip of the bulge:

atide = − GM

(a− (R+ h))2− (−GM

a2)

' − GM

(a−R)2+GM

a2

= − GM

a2(1− (R/a))2+GM

a2

' −GMa2

(1 +

2R

a

)+GM

a2

' GM

a2R

a(1)

where the second equality is taking h� R approximation, fourth equality is taking

R� a approximation and binomial 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

agrav =GMbulge

R2

' GρR2h

R2

= Gρh

' GMh

R3(2)

where the second equality is assuming M to be of uniform density ρ and taking

the bulge to be approximatly a cylinder of radius R and height h and ρ 'M/R3

in the last equality.

Equate atide with agrav to get

h =m

M

(R

a

)3

R (3)

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2. Which is faster? Earth’s spin or moon’s orbital frequency? In this set-up, draw

the tidal bulge relative to moon’s orbit, and discuss in which direction angular

momentum transfer occurs. Comment on the behaviour of the following properties:

Earth’s spin, moon’s orbital distance, moon’s orbital frequency.

Earth’s spin period is about a day while the moon’s orbital period is about a

month 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

moon. In this scenario, the moon is pulling the tidal bulge back so the angular

momentum is transferred from the spin to the orbit. Therefore, Earth’s spin slows

down while the moon’s orbital distance increases. By Kepler’s third law, P ∝a3/2 and P ∝ ω−1 so ω ∝ a−3/2. Consequently, the moon’s orbital frequency decreases.

2 Death or Spaghettification?

1. A star of mass m and radius R floats toward a black hole of mass M. The current orbitaldistance is a. What is the tidal force on this star? Don’t just look it up but derive it. Recallthat “tidal” is just a another word for “differential”.

The derivation is exactly the same as the derivation of atide 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 mass m apart, consider splitting

it in half. Take the differential gravitation force at the tip of these halved

pieces of the star:

Ftide = −GM(m/2)

(a−R)2− (−GM(m/2)

(a+R)2) (4)

I’ll leave the math to you but in the end, you should get (within a few factors)

Ftide = −GMm

a2R

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 magnitude estimate

is fine.

Gravity is the main binding force of stars, and it is

Fgrav = −3G(m/2)2

5R2(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 comes from.

In this order of magnitude estimate, however, it really doesn’t matter. If you’re

curious, I suggest you to look at Caroll & Ostlie or ask GSIs.

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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 same thing as the

tidal bulge claculation except h is R. So you could’ve just started from 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 maximum possible bulge height for the star to be remained as one single

object. In any case, you should find

a =

(M

m

)1/3

R (7)

Rearrange this to findm

R3=M

a3(8)

so the tidal disruption is the radius up to which if you smear out the mass of

the ‘‘disruptor’’ uniformtly, the resultant density is equal to the density of

the star that is being disrupted.

4. Given a star of mass m, what is the critical mass of the black hole that causes

the star to plunge towards the black hole and die instead of being spaghettified?

Assume the black hole to be non-spinning. If you need a formula that is not given

on the formula sheet, you should derive it.

Once your’e inside RISCO, you’ll be spiralling towards the black hole. So equate

tidal disruption radius to RISCO, and recall that for non-spinning black holes,

RISCO = 6Rgrav and Rgrav = GM/c2:

(M

m

)1/3

R =6GM

c2(M

m

)1/3

=6GM

c2R

M−2/3 =6Gm1/3

c2R

M =

(c2R

6Gm1/3

)3/2

(9)

3 Lagrange Points

1. Consider a three-body system: 2 stars of mass M and radius R and m∗ with m∗ � M andradius r∗ and a test particle. The separation between the two stars is a. How many Lagrangepoints are there? In other words, how many equilibrium points are there (positions at whichthe particle will stay stationary unless perturbed)? Are any of the points stable equilibrium?

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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 thekind of forces that balance there.

There are 5 Lagrange points and only the fourth and the fifth points are dynamically

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∗ �M is critical in the

derivation.

It’s the first Lagrange point. Understand that conceptually, Roche lobe radius

is where the motion of the test particle start to be dominated by the gravitational

pull of M rather than m. The force balance, then, is between the tidal force

from M trying to pull matter away from m and the force that gravitationally binds

the matter with m. I’m going to call the Hill radius rH measured from the position

of m. The tidal acceleration is then

atide = −GMa2− (− GM

(a− rH)2)

' GM

a2rHa

(10)

where I skipped some algebra. Note that in your derivation, m∗ �M is critical

as this allows you to make the approximation rH � a. The binding acceleration

(really, the binding force per unit mass) is

abind = −Gm∗r2H

. (11)

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Equate the two to get

rH =(m∗M

)1/3a (12)

3. Discuss what happens to the test particle as you perturb it from the equilibrium

points ever so slightly. For instance, if you deflect the particle from the Hill

radius upwards or downwards, what happens to the motion of the particle?

L1,2,3 are saddle nodes so any perturbation will make the particle run away from

the equilibrium point. You’d think that if you deflect them directly vertically,

they should oscillate about the equilibrium points but the coriolis force will

deflect them away from this directly vertical line making the test particle run

away. On the other hand, at L4,5, the Coriolis force will help them draw out

some sort of closed orbit around the equilibrium points as the perturbed test

particle moves outwards.

4 Accretion Disk

1. In the evolution of binary star system (be it red giant & main sequence star, compact object& main sequence star, etc.) an accretion disk can form. When does this happen?

This happens when an object fills its Roche lobe. In other words, mass transfer

occurs through the Roche lobe radius.

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 from the spin of the

stars.

3. Initially, disk material starts out on an eccentric orbit but in the end it circularizes.

How?

As the accreted material form a disk, to close the orbit, it must hit itself to

the stream of new accreted material. This causes energy dissipation via heat,

etc. Recall that E = −GMm/2a. Loss of energy means energy is more negative;

this means the orbital semi-major axis a is decreasing. On the other hand, the

angular momentum L = µ√G(M +m)a(1− e) is conserved. Since a is dropping, (1−

e) must be increasing to compensate. This means e has to approach zero; in other

words, the orbit becomes 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 luminosity? Can you explain

or derive how this is so?

Equal budget so half and half. Since luminosity is ∆E/∆t, we only need to compare

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

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energy before accretion is −GMm/2a whereas the orbital energy after accretion

is −GMm/2R. Consider R� a, then

∆Edisk = | − GM∆m

2R− (−GM∆m

2a)|

' GM∆m

2R(13)

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 simply

equal to the gravitational potential −GM∆m/R. The initial energy is the total

orbital energy at R. The change in energy is then

∆Eobj = | − GM∆m

R+GM∆m

2R|

=GM∆m

2R(14)

You can see that ∆Edisk = ∆Eobj.

5. For an accretion disk around black hole, what is the relative budget between disk

luminosity and object luminosity?

You can’t really define the surface of the black hole so there’s only disk luminosity.

6. Take an annulus of the disk at some arbitrary radius r and assume this annulus

to be a blackbody. What is the temperature 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 luminosity is GMM/2r but luminosity is also F × Areawhere F = σT 4, the blackbody flux at the surface and area is π(2r)2 − πr2. To

account for total luminosity, you’ll need to multiply by 2 since the light is

emitted from both sides of the disk. Equate this two expressions:

GMM

2r= σT 4[π(2r)2 − πr2]× 2

= 6πσT 4r2

T 4 =1

12πσ

GMM

r3

T ∝ r−3/4 (15)

Therefore, inner disk is hotter than the outer disk. Your disk spectra should

simply be superposition of blackbody spectra of different temperature. Recall

that from Wien’s peak law, T ∝ ν. Also, what you observe is specific flux Fν ∼F/ν. Luminosity is 6πσT 4r2 but it is also F (4πd2) where d is the distance between

the disk and the observer. This means F ∝ T 4r2 but T ∝ r−3/4 so F ∝ r−1. You

know that r ∝ T−4/3 and T ∝ ν; this means r ∝ ν−4/3 so F ∝ ν4/3. Finally, Fν ∝ν1/3 which should be the result of the superposition of blackbody spectra you sketched

earlier.

Good luck on the midterm!

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