AST 341 - Homework V - Solutions

TA: Marina von Steinkirch, steinkirch@gmail.com

State University of New York at Stony Brook

November 22, 2010

1 (1 point) Derive the homologous form of theluminosity equation df

dx = Dp2tν−2x2.

The homologous relations are

x =r

R,

f =L(r)

L0,

p =P (r)

P0, where P0 =

GM2

4πR4

t =T (r)

T0, where T0 =

µGM

RRg.

We know also the following identities:

ε = ε0ρTν ,

ρ =µP (r)

RgT (r),

D = D0ε0µν M

ν+2

LRν+3,

dL

dr= 4πr2ρε.

Now, from class we have

dp

dx=dp

dL

dL

dr

dr

dx,

following immediately

df

dx=

df

dL× dL

dr× dr

dx,

=1

L× 4πr2ρε×R

=1

L× 4πr2ρ2εT (r)µ ×R

1

Plugging the identities into this last equation and evaluating correctly ρ2

and D givesdf

dx= Dp2tν−2x2.

2 11.2 (1 point)

L� 3.84× 1026 J/s∼ 1.2× 1034 J/year

M� 1.988× 1030kg

2.1 At what rate is the Sun’s mass decreasing due to nu-clear reactions? Express in solar masses per year.

From the previous chapter, example 10.3.2, we can just use the Einstein’senergy-mass relation, E = mc2, we have M� = 6.67× 10−14M�/year.

2.2 Compare you answer with the mass loss rate due tothe solar wind.

From example 11.2.1 in our book, loss rate due to the solar wind is giving myM� = 3× 10−14M�/year, which is half of what we have found before.

2.3 Assuming that the solar wind mass loss rate remainsconstant, would either mass loss process significantlyaffect the total mass of the Sun over its entire main-sequence lifetime?

Let us say that Sun will stay in its main-sequence lifetime for ∼ 1010 year.These both quantities of mass loss, ∼ 10−14M�/year ×1010 year ∼ 10−4M�,are negligible.

3 11.5 (1 point)

3.1 Using Eq. 9.63

(∆λ)1/2 =2λ

c

√(2kT

m+ v2turb

)ln 2,

2

and neglecting turbulence, estimate the full width athalf-maximum of the Hydrogen Hα absorption linedue to random thermal motions in the Sun’s photo-sphere. Assume that the temperature is the Sun’seffective temperature.

For Hα absorption,

λα 656.280 nmT 5777 Kkb 1.38× 10−23 J K−1

m mH ∼ mp ∼ 1.67× 10−27 kg.

Plugging back in 9.63, we have (∆λ)1/2 = 0.03558 nm.

3.2 Using Hα redshift data for solar granulation, estimatethe full width at half-maximum when convective tur-bulent motions are included with thermal motions.

From data we have vturb = 0.4kms−1, and we find a very similar result (∆λ)1/2 =0.03562 nm.

3.3 What is the ratio of v2turb to 2kT/m?

The ratio is ∼ 10−3, meaning that they are very similar.

3.4 Determine the relative change in the full width athalf-maximum due to Doppler broadening when tur-bulence is included. Does turbulence make a signifi-cant contribution to (∆λ)1/2 in the solar photosphere?

δ(∆Λ)1/2

(∆Λ)1/2∼ 0.1%,

which is negligible.

3

4 11.12 (1 point) Calculate the magnetic pres-sure in the center of the umbra of a largesunspot. Assume that the magnetic field strengthis 0.2 T. Compare your answer with a typicalvalue of 2 × 104N.m−2 for the gas pressure atthe base of the photosphere.

µ0 4π × 107N.A−2

| ~B| 0.2 T

The magnetic pressure in the center of the umbra can be calculated by

PB =B2

2µ0= 1.5916× 104N.m−2,

which is the same order of the typical value at the base of the photosphere.

5 12.10 (1 point) Calculate the Jeans length forthe dense core of the giant molecular cloud inexample 12.2.1.

T 10K

ρ0 3× 10−17 kg m−3

µ 2MH ∼ mp 1.67× 10−27 kg

G 6.67× 10−11 N.m2 kg−2

kb 1.38× 10−23 J K−1

The Jeans length is given by

RJ ∼

(15kbT

4πGµmHρ0

)1/2

,

∼ 5× 1015 m

4

6 12.18 (1 point)

6.1 Beginning with Eq. 12.19,

d2r

dt2= −GMr

r2,

adding a centripetal acceleration term and using con-servation of angular momentum, show that the col-lapse of a cloud will stop in the plane perpendicularto its axis of rotation when the radius reaches

rf =ω20r

40

2GMr

,

where Mr is the interior mass and ω0 and r0 are theoriginal angular velocity and radius of the surface ofthe cloud, respectively. Assume that the initial radialvelocity of the cloud is zero and rf � r0.

Adding a centripetal acceleration term we have

d2r

dt2= −GMr

r2+ ω2r = U

dU

dr,

we then multiply both size by dr integrate in the limit,resulting

GMr

r+ω2r2

2=U2(r)

2.

Making r = rf and putting it back gives

ωF =

√2GMr

r3f→ rf =

ω20r

40

2GMr.

6.2 Assume that the original cloud has a mass of 1 M�and an initial radius of 0.5 pc. If collapse is haltedat ∼ 100 AU, find the initial angular velocity of thecloud.

ω0 =

√2GMrrf

r0= 2.8× 10−16 rad/s.

6.3 What was the original rotation velocity (in m s−1) ofthe edge of the cloud?

ω0 =v0r0→ v0 = 4.2 m/s.

5

6.4 Assuming that the moment of inertia is Is = 25Mr2,

when the collapse begins and Id = 12Mr2 when it stops,

determine the rotation velocity at 100 AU.

Here we can use conservation of momentum,

I0ω0 = Ifωf → ωfrf = vf = 3.4 km/s.

7 (2 points) T Tauri stars collapse quasi-statically,converting gravitation potential energy intoheat. If you could measure the radius of a TTauri star very accurately, how much smallerwould it be after a year? How long would ittake before you noticed the star fading?

Teff 4500 KRπ 3R�Mπ 1M�

This problem is partially worked out at page 426 of the textbook. First,

L = 4πR2σT 4 = 1.3× 1027 W,

and it is related to the gravitational potential as

L =GM

R2dR

giving

dR =L/R2

GM2∼ 660 m/ year.

To notice the fading we should be able to notice the change of magnitude:

δm =10−3

2.5= 2.5 log (

L+ δL

L),

giving δL = 1.2× 1024 W.From the luminosity δL = 8πRdRσT 4 = 8 × 1020 W/ year, therefore we

would need more than thousand years to notice it.

6

8 13.1 (1 point)

8.1 How long does it takes the 5 M� star to cross theHertzsprung gap relative to its main-sequence life-time?

∆t = 0.46 million of years (Myr), 0.5%.

8.2 How long does the 5 M� star spend on the bluewardportion of the horizontal branch relative to its main-sequence lifetime?

∆t = 1.50 million of years (Myr), 1.6%.

8.3 How long the 5 M� star spend on the redward portionof the horizontal branch relative to its main-sequencelifetime?

∆t = 6.32 million of years (Myr), 6.8%.

9 13.5 (1 point) Use equation 10.27,

Tquantum =Z2

1Z22e

4µm12π2ε20h

2k,

to show that the ignition of the triple alphaprocess at the tip of the red giant branchought to occur at more than 108 K.

Z1 2Z2 2e 1.6× 10−14C

First of all, the reduced mass is µ ∼ 2mp ∼ 3.34 × 10−27 kg. Plugging ineverything into the equation gives

Treaction = 6.2× 108K.

7