2019 National Astronomy Olympiad (NAO)- Solutions

1 Short Questions

1. (7 points) Assuming that the present density of baryonic matter is ρb0 “ 4.17ˆ10´28 kg m´3, what wasthe density of baryonic matter at the time of Big Bang nucelosynthesis (when T „ 1010 K)? Assume thepresent temperature, T0 to be 2.7 K.

Solution: Answer: ρBBN “ 21 kg m´3.ρb,BBN “ ρb0a

´3, where a is the scale factor.We also know that T0 “ aT paq, giving us:ρb,BBN “ ρb0p

TBBNT0

q3.

Using ρb0 “ 4.17ˆ 10´28 kg m´3, TBBN “ 1010 K and T0 “ 2.7 K, we get ρBBN “ 21 kg m´3.

2. Note: constants required are the semimajor axis of the moon’s orbit (384399 km), thesemimajor axis of the earth’s orbit (1.4960 ˆ 108 km), the radius of the sun (695700 km),the radius of the earth (6371 km), the radius of the moon (1737 km), the mass of the earth(5.972ˆ 1024 kg), the mass of the sun (1.989ˆ 1030 kg), and G.

(7 points) On the night of January 21st, 2019, there was a total lunar eclipse during a supermoon. At thetime, the moon was close to perigee, at a distance of 351837 km from the earth, which was 1.4721ˆ 108

km from the sun. The gamma (γ) of a lunar eclipse refers to the closest distance between the center ofthe moon and the center of the shadow, expressed as a fraction of the earth’s radius. For this eclipse,γ “ 0.3684. Given this information, find the closest estimate for the duration of totality of the eclipse.

Solution: First, calculate the radius of the shadow cast by earth. The size of the umbra at adistance of 351837 km is rumbra “ 6371 ´ 695700´6371

1.4721ˆ108 ˆ 351837 “ 4723 km. However, the distancetravelled by the moon through the umbra is far less than twice the umbral radius. Firstly, themoon does not pass through the center of the shadow; secondly, totality occurs when the moon iscompletely within the shadow, not when the center of the moon is in the shadow. So, the distancetravelled by the moon during totality is d “ 2 ˆ

a

prumbra ´ 1737q2 ´ p0.3684ˆ 6371q2 “ 3693km. The velocity of the moon relative to the earth can be calculated using the vis-viva equation:

vmoon “b

Gˆ 5.972ˆ 1024 ˆ`

2351837000 ´

1384399000

˘

“ 1.108 km/s. However, the earth is also

moving around the sun, so the umbra is also moving relative to the earth. The velocity of the earth

is vearth “

c

Gˆ 1.989ˆ 1030 ˆ

´

21.4721ˆ1011 ´

11.4960ˆ1011

¯

“ 30.259 km/s. Therefore, the velocity

of the umbra is vumbra “vearth

1.4721ˆ108 ˆ 351837 “ 0.072 km/s. The earth and the moon orbit in thesame direction, so the velocity of the moon relative to the umbra is v “ vmoon ´ vumbra “ 1.036km/s. Finally, the duration of totality is ttotality “

dv “ 59.426 minutes.

3. (7 points) You are in the northern hemisphere and are observing rise of star A with declination δ “ ´8o,and at the same time a star B with declination δ “

Ř

16o is setting. What will happen first: nextsetting of the star A or rising of the star B?

Solution: Star B will rise first. At the same time as the star A rise, a point on the opposite side ofthe sky with declination

Ř

8o is setting. Star B sets at the same time as that point, but having ahigher declination, and being in the north hemisphere, will spend less time under the horizon.

4. (7 points) Consider a star with mass M and radius R. The star’s density varies as a function of radiusr according to the equation ρprq “ ρcenterp1 ´

a

rRq, where ρcenter is the density at the center of thestar. Derive an expression for dP dr in terms of G, M , R, and r, where P is the pressure at a givenradius r

Solution: The mass inside a given radius r is mprq “şr

04πr2ρprqdr. Thus, we have

mprq “4

21πρcenterr

3p7´ 6

c

r

Rq

Note that M “ mpRq, so we can solve for ρcenter “21M4πR3 . Hydrostatic equilibrium requires that

dP dr “ ´ρprqGmprqr2 . Thus, substituting, we have

dP dr “ ´21GM2r

4πR6p7´ 13

c

r

R` 6

r

Rq

2 Medium Questions

5. (15 points) An alien spaceship from the planet Kepler 62f is in search of a rocky planet for a remotebase. They’re attracted to Earth because of a fortunate coincidence: its axis of rotation points directlyat their home planet. That means they can have uninterrupted communication with home by plantingfixed transmitters on The North Pole. But first, they need to find out if Earth’s axis will always pointin the same direction or if it undergoes precession. They can’t know without years of observation, sothey hope that we, its now-extinct intelligence, have left behind the answer.

While orbiting Earth, they see a few remarkable structures, including the Hoover Dam in Nevada.Zooming in on the dam, a colorful plaza with peculiar markings on its floor catches their attention.Descending on the plaza, they realize the markings are a map of the sky when the dam was built, left toindicate the date to posterity. Figure 1 is an overhead architectural map of this plaza. The center-pointdepicts the north ecliptic pole, and the large circle represents the path of the Earth’s axis throughoutits counter-clockwise procession. As they interpret the map, they’re dismayed to realize that their starhas not been and will not be Earth’s north star for very long.

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Figure 1: Overhead architectural plan of the Hoover Dam plaza depicting Polaris as north star

For the purpose of this question, assume that the Earth’s axial tilt is a constant i “ 23.50 and its axisprecesses at a constant rate.

a) Using the values on the map, and knowing that the aliens used carbon-aging to determine that thedam is 12,000 years old, find all possible values for the period of Earth’s axial precession.

b) Using the most optimistic answer (longest period) from part (a), calculate how many arcsecondsthe Earth’s axis precesses each day. Use the period you calculate here in the next two sections.

c) If they hadn’t been lucky enough to come across the star map and decided to build a radio interfer-ometer to observe the movement of the celestial pole over the course of 30 days instead, how manykilometers would the baseline of their telescope array have to be, assuming it operated at a 20cmwavelength?

d) As a last resort, to keep Earth’s axis fixed, the aliens decide to counter the forces that cause theEarth’s precession by building giant nuclear thrusters on the Earth’s surface. Assume Earth’s pre-

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3.6 and 4.5 µm Phase Curves for HD 189733b 7

0.975

0.980

0.985

0.990

0.995

1.000

Rel

ativ

e Fl

ux

-0.6 -0.4 -0.2 0.0 0.2 0.4 0.6Orbital Phase

0.997

0.998

0.999

1.000

1.001

Rel

ativ

e Fl

ux

Figure 6. Final 3.6 µm photometry after correcting for in-trapixel sensitivity variations and stellar activity (black filled cir-cles), binned into four minute intervals. The best-fit phase, transit,and eclipse curves are overplotted as a red line. The lower panelshows the same data as the upper panel, but with an expanded yaxis for a better view of the phase curve.

local maximum just before the secondary eclipse, aroundthe same time where we would expect to see a maxi-mum in our best-fit phase curve. We therefore concludethat the quadratic fit is degenerate with our phase curvefunction in this bandpass and opt to limit the quadraticcoefficient in our subsequent fits to positive values (i.e.,a linear function or a local flux minimum) in order toavoid this degeneracy and ensure good agreement withour visible-light spot models. Our best-fit solution in thisband is very close to linear and indicates that the starincreased in brightness by 0.262% ± 0.016% during ourobservations (see Table 1, in good agreement with ourprediction of 0.1% − 0.4% from the visible-light photom-etry.

3. RESULTS

We fit the trimmed data and calculate uncertaintiessimultaneously for the transit, secondary eclipses, phasecurve, stellar activity, and intrapixel sensitivity cor-rections using a Markov Chain Monte Carlo (MCMC)method (see, for example Ford 2005; Winn et al. 2007b)with a total of 105 steps and either 16 (3.6 µm) or15 (4.5 µm) free parameters. These parameters in-clude: the four phase function coefficients c1 − c4, a/R⋆,i, RP /R⋆, transit time, two secondary eclipse depths,two secondary eclipse times, either a linear (4.5 µm) orquadratic (3.6 µm) function of time to account for stel-lar variability, and two noise parameters discussed be-low. We plot the normalized time series after the best-fitintrapixel sensitivity variations and stellar trends havebeen removed in Fig. 6 and 7.

We find that there is still some remaining time-correlated noise in the residuals from our best-fit solu-tion (Fig. 5); we account for this time correlation byimplementing a wavelet-based MCMC fit as describedin Carter & Winn (2009) and compare the results ofthis fit to the standard χ2-based methods described inFord (2005) and Winn et al. (2007b). For the standard

0.975

0.980

0.985

0.990

0.995

1.000

Rel

ativ

e Fl

ux

-0.6 -0.4 -0.2 0.0 0.2 0.4 0.6Orbital Phase

0.997

0.998

0.999

1.000

1.001

Rel

ativ

e Fl

ux

Figure 7. Final 4.5 µm photometry after correcting for intrapixelsensitivity variations and stellar activity (black filled circles); seeFig. 6 for a complete description.

MCMC fit, which assumes that the uncertainties on in-dividual points are Gaussian and time-independent, weset the per-point uncertainty equal to the value needed toproduce a reduced χ2 equal to one for the best-fit solution(0.333% at 3.6 µm and 0.466% at 4.5 µm). In our waveletMCMC we maximize the likelihood function instead ofminimizing χ2, which allows us to fit for the white (σw)and red (σr) noise contributions to the per-point uncer-tainties. The wavelet transform routine requires our datato have a length equal to a power of two, which we achieveby subdividing each phase curve into thirteen segmentsof equal length and zero-padding by either 6% (3.6 µm)or 3% (4.5 µm). We prefer zero-padding over trimmingas it allows us to include all available data in our fits,and we find that this approach has a negligible effect onour best-fit noise parameters.

We initialize each chain at a position determined byrandomly perturbing the best-fit parameters from aLevenberg-Marquardt minimization. After running thechain, we search for the point where the χ2 value firstfalls below the median of all the χ2 values in the chain(i.e. where the code had first found the optimal fit), anddiscard all steps up to that point. For the wavelet MCMCwe perform a similar trim where the likelihood first risesabove the median value. We set our best-fit parametersequal to the median of each distribution and calculatethe corresponding uncertainties as the symmetric rangeabout the median containing 68% of the points in thedistribution. The distribution of values was very closeto symmetric for all parameters, and there did not ap-pear to be any detectible correlations between variablesaside from i and a/R⋆. Our wavelet analysis producesmodestly higher uncertainties for most parameters, withthe greatest increases in the planet-star radius ratio andsecondary eclipse depth errors, which were a factor oftwo larger than in the standard Gaussian χ2 fits. We listthe best-fit parameters and corresponding wavelet-baseduncertainties in Table 1.

For both fits the standard deviation of our best-fitresiduals is a factor of 1.12 higher than the predicted

Figure 2: Full-phase light curve of HD 189733b. From Knutson et al. (2012).

cession is caused by external forces alone and calculate the average force (in kN) that a strategicallypositioned thruster would have to exert to counter them.

Solution:

(a) We know that Pp “3600

ω “ 360tθtotal

where t “ 12, 000 years and θtotal “ 167.60 or 3600 ` 167.60

or ... which gives: Pp “ 25775.6 years or Pp “ 8188 years or Pp “ 4867 years etc.

(b) The axis is in the direction of the angular momentum vector, which traverses a path on a smallcircle of the celestial sphere, traveling a total of 2π sin i radians in each period. So the axismoves 2π sin i

25775.6ˆ365 « 0.052 per day.

(c) R “ λD so D “ 0.2ˆ3600ˆ180

0.05ˆ30ˆπ « 25 km

(d) ~τ “ ~r ˆ ~F so∥∥∥~F∥∥∥ “ ‖~τ‖

R “

∥∥∥ ~dLdt

∥∥∥R “

‖~L‖ dθdtR “

I‖~ω‖ dθdtR “

8.01ˆ1037ˆ 2π

24ˆ3600ˆ0.0552

3600 ˆ1

24ˆ3600π

180

6.37ˆ106 «

2.82ˆ 1012 kN

6. (15 points) Figure 2 shows a full-phase light curve (“phase curve”) of the exoplanet HD 189733b takenby the Spitzer space telescope. Use this figure to answer the following questions. The star HD 189733has an effective temperature of 4785 K and a radius of 0.805 Solar radii.

a) Use the depth of the planet’s transit to estimate the radius of HD 189733b, in Jupiter radii.

b) Use the depth of the eclipse of the planet by the host star to estimate the ratio of the flux of theplanet HD 189733b to that of the host star HD 189733.

c) HD 189733b is so close-in to its host star that it is expected to be tidally locked. Use the phasecurve to estimate the ratio of the dayside flux emitted by the planet to the nightside flux emittedby the planet.

d) This phase curve also noticeably has a phase curve offset, that is, the maximum in planet andstar flux does not occur exactly at secondary eclipse. What process that occurs in a planetaryatmosphere could cause such a phase curve offset?

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Solution: a) The transit depth (reading off top plot) is « 0.025. Know that transit depth

Td9pRpRsq2 Ñ Rp “ Rs

?Td “ 0.805˚6.95ˆ 108 mSolar radius

?0.0256.91ˆ107 mJupiter radius “

1.28 Jupiter radii.b) Eclipse depth is « 0.002 in normalized flux. The bottom of the eclipse occurs at a normalizedflux of 0.9985. FpFs « 0.0020.9985 « 2 ˆ 10´3.c) Max normalized flux « 1.0005 Ñ max normalized planet flux « 1.0005´0.9985 “ 2 ˆ10´3. Minnormalized flux « 0.9996 Ñ min normalized planet flux « 0.9996´ 0.9985 “ 1.1 ˆ 10´3. Ratio ofmax to min planet flux“ 2 ˆ 10´31.1 ˆ 10´3 “ 1.8.d) Atmospheric circulation that is driven by the large dayside-to-nightside pressure gradient due totidal locking.

7. (15 points)

a) Mass-Radius Relation Stellar physics often involves guessing the equation of state for stars,which is typically a relation between the pressure P and the density ρ. A family of such guessesare known as polytopes and go as follows-

P “ K ργ (1)

where K is a constant and the exponent γ is fixed to match a certain pressure and core temperatureof a star. Given this, show that one can obtain a crude power-law scaling between the mass M of apolytopic star and its radius R of the form M9Rα. Find the exponent α for polytopic stars (justifyall steps in your argument). Also, indicate the exponent γ for which the mass is independent of theradius R. Bonus: Why is this case interesting?

b) Black Holes as Blackbodies The mass radius relation for ideal non-rotating, uncharged blackholes is known from relativity to be

R “2GM

c2(2)

Moreover, Stephen Hawking showed that a black hole behaves like a blackbody, where its temper-ature (known as the Hawking temperature) is given by

T “~c3

8πkBGM(3)

Given this information, show that the lifetime of a black hole (justify this phrase!) t˚ scales withits mass M as

t˚9Mβ (4)

where you should find the exponent β

c) Minimal Black Holes Using the information of the previous part, and Wien’s displacement law,estimate the smallest possible mass of a black hole. State any possible flaws with this estimate.

Solution:

(a) To maintain hydrostatic equilibrium in a star, we need the pressure to balance the gravitationalpull. This translates to the radial force balance equation

dP

dr“ ´

GMρ

r2(5)

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(Consider the forces on a thin shell of gas in the stellar interior to obtain this). Now to obtaina crude estimate, we do the following

dP

dr«

∆P

∆R“´P

R« ´

GMρ

R2(6)

where P is the pressure in the core of the star and R is the radius of the star. Thus, we obtainthe scaling

P „Mρ

R„ ρ2R2 (7)

Using (1), we obtain then

ρ „ R2γ´2 (8)

Thus, we finally arrive at

M „ R3γ´4γ´2 (9)

ùñ α “3γ ´ 4

γ ´ 2(10)

Thus, if γ “ 43, the mass of the polytopic star is independent of the radius R. This is aninteresting case, since this has connections to the equation of state for a relativistic degeneratefermion gas, which is used to model white dwarves. This independence of the mass and radiuscan be thought of as a crude way to understand the Chandrashekar limit.

(b) We have that

R “2GM

c2(11)

T “~c3

8πkBGM(12)

Using the assumption that a black hole is a blackbody, we apply Stefan’s law to find the powerof radiation emitted by a black hole

dE

dt“ 4πR2σT 4 “

~4c8σ

256π3G2k4B

1

M2(13)

Since this energy has to arise from somewhere, we take this to emerge from the mass-energy ofa black hole, giving us

dE

dt“ ´c2

dM

dt9

1

M2(14)

Thus, we get

M2dM “ ´Kdt (15)

where K is a constant of proportionality involving the fundamental constants and σ. Integratingthe above equation, we see that the blackhole loses mass and its lifetime scales with its massas follows

t˚9M3 (16)

ùñ β “ 3 (17)

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(c) Under the assumption that a black hole is a blackbody, we are justified in thinking that itsspectrum has the characteristic Planckian spectrum with the Wien’s law peak for the emittedphoton being given by:

λ “b

T, b “ 2.9ˆ 10´3m K (18)

If we estimate that one photon (or any Op1q photons, as required to preserve momentum)having this wavelength carries away the entire mass energy of the black hole, we estimate that

Mc2 «2π~cλ

(19)

ùñ M2 «

ˆ

~c2?GkBb

˙2

(20)

M «

ˆ

~c2?GkBb

˙

« 9.68ˆ 10´9 kg (21)

M « 4.84ˆ 10´39Md (22)

There are several problems with this method of estimation. Firstly radiation is a fully quantumprocess, so our assumption of radiating a single photon is not really correct. Moreover, we useWien’s law in its wavelength form, to obtain energy but we really need it in the frequency formto actually obtain the peak of the energy of the photon spectrum. This is an order of magnitudeestimation, so any reasonably justified argument should be given full credit.

8. (15 points) In a rather weird universe, the gravitational constant G varies as a function of the scalefactor aptq.

G “ G0fpaq (23)

Consider the model fpaq “ ebpa´1q where b “ 2.09.

a) Assuming that the universe is flat, dark energy is absent, and the only constituent is matter, estimatethe present age of this weird universe according to this model. Assume that the Friedmann equation:

Hpaq2 “ H20 pΩm ` Ωr ` Ωk ` ΩΛq (24)

still holds in this setting.

b) What is the behaviour of the age of the universe t as the scale factor aptq Ñ 8 ?

Note that all parameters with subscript 0 indicate their present value. Take the value of Hubble’sconstant as H0 “ 67.8 kms´1Mpc´1

Hint : You might need the following integrals

ż 8

0

x2e´x2

dx “

?π

4

ż 1

0

x2e´x2

dx « 0.189471 (25)

Solution:

(a) We have from the Friedmann equations that in such a matter-only universe,

Hpaq2 “ H20 Ωm (26)

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whereΩm “

ρmρc

(27)

ρc “3H2

0

8πG(28)

Using eqs. (26) to (28) and the relation ρm “ ρm0a´3 we infer that

Ωm “ Ωm0fpaqa´3 “ fpaqa´3 (29)

where for a matter-only universe Ωm0“ 1. Using now the relation Hptq “ 9a

a , we obtain that

t “

ż t

0

dt1 “

ż aptq

0

dt1

da1da1 “

ż aptq

0

a1

9a1a1da1 “

ż aptq

0

da1

Hpa1qa1(30)

t “1

H0

ż aptq

0

da1

a1a

fpa1qa1p´3q(31)

Using fpaq “ ebpa´1q, and the substitution x “a

a1b2 simplifies this to

t “4?

2eb2

b32H0

ż

?aptqb2

0

dx x2e´x2

(32)

Using the given value of H0 and the integrals in the hint, we arrive at the present age (aptq “ 1)of such an obscure universe to be

t « 15.0 billion years (33)

which is surprisingly close to the current estimate of our universe’s age.

(b) What is even more surprising in this model, is that if one were to set aptq “ 8 one would findthat the integral above is finite, as shown in the hint.

t “

?2πeb2b´32

H0(34)

The value b “ 2.09 gives us t « 34.1 billion years. Thus, the scale factor blows up in a finiteamount of time, which is a weird feature of this model given this simplistic treatment.

(The above model appeared in the paper Varying-G Cosmology with Type IaSupernovae by Rutger Dungan and Harrison B. Prosper (arXiv:0909.5416v2) where the authorsshowed that the type Ia supernovae data alone wasn’t sufficient to conclude the existence ofdark energy. The value b “ 2.09 was obtained by fitting the experimental luminosity distanceto the theoretical luminosity distance obtained in the model.)

9. (15 points)

a) Find the shortest distance from Boston (42.36010 N, 71.05890W ) to Beijing (39.90420 N, 116.40740 E)travelingalong the Earth’s surface. Assume that the Earth is a uniform sphere of radius 6371 km.

b) What fraction of the path lies within the Arctic circle (north of 66.56080 N)?

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Solution:

(a) We choose a coordinate system with Boston lying on the x-axis and the axis of the Earth’s rota-tion corresponding to the z-axis. The two cities are separated by 360´ p116.4074` 71.0589q “172.53370 of longitude. Now, in spherical coordinates, we get that pθ, φq is given by p90 ´42.3601, 0q “ p47.63990, 00q for Boston and by p90´ 39.9042, 172.5337q “ p50.09580, 172.53370qfor Beijing. Converting to cartesian coordinates on the unit sphere (we multiply by the radiuslater to get the actual distance), we get that psinpθq cospφq, sinpθq sinpφq, cospθqq are given by~r1 “ p0.738925, 0, 0.673788q and ~r2 “ p´0.760614, 0.0996817, 0.641506q. The cosine of the angleθ between them is given by cospθq “ ~r1 ¨ ~r2 “ ´0.129798 ùñ θ “ 97.4580 “ 1.701. Noting thatthe distance is given by D “ REarthθ when θ is measured in radians, we get D “ 10840 km.

(b) The equation of the plane passing through the origin and the two cities is given by p~r1ˆ ~r2q¨~x “0. Now, solving the equations ´0.0671643x ´ 0.986517y ` 0.0736573z “ 0. Setting z “sinp66.56080q “ 0.917483, we get 0.0671643x` 0.986517y “ 0.0675793 and x2 ` y2 “ 0.158225.

Writing y in terms of x, we get x2`p0.0675793´0.0671643xq2

0.9865172 ´0.158225 “ 0. Thus, the product ofx values that satisfy this equation is ´0.152824. Doing the same with the variables switched,

we get y2 `p0.0675793´0.986517yq2

0.06716432 ´ 0.158225 “ 0. The product of y-values is thus given by0.00394098. Thus, the dot product of the points of contact with the Arctic circles is given by0.692892. This gives an angle of 46.1410. Therefore, the final ratio is 46.14

97.458 “ 0.47.

3 Long Questions

10. (25 points) In this problem, we will try to understand the relationship between magnetic moments andangular momenta, first for charged particles and how this can be extended to planetary objects.

(a) (5 points) Consider a charge e and mass m moving in circular orbit of radius r with constant speedv. Write down the angular momentum L of the charge and magnetic moment µ of the effectivecurrent loop. Recall that the magnetic moment of a current loop with current I and radius r isgiven as µ “ IA where A is the area of the loop.

(b) (3 points) Use the above results to find a relationship between the magnetic moment µ and angularmomentum L in terms of intrinsic properties of the particle (charge,mass).

(c) (2 points) The relationship from part (b) can be expressed as µ “ γL. γ is usually referred to as theclassical gyromagnetic ratio of a particle. Evaluate the classical gyromagnetic ratio for an electronand for a neutron in SI units.

(d) (7 points) For extended objects such as planets, the magnetic dipole moment is not directly accessiblewhereas the surface magnetic field can be measured. Assuming a magnetic dipole of magneticmoment µ located at the center of a sphere of radius r, write down the expression for the surfacemagnetic field Bsurf and the surface magnetic moment defined as Msurf “ Bsurfr

3. You may usethe value of the angular dependence at the magnetic equator for the following parts.

(e) (3 points) Assuming a gyromagnetic relationship exists between magnetic moment µ and angularmomentum L of an extended object, write down the relationship between the surface magneticmoment Msurf and angular momentum L as Msurf “ κL. You will observe that κ depends onlyon fundamental constants and intrinsic properties of the extended object.

(f) (3 points) The surface magnetic moments for Mercury and Sun are 5ˆ 1012 T m3 and 3ˆ 1023 Tm3 respectively. Assuming the bodies are perfect spheres, evaluate the constant κ for Mercury andthe Sun. Comment on values obtained and if they fit into the model developed in parts (c) and (d).

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(g) (5 points) The surface magnetic moments Msurf and angular momenta L of various solar systembodies are plotted in the figure 3. Justify that the data implies Msurf „ Lα and calculate theconstant α. What is the expected value of α from the model developed in parts (c) and (d)?

Figure 3: Surface magnetic moment vs angular momentum for solar system objects. Figure taken fromVallee, Fundamentals of Cosmic Physics, Vol. 19, pp 319-422, 1998.

(h) (2 points) Certain bodies such as Venus, Mars and the Moon are remarkably separated from thetrend observed for other bodies. What can you say about magnetism in these bodies when comparedto the others?

Solution:

(a) L “ mvr and µ “ IA “ ev2πr pπr

2q “ evr2

(b) µ “ e2mL

(c) γe “ 8.794ˆ 1010 C/kg and γ “ 0 for a neutron, since the charge is zero.

(d) Use the magnetic field of a dipole µ as Bsurf “µ0

4πµr3 p2 cospθqr ` sinpθqθq. The r3 dependence

is crucial here which follows for any dipole type distribution; partial credit if that is shown.Msurf “

µ0

4πµp2 cospθqr ` sinpθqθq

(e) Neglecting the angular dependence, Msurf “`

µ0

4πγ˘

L “ κL

(f) For Mercury, angular momentum L “ 1030 kg m2/s, so κ “ 5 ˆ 10´18 T s2/(kg m). ForSun,angular momentum L “ 1042 kg m2/s, so κ “ 3ˆ10´19 T s2/(kg m). These two values areapproximately close to within an order of magnitude. We therefore expect that the mechanismsfor magnetic dipole generation are similar in Mercury and the Sun.

(g) Observe that the plot is on log-log scale. A linear relationship on this plot implies a power-lawform. From the slope of linear trend-line on the graph, α « 0.8. Full credit for realizing α ă 1.0; if α “ 1.0 is written, partial credit.

(h) Venus, Mars and Moon do not have a active dynamo and hence the magnetism present withthem is of a different nature from the other bodies. These bodies had a magnetic dynamo inthe past which has since died down. The magnetism is only remnant from this past dynamo.

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11. (25 points) Cygnus X-1/HDE 226868 is a binary system consisting of a black hole Cygnus X-1 and bluesupergiant HDE 226868. The mass of HDE 226868 is 30M@ and the period of the binary system is 5.6days. Radial velocity data reveals that the orbital velocity of HDE 226868 is 116.68 km/s at apoapseand 123.03 km/s at periapse.

(a) (5 points) Determine the eccentricity of the orbit of HDE 226868.

(b) (5 points) Determine the length of the semimajor axis of the orbit of HDE 226868.

(c) (5 points) Determine the mass of Cygnus X-1, to at least 3 significant figures.

The peak blackbody temperature of an accretion disk occurs at a distance of rpeak and a temperature ofTpeak. One can determine the peak blackbody temperature by assuming that it corresponds to the peakin the x-ray spectrum. Due to relativistic effects, the actual peak blackbody temperature Tpeak is relatedto the peak color temperature Tcolor derived from observed spectral data by Tcolor “ fGRfcolTpeak, wherefGR « 0.510 and fcol « 1.7. Three x-ray spectra of Cygnus X-1 are shown in Figure 4.

Figure 4: Three x-ray spectra from Cygnus X-1. From Gou et al. (2011).

(d) (4 points) Using spectrum SP2, determine the peak blackbody temperature Tpeak of the accretiondisk around Cygnus X-1.

The total luminosity of the blackbody component of the accretion disk can be estimated by Ldisk «4πσr2

peakT4peak (Makishima et al. 1986). The radius rlast of the innermost edge of the accretion disk is

related to the radius rpeak of the peak blackbody temperature by rpeak “ ηrlast, where η « 0.63. In1996, the blackbody luminosity of the accretion disk around Cygnus X-1 was estimated to be 2.2ˆ 1037

ergs/s.

(e) (4 points) Determine the radius rlast of the innermost edge of the accretion disk around CygnusX-1.

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Assume that the innermost edge of the accretion disk is located at the innermost stable circular orbit(ISCO), whose radius risco is a function of the spin of the black hole. The relationship between riscoand a˚, the spin parameter of the black hole, can be estimated by:

risco “GM

c2

´

a

8.354 ¨ rp2´ a˚q2 ´ 1s ` 1¯

(f) (2 points) Determine the spin parameter a˚ of Cygnus X-1.

Solution:

(a) In the absence of external forces, the total angular momentum of the binary system is conserved.The angular momentum l “ mr2ω “ mrv of each component must also be conserved. Therefore,

m1r1,av1,a “ m1r1,pv1,p ùñv1,pv1,a

“r1,ar1,p

“ap1`εqap1´εq ùñ ε “

v1,p´v1,av1,p`v1,a

“ 123.03´116.68123.03`116.68 “ 0.0265.

(b) The area swept by the vector r1 (from the center of mass to the primary) over time interval∆t is approximately a right triangle of base r1 and height v1∆t, and therefore area 1

2r1v1∆t.

Over a single orbit, the area swept is A “ 12r1v1T “ πa2

1

?1´ ε2, for pr1, v1q at any point

in the orbit (by Kepler’s second law, the rate of area swept is constant). So, πa21

?1´ ε2 “

12a1p1` εqv1,aT ùñ a1 “

b

1`ε1´ε ¨

v1,aT2π “

b

1`0.02651´0.0265 ¨

116680¨5.6¨24¨36002π “ 9.23ˆ 109 m.

(c) Kepler’s third law states that pa1 ` a2q3 “

Gpm1`m2q

4π2 T 2, so a2 “

”

Gpm1`m2q

4π2 T 2ı13

´ a1.

Additionally, the definition of center of mass is that a1m1 “ a2m2, so a2 “a1m1

m2. Putting these

two equations together yields”

Gpm1`m2q

4π2 T 2ı13

´a1 “a1m1

m2, or

”

Gpm1`m2q

4π2 T 2ı13

´a1´a1m1

m2“

0. Solving by iteration gives m2 “ 2.40ˆ 1031 kg, or m2 “ 12.1M@.

(d) The spectrum peaks at about 1.5 keV, or 2.4ˆ10´16 J. The energy of a photon is related to itswavelength by E “ hc

λ , which is related to the peak blackbody temperature by Tcolor “bλ , where

b is Wein’s constant. Therefore, E “ hcTcolorb , or Tcolor “

Ebhc “

2.4ˆ10´16¨2.898ˆ10´3

6.626ˆ10´34¨3.00ˆ108 “ 3.50ˆ106

K. Finally, Tpeak “TcolorfGRfcol

“ 4.04ˆ 106 K.

(e) Ldisk “ 2.2 ˆ 1037 ergs/s, or 2.2 ˆ 1030 W. Ldisk “ 4πσr2peakT

4peak, so rpeak “

1T 2peak

b

Ldisk4πσ “

1p4.04ˆ106q2

b

2.2ˆ1030

4π¨5.67ˆ10´8 “ 107.66 km. rlast “ ηrpeak “ 67.83 km.

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