usaao 2015 (second round)

8/20/2019 USAAO 2015 (Second Round)

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USAASTRONOMY & ASTROPHYSICS

OLYMPIAD

2015

SECOND ROUND

S i Ol i d Bl


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USAAAO 2015 Second Round Solutions

Problem 1

T = 10000 K, m = 5, d = 150 pcm − M = 5log(d/10)M = m − 5 log(d/10) = 5 − 5 log(15) = −0.88We compare this with the absolute magnitude of the sun, 4.83.

LLsolar

= 100(4.83−−0.88)/5 = 192LsolarFrom the Stefan-Boltzmann Law, L = 4πR2σT 4

Finding the ratio with the sun, we get:

LLsolar

=

RRsolar

2 T T solar

4

RRsolar

=

LLsolar

1/2 T solarT

2= 4.6Rsolar

Problem 2

T = 3 days, K = 50 m/s, M = 1M solarT 2 = a

3

M 3365

2/3= 0.041AU for the planet’s orbit.

For the star’s orbit:T = 2πrvr = T v

2π = 1.38 ∗ 10−5AU

Relating the two, we have:m∗r∗ = m pr p, so we have m p =

r∗rpm∗

Plugging in, we find m p = 3.36 ∗ 10−4M solar

Problem 3Solar rotation rate 24.5 days, M J = 9.54 ∗ 10−4M solar, a = 5.2 AU, solar

radius 695,000 km.L = IωLsolar =

25MR

2∗ 2π24.5 = 5 ∗ 10

10

For Jupiter, L = mrvr = 5.2 ∗ 149.6 ∗ 106 = 7.77 ∗ 108kmv = 2πr

5.22/3∗365 = 1.13 ∗ 106 km/day.

L = 9.54 ∗ 10−4 ∗ 7.77 ∗ 108 ∗ 1.13 ∗ 06 = 8.38 ∗ 1011

So Jupiter has the greater angular momentum.

Problem 4

m = 10, T = 6000 K at the main sequence turnoff.

Oldest main sequence stars are 6000 K, which is approximately sun-like. Wetherefore assume the absolute magnitude of the stars at the turnoff point is 4.83.

m − M = 5log(d/10)d = 10 ∗ 10(m−M )/5 = 108 pcThe stars at the turnoff point are sunlike, so we expect them to have a

lifetime of 10 Gyr. Since these are the oldest main sequence stars in the cluster,the cluster has an age of approximately 10 Gyr.

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Problem 5

Question not graded

Problem 6

m = 8, p = 0.003”, and T = 6000 K.d = 1/p = 333 pcm − M = 5log(d/10)M = m − 5 log(d/10) = 0.385

LLsolar

= 100(M solar−M )/5 = 59.9LsolarThe star’s temperature is approximately sunlike, suggesting class G, but it

is significantly more luminous, suggesting a giant. G0III would be a reasonablepossible spectral type.

Problem 7

From Wein’s Law, λ = bT . From the definition of Redshift, z = λ−λ0λ0 , whereλ0 is the emitted wavelength. Solving for λ0, we get λ0 = λ/(z + 1)

Combining this with Wein’s Law, we get:

T = b∗(z+1)lambdaUsing Wein’s Law and the given temperature of 2.73 K, we find that the

recieved wavelength is 1.06 mm. Plugging this into the above expression, weobtain at temperature of 30.03 K for z = 10.

Problem 8

At blackbody equilibrium, power in is equal to power out, so we have:P out = 4πR2 pσT

4 p

P in = A ∗ (1 − α) ∗ L4πD2 = πR

2 p(1 − α)

4πR2∗σT 4

∗

4πD2

P in = P out4πR2 pσT

4 p = πR

2 p(1 − α)

4πR2∗σT 4

∗

4πD2

T 4 p = (1 − α)T 4∗

R2∗

4D2

T p = (1 − α)1/4T ∗

R∗2D

Problem 9

7.2 micron pixels, f/10, D = 0.256 m.f/10, so focal length is 2.56 m.Angular resolution is given by pixelsizefocallength ∗ 206265

Plugging in, we get a resolution of 0.58 arcseconds/pixel

Problem 10A Hohmann transfer orbit is being used to go from the 1 AU orbit of the

Earth to Saturn’s orbit at 9.6 AU. The semimajor axis of the transfer orbit isthus 5.3 AU.

Using the vis-viva equation, v2

2 −

µr = −

µ2a

, this corresponds to a velocity of 40,080 m/s at the Earth’s orbital distance.

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When it starts, however, the spacecraft is already in Earth parking orbit, so

benefits from both its orbital velocity around the Earth and the Earth’s orbitalvelocity around the Sun.When in the parking orbit, the probe has a maximum velocity relative to

the Sun of

GM erorb

+

GM srearth

= 37540m/s

The difference between the spacecraft’s current maximum velocity relativeto the Sun and the orbital velocity of the transfer ellipse is equal to the required∆v, since we’re neglecting further attraction from the Earth. The necessary ∆vis thus 2550 m/s, and the burn must occur on the night side of the Earth inorder to increase the orbital radius and match Saturn’s orbit (since the parkingorbit is prograde).

We use the vis-viva equation again to calculate the probe’s velocity at Sat-urn’s orbital distance, and compute Saturn’s orbital velocity using the same

technique we used for Earth. We then subtract Saturn’s velocity from theprobe’s velocity to find the probe’s velocity relative to Saturn.We then calculate the velocity relative to Saturn for a 100,000 km circular

orbit. The difference between this value and the value above is the ∆v requiredto make Saturn orbit, coming out to 24900 m/s. The burn must occur on thenight side of Saturn in order to enter a prograde orbit.

Problem 11

M = 0.54M s, P = 6 years, perihelion distance 0.537 AU, parallax 0.05”.From the parallax equation, d = r/θ, where d is the distance, r is the baseline,

and θ is the parallax angle. We therefore need to find the baseline, XY, whichis the latus rectum of the ellipse.

Using Kepler’s Third Law, we find that the semimajor axis is 2.69 AU. This

corresponds to an aphelion of 2.135 AU, and therefore an eccentricity of e =0.3.

The latus rectum of an ellipse is given by l = a(1 − e2). Plugging in, we getl = 2.45 AU, yielding a distance d = 49.0 pc

Problem 12

Simply dividing the length of the year by 6 is incorrect. Because of theinclination of the Earth, the Sun also varies in declination (from -23.5 to +23.5degrees), but maintains (for this problem) a constant angular velocity. Sphericaltrigonometry is therefore required to determine the actual angle that the Suntraverses going from 0 to 4 hours (62.1 degrees). Since the Sun has constantangular velocity, traversing 360 degrees per year, the number of days can be

expressed as (62.1/360)*365 = 63.0 days.

Problem 13

This question also requires spherical trigonometry. Picking RA = 0, Dec =90 is likely the easiest third point. Now that we have a spherical triangle, wecan use the spherical law of sines or cosines to find the separation angle (18.59degrees).

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The position angle is measured east of north. For our chosen triangle, this

corresponds to the angle with Betelgeuse as its vertex. Again applying thespherical trigonometric relationships, we find the position angle to be 33.12degrees.

To cover both stars, the picture must cover 18.6 degrees of the sky, whichcorresponds to 66960 arcseconds. Plate scale, in arcseconds/mm, is given by206265/focal length. Solving for the focal length, we get a value of 3.08*filmsize, which is a focal length of 108 mm on 35 mm film or 216 mm on 70 mm.

Long Problem

a. Drawing the transit light curve, we can see that between the first andsecond contacts, the relative motion of the stars is equal to twice the radius of the transiting star. Similarly, the time between second and third contacts isproportional to twice the primary star’s radius. Dividing, we get r1r2 =

t2−t1t3−t2

=

13.67We can use Wein’s Law and the provided blackbody peaks to calculate the

temperature of each star. We can now use the Stefan-Boltzmann Law to findthe ratio of the luminosities:

L1L2

=R1R2

2 T 1T 2

4= 934

b. We use the provided chart to determine the orbital period of the binarysystem (approximately 5.7 days) as well as the semi-amplitude (using ∆λλ =

vc ).

Using the period and velocity of each star, we determine the semimajor axesaccording to r = vT

2π, for semimajor axes of 0.0450 and 0.0771 AU for stars 1

and 2 respectively.Applying Kepler’s Third Law for the entire system, we get a total mass of

1.90 solar masses. Since m1r1 = m2r2, we can determine the individual mass of

each star, 1.20 solar masses for star 1 and 0.70 solar masses for star 2.c. Using the velocity of each star, we can find the relative velocity of the

two stars (just sum). This can then be used to solve the equations from part adirectly, giving radii of 1.35 and 0.099 solar radii for stars 1 and 2, respectively.We can now apply the Stefan-Boltzmann Law, dividing by the solar expression,to determine the luminosity of each star (2.54 solar luminosities for star 1, 0.0027for star 2).

d. Apparent magnitude can be found by finding the total flux of the systemand using the Sun’s apparent magnitude and the solar flux as a standard can-dle, applying ∆M = 2.512 log F 1F 2 . Similarly, using the luminosity of the systemwith the Sun’s absolute magnitude as a standard candle can provide an abso-lute magnitude for the system. Now that we have an apparent and absolute

magnitude, we apply the distance modulus to get d = 5.94 pc.e. Applying the small angle formula, we get a maximum angular separationof 0.021”. The best possible resolution of the telescope is given by θ = 1.22 λD ∗206265 = 0.017”, so the stars are distinguishable. The smallest visible size canbe found by applying the small angle formula with the limiting resolution, andis 0.10 AU.

f. Examining the given plot, there appears to be a longer period variation

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with a period of approximately 214 days. Based on the change in wavelength,

the variation has an amplitude of approximately 83.2 km/s. Using this velocityand the period of oscillation, we can determine the semimajor axis of the binarystars’ orbit (1.62 AU)

We now apply the fact that m1a1 = m2a2 and Kepler’s third law for thebinary-unknown system, solving for a2 and m2. We find that the unknownobject has a mass of 16.2 solar masses and a semimajor axis of 0.19 AU. Giventhe high mass and lack of a visible counterpart, the object is likely a stellarmass black hole.

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USAAAO First Round 2015

This round consists of 30 multiple-choice problems to be completed in 75 minutes. You

may only use a scientific calculator and a table of constants during the test. The top 50%

will qualify for the Second Round.

1. At arms length, the width of a fist typically subtends how many degrees of arc?

a. 1 o

b. 5 o

c. 10 o

d. 15 o

e. 20 o

2. To have a lunar eclipse, the line of nodes must be pointing at the sun. The moon must

also be in what phase?

a. New

b. First Quarter

c. Waxing Gibbous

d. Full

e. Waning Crescent

3. Mars orbits the sun once every 687 days. Suppose Mars is currently in the constellation

Virgo. What constellation will it most likely be in a year from now?

a. Virgo

b. Scorpiusc. Aquarius

d. Taurus

e. Cancer

4. To calculate the field of view of a telescope, you measure the time it takes Capella

(RA:5.27h, dec:45.98 o ) to pass across the eyepiece. If the measured time is 2 minutes and

30 seconds, what is the field of view in arcseconds?

a. 11.6’

b. 26.5’

c. 37.5’d. 52.5’

e. 66.8

5. A telescope with focal length of 20 mm and aperture of 10 mm is connected to your

smartphone, which has a CCD that measures 4.0mm by 4.0mm. The CCD is 1024 by

1024 pixels. Which is closest to the field of view of the telescope?

a. 1 o


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b. 5 o

c. 10 o

d. 15 o

e. 20 o

6. What is its the resolution in arcseconds per pixel?

a. 10”/pixel

b. 40”/pixel

c. 120”/pixel

d. 1200”/pixel

e. 3600’/pixel

7. Comet 67P/Churyumov–Gerasimenko has an orbital period around the Sun of 6.44 years.

What is its semimajor axis, in AU?

a. 41.47

b. 16.34

c. 6.44

d. 3.46

e. 1.86

8. Which of the following techniques most directly constrains the mass of an exoplanet?

a. Radial Velocity

b. Transit Timing

c. Microlensing

d. Direct Imaging

e. Proper Motion

9. Which two properties of galaxies does the Tully-Fisher relation utilize a correlation between?

a. Luminosity and velocity dispersion

b. Luminosity and rotational velocity

c. Radius and metallicity

d. Luminosity and metallicity

e. Mass and surface brightness

10. A binary star system has two components: Star A and Star B. Star A has a mass of 5 solar

masses, and Star B has the same mass as our Sun. Assuming circular orbits, how many

times closer to the center of mass of the system is Star A than Star B?a. 1

b. 3

c. 5

d. 10

e. 25


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16. A and B, two main sequence stars of the same spectral class, have apparent magnitudes

of 17 and 12, respectively. If star A is 1 kpc away, what is the distance to star B?

a. 10 pc.

b. 100 pc.

c. 10 kpc.

d. 50 pc.

e. 100 kpc.

17. Given that dark energy is vacuum energy, and that the densities of dark energy, dark

matter and normal matter in the universe are currently = 6.7 , =ρΛ

10 g /cm ×−30 3

ρ DM

2.4 and = 0.5 , what is the ratio of the density of dark 10 g /cm ×−30 3

ρ Λ 10 g /cm ×−30 3

energy at the time of the cosmic microwave background emission, to the current density

of dark energy?

a. 0.432

b. 2.31

c. 1

d. 2.5

e. 0.5

18. A type Ia supernova was observed in a galaxy with a redshift of 0.03. The supernova was

determined to be 1.3 pc away from Earth. Determine the Hubble time using this10×8

observation.

a. 1.41 years0× 1 10

b. 1.41 seconds0× 1 10

c. 1.33 years0× 1

9

d. 47.1 years

e. 1.33 seconds0× 1 9

19. In a main sequence star, gravitational collapse is counteracted by:

a. Radiation pressure

b. Heat

c. Neutrinos

d. Electron degeneracy pressure

e. Neutron degeneracy pressure

20.If the hydrogen alpha line of a star, normally 656.3 nm, is observed to be 662.5nm, what is the star’s radial velocity relative to the Earth?

a. 2.83*10 6 m/s

b. -2.83*10 6 m/s

c. 0.00945 m/s

d. -0.00945 m/s

e. -2.83*10 3 m/s


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21. Within M-type stars, heat transfer occurs primarily through:

a. radiation

b. conduction

c. convection

d. contraction

e. collapse

22.If a 1.2 solar mass star shows a radial velocity variation with a period of 9.2 days

and amplitude of 32 m/s , estimate the minimum mass of the companion:

a. 7.5*10 26 kg

b. 1.2*10 26 kg

c. 6.9*10 27 kg

d. 5.1*10 15 kg

e. 3.3*10 27 kg

23.Calculate the planetary phase angle (counterclockwise from Earth, a = 1.0 AU)

that a probe may correctly complete a Hohmann transfer orbit to Venus (a = 0.7

AU)

a. 141 degrees

b. 17.5 degrees

c. 121 degrees

d. 241 degrees

e. 343 degrees

24.Calculate the blackbody equilibrium temperature of Mars. Take Mars’s albedo to

be 0.25 and semimajor axis to be 1.5 AU

a. 300 K b. 212 K

c. 161 K

d. 228 K

e. 260 K

25. Calculate the semimajor axis of a satellite orbiting the Earth with a velocity of 8.3

km/s at a distance of 300 km from the Earth’s surface.

a. 154 km

b. 308 km

c. 15800 kmd. 7900 km

e. 3950 km

26.On the night of December 23rd-24th 2015, an occultation of a bright star by the moon

will be visible from Britain to Japan. Given that the moon is in full phase on December

25th, which star does the moon occult?

a. Aldebaran (RA 4h 37m, Dec 16 o 31’)


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b. Pollux (RA 7h 45m, Dec 28 o 2’)

c. Regulus (RA 10h 8m, Dec 11 o 58’)

d. Spica (RA 13h 25m, Dec -11 o 14’)

e. Antares (RA 16h 29, Dec -26 o 26’)

27. A synodic day on Mars is 24 hours and 40 minutes. If one Martian year is 687 earth-days,

which of the following is closest to a sidereal day on Mars?

a. 23h 56m

b. 24h 15m

c. 24h 37m

d. 24h 40m

e. 24h 42m

28. Suppose at the equator, a star passes through the zenith at local noon on the summer

solstice. What is the right ascension and declination of the star?

a. 0h 0 o

b. 0h 90

o

c. 6h 0 o

d. 12h 0 o

e. 12h 90 o

29. 40 light years away, an exoplanet orbits a star of 5 solar masses every 14 years.

Assuming this system has an inclination of 90˚ as viewed from Earth, what is the

projected diameter of the exoplanet’s orbit as viewed from Earth?

a. 0.3”

b. 0.8”

c. 1.6”d. 2.5”

e. 1.2”

30. A planet orbits a star with a projected semimajor axis of 0.24”. What is the necessary

aperture size of a telescope than can resolve this orbit using 1000 nm light?

a. 0.13 m

b. 0.52 m

c. 1.05 m

d. 3.10 m

e. 2.04 m


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USAAAO First Round 2015

Answers1. C

2. D

3. C

4. B

5. C

6. B

7. D

8. A

9. B

10. C

11. B

12. B

13. D

14. C

15. E

16. B

17. C

18. A

19. A

20.B21. C

22.A

23.D

24.B

25.D

26.A

27. C

28.C

29.C30.B

usaao 2015 (second round)

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