SECONDROUND
2015
USA ASTRONOMY & ASTROPHYSICS OLYMPIAD
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USAAAO 2015 Second Round Solutions Problem 1
T = 10000 K, m = 5, d = 150 pc m − M = = 5log(d/ 5log(d/10) 10) M = m − 5 log( log(d/ d/10) 10) = 5 − 5 log log(15 (15)) = −0.88 We compare this with the absolute magnitude of the sun, 4.83. L (4.83 0.88)/5 = 192L 192Lsolar Lsolar = 100 From the Stefan-Boltzmann Law, L Law, L = 4πR 2 σT 4 Finding the ratio with the sun, we get: −−
L Lsolar R Rsolar
= =
R Rsolar L Lsolar
2
4
T
T solar solar
1/2
2 T solar solar T
= 4.6Rsolar
Problem 2
T = 3 days, K = 50 m/s, M = 1M 1M solar solar a 2 T = M 3
3 2/3 365
= 0. 0 .041 041AU AU for for the planet’s orbit. For the star’s orbit: T = 2πr v r = T2πv = 1.38 ∗ 10 5 AU Relating the two, we have: m r = m = m p r p , so we have m p = rrp m Plugging in, we find m p = 3.36 ∗ 10 4 M solar solar −
∗
∗
∗
∗
−
Problem 3 4 M Solar rotation rate 24.5 days, M J J = 9.54 ∗ 10 solar solar , a = 5.2 AU, solar radius 695,000 km. L = I = I ω 2π 10 ∗ Lsolar = 25 M R2 ∗ 24 .5 = 5 10 For Jupiter, L Jupiter, L = = mrv mrv r = 5.2 ∗ 149 149..6 ∗ 106 = 7.77 ∗ 108 km v = 5.2 2/πr 365 = 1.13 ∗ 106 km/day. L = 9.54 ∗ 10 4 ∗ 7. 7.77 ∗ 108 ∗ 1. 1.13 ∗ 06 = 8. 8 .38 ∗ 1011 So Jupiter has the greater angular momentum. −
2
3
∗
−
Problem 4
m = 10, T = 6000 K at the main sequence turnoff. Oldest main sequence stars are 6000 K, which is approximately sun-like. We therefore assume the absolute magnitude of the stars at the turnoff point is 4.83. = 5log(d/ 5log(d/10) 10) m − M = m M )/5 ( d = 10 ∗ 10 = 108 pc 108 pc The stars at the turnoff point are sunlike, so we expect them to have a lifetime lifetime of 10 Gyr. Since these these are the oldest oldest main sequence stars in the cluster, the cluster has an age of approximately 10 Gyr. −
1
Problem 5
Question not graded Problem 6
m = 8, p = 0.003”, and T = 6000 K. d = 1/p = /p = 333 pc 333 pc m − M = = 5log(d/ 5log(d/10) 10) M = m − 5 log( log(d/ d/10) 10) = 0. 0.385 L M solar ( solar M )/5 = 100 = 59. 59 .9Lsolar Lsolar The star’s temperature is approximately sunlike, suggesting class G, but it is significantly more luminous, suggesting a giant. G0III would be a reasonable possible possible spectral spectral type. −
Problem 7
From Wein’s Law, λ Law, λ = T b . From the definition of Redshift, z Redshift, z = λ λ λ , where λ0 is the emitted wavelength. Solving for λ for λ 0 , we get λ 0 = λ/ = λ/((z + 1) Combining this with Wein’s Law, we get: (z +1) T = blambda Using Wein’s Law and the given temperature of 2.73 K, we find that the recieved recieved waveleng wavelength th is 1.06 mm. Plugging Plugging this into the above expression, expression, we obtain at temperature of 30.03 K for z = 10. −
0
0
∗
Problem 8
At blackbody equilibrium, power in is equal to power out, so we have: 2 4 P out out = 4πR p σT p πR 2 σT 4
L P in α) ∗ 4πD = πR p2 (1 − α) α) 4 4πD in = A ∗ (1 − α) P in in = P out out πR σT 2 4πR p σT p4 = πR π R p2 (1 − α) α) 4 4πD ∗
2
2
∗
2
4
∗
∗
2
R2
T p4 = (1 ( 1 − α) α)T 4 4D
∗
∗
T p = (1 − α) α)1/4 T
∗
2
R 2D ∗
Problem 9
7.2 micron pixels, f/10, D = 0.256 m. f/10, so focal length is 2.56 m. pixelsize Angular resolution is given by focallength ∗ 206265 Plugging in, we get a resolution of 0.58 arcseconds/pixel Problem 10
A 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 is thus 5.3 AU. Using the vis-viva equation, v2 − µr = − µ , this corresponds to a velocity of 2a 40,080 m/s at the Earth’s orbital distance. 2
2
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 orbital velocity around the Sun. When in the parking orbit, the probe has a maximum velocity relative to GM s e the Sun of GM 37540m/s rorb + rearth = 37540m/s The difference between the spacecraft’s current maximum velocity relative to the Sun and the orbital velocity velocity of the transfer ellipse ellipse is equal to the required required ∆v , since we’re neglecting further attraction from the Earth. The necessary ∆ v is thus 2550 m/s, and the burn must occur on the night side of the Earth in order to increase increase the orbital orbital radius radius and match Saturn’s orbit (since the parking parking orbit is prograde). We use the vis-viva equation again to calculate the probe’s velocity at Saturn’s orbital distance, distance, and compute Saturn’s orbital velocity velocity using the same techni technique que we used used for Earth. Earth. We then then subtra subtract ct Saturn Saturn’s ’s veloci velocity ty from the probe’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 difference between between this value value and the value above is the ∆v required to make Saturn orbit, orbit, coming out to 24900 m/s. The burn must occur on the night side of Saturn in order to enter a prograde orbit.
Problem 11
M = 0.54M 0.54M s , P = 6 years, perihelion distance 0.537 AU, parallax 0.05”. From the parallax equation, d equation, d = = r r/θ /θ,, where d is the distance, r is the baseline, and θ is the parallax parallax angle. We therefore need to find the baseline, baseline, XY, which is 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 by l = a = a(1 (1 − e2 ). Plugging in, we get l = 2.45 AU, yielding a distance d = 49.0 pc Problem 12
Simply Simply dividin dividingg the length length of the year year by 6 is incorrec incorrect. t. Becaus Becausee of the inclination of the Earth, the Sun also varies in declination (from -23.5 to +23.5 degrees), but maintains (for this problem) a constant angular velocity. Spherical trigonometry is therefore required to determine the actual angle that the Sun trave traverse rsess goi going ng from 0 to 4 hours hours (62.1 degrees) degrees).. Since Since the Sun has consta constant nt angular 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 likely the easiest easiest third p oint. oint. Now that we have a spherical triangle, triangle, we can use the spherical law of sines or cosines to find the separation angle (18.59 degrees).
3
The position angle is measured measured east of north. north. For our chosen chosen triangle, triangle, this corres correspond pondss to the angle angle with with Betelg Betelgeus eusee as its verte vertex. x. Aga Again in applying applying the spheri spherical cal trigon trigonome ometri tricc relati relations onship hips, s, we find the positio position n angle angle to be 33. 33.12 12 degrees. To cover both stars, the picture must cover 18.6 degrees of the sky, which corresponds corresponds to 66960 arcseconds. arcseconds. Plate scale, scale, in arcsecond arcseconds/mm, s/mm, is given by 206265/foca 20626 5/focall length. Solving Solving for the focal length, we get a value of 3.08*film size, which is a focal length of 108 mm on 35 mm film or 216 mm on 70 mm. Long Problem
a. Drawin Drawingg the transit transit light light curve curve,, we can see that that betwe between en the first first and second contacts, the relative motion of the stars is equal to twice the radius of the transiting transiting star. Similarly Similarly,, the time b etween etween second and third contacts is proportional to twice the primary star’s radius. Dividing, we get rr = tt tt = 13..67 13 We can use Wein’s Law and the provided blackbody peaks to calculate the temperature temperature of each star. We can now use the Stefan-Boltzman Stefan-Boltzmann n Law to find the ratio of the luminosities: 2 4 L R T = = 934 L R T b. We use the provided provided chart chart to determine determine the orbital period of the binary system (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 axes according to r = vT , for semimajor axes of 0.0450 and 0.0771 AU for stars 1 2π and 2 respectively. Applying Kepler’s Third Law for the entire system, we get a total mass of 1.90 solar masses. Since m Since m 1 r1 = m = m 2 r2 , 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 Using the veloci velocity ty of each each star, star, we can find the relativ relativee veloci velocity ty of the two stars (just sum). This can then be used to solve the equations from part a directly, 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, 0.0027 for star 2). d. Apparent magnitude can be found by finding the total flux of the system and using the Sun’s apparent magnitude and the solar flux as a standard candle, applying ∆M ∆M = 2.512 log F F . Similarly, using the luminosity of the system with the Sun’s absolute magnitude as a standard candle can provide an absolute lute mag magnit nitude ude for the system. system. No Now w that that we have have an apparen apparentt and absolute absolute magnitude, we apply the distance modulus to get d = 5.94 pc. e. Applying Applying the small angle formula, formula, we get a maximum maximum angular separation separation ∗ of 0.021”. The best possible resolution of the telescope is given by θ = 1.22 λ D 206265 = 0. 0.017”, so the stars are distinguishable. The smallest visible size can be found by applying the small angle formula with the limiting resolution, and is 0.10 AU. f. Examining Examining the given plot, there appears to be a longer period variation variation 1 2
1
1
2
2
1 2
4
1
2− 1
2
3− 2
with a period of approximat approximately ely 214 days. days. Based on the change change in wavele wavelength, ngth, the variation variation has an amplitude of approximate approximately ly 83.2 km/s. Using this velocity velocity and the period of oscillation, we can determine the semimajor axis of the binary stars’ orbit (1.62 AU) We now apply the fact that m1 a1 = m2 a2 and Kepler’s third law for the binary-unkn binary-unknown own system, system, solving solving for a2 and m2 . We find that that the unknow unknown n object has a mass of 16.2 solar masses and a semimajor axis of 0.19 AU. Given the high mass and lack of a visible counterpart, the object is likely a stellar mass black hole.
5
USAAAO First Round 2015 This round consists of 30 multiple-choice 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 the Second Round. 1. At arms length, length, the width of a fist typically subtends how many degrees of arc? a. 1o b. 5o c. 10o d. 15o e. 20o 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 New b. First Quarter c. Waxing Waxing Gibbous Gibbous d. Full Full e. Waning Waning Cresce Crescent nt 3. Mars orbits the sun once every 687 days. Suppose Mars is currently in in the constellation Virgo. What constellation will it most likely be in a year from now? a. Virg Virgo o b. Scorpius c. Aqua Aquari rius us d. Taur Taurus us e. Canc Cancer er 4. To calculate the field field of view of a telescope, you measure the time time it takes Capella (RA:5.27h, dec:45.98o) 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 11.6’’ b. 26.5’ c. 37.5 37.5’’ d. 52.5 52.5’’ e. 66.8 5. A telescope with focal length of 20 mm and aperture aperture of 10 mm mm is connected to 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. 1o
b. 5o c. 10o d. 15o e. 20o 6. What is its the the resolution in arcseconds per pixel? a. 10”/ 10”/pi pixe xell b. 40”/pixel c. 120”/ 120”/pi pixe xell d. 1200”/ 1200”/pix pixel el e. 3600’/ 3600’/pix pixel el 7. Comet 67P/Churyumov–Gerasimenko has an orbital period around the Sun of 6.44 years. What is its semimajor axis, in AU? a. 41.4 41.47 7 b. 16.34 c. 6.44 6.44 d. 3.46 3.46 e. 1.86 1.86 8. Which of the following techniques most directly constrains the mass of an exoplanet? a. Radial Radial Veloci Velocity ty b. Transit Timing c. Microl Microlens ensing ing d. Direct Direct Imaging Imaging e. Proper Proper Motion Motion 9. Which two properties of galaxies does the Tully-Fisher relation utilize a correlation between? a. Luminosity Luminosity and velocity velocity dispers dispersion ion b. Luminosity and rotational velocity v elocity c. Radius and metallicity metallicity d. Luminosity Luminosity and metallicity metallicity e. Mass and surface surface brightness 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
11. What is, approximately, the peak wavelength of electromagnetic radiation emitted by a star at a temperature of 5,000 K? a. 580 Angstro Angstroms ms b. 5,800 Angstroms c. 4,600 4,600 Angstr Angstroms oms d. 2,900 Angstroms Angstroms e. 58,000 58,000 Angstr Angstroms oms 12. Stars A and B are observed over a period of 1 year. Both stars appear to move with with respect to the background stars from the position indicated on the left in the diagram below, to the position indicated on the right, and then back to the position on o n the left over ov er the full year. y ear. Which star is further from the Earth? a. Star Star A b. Star B c. Both stars stars are are the same same distance distance from from the Earth d. Not enough informatio information n given given 13. Suppose that you measure the parallax angle for a particular star to be 0.25 arcsecond. The distance to this star is a. 2 pc b. 0.5 ly c. 2 ly ly d. 4 pc e. 0.5 0.5 pc 14. On the main sequence, stars obtain their energy a. from chemical chemical reactions. reactions. b. from gravitational contraction. c. by converting converting hydrogen hydrogen to to helium. helium. d. by converting converting helium helium to carbon, carbon, nitrogen, nitrogen, and and oxygen. e. from nuclear nuclear fission. fission. 15. Star A has a radius that is 2 times larger than the radius of star B, and a surface temperature that is 2 times smaller than the surface temperature of star B. Therefore, star A is a. 4 times times more more luminou luminouss than than star star B. b. 16 times less luminous than star B. c. 16 times times more more luminous luminous than star B. d. as luminous luminous as star B. e. 4 times times less less luminous luminous than than star star B.
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. pc. b. 100 pc. c. 10 kpc. kpc. d. 50 pc. pc. e. 100 100 kpc. 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 ρ Λ = 6.7 × 10 2.4 × 10
−30
g /cm
3
and ρ Λ = 0.5 × 10
−30
g /cm
3
−30
g /cm
3
, ρ DM =
, what is the ratio of the density of dark
energy at the time of the cosmic microwave background emission, to the current density of dark energy? a. 0.43 0.432 2 b. 2.31 c. 1 d. 2.5 2.5 e. 0.5 18. A type Ia supernova was observed in a galaxy with a redshift of 0.03. The supernova was 8
determined to be 1.3 × 10 pc away from Earth. Determine the Hubble time using this observation. a. 1.41 × 10 b. 1.41 × 10
10 10
years seconds
9
c. 1.33 × 10 years d. 47.1 47.1 years years 9
e. 1.33 × 10 seconds 19. In a main sequence star, gravitational collapse collapse is counteracted counteracted by:
a. Radi Radiat atio ion n pres pressu sure re b. Heat c. Neutrinos d. Electron Electron degenerac degeneracy y pressure pressure e. Neutron de degeneracy pr pressure 20.If the hydrogen alpha line of a star, normally 656.3 nm, is observed to be 662.5 nm, what is the star’s radial velocity relative to the Earth? a. 2.83*106 m/s b. -2.83*106 m/s c. 0.00 .00945 m/s d. -0.0 -0.009 0945 45 m/s m/s 3 e. -2.83*10 m/s
21. Within M-type stars, heat heat transfer occurs primarily primarily through: a. radiation b. conduction c. conve onvect ctio ion n d. cont contra ract ctio ion n e. collapse 22.If 22. If a 1.2 solar solar mass star star shows a radial velocity velocity variation with a period of 9.2 9.2 days and amplitude of 32 m/s , estimate the minimum mass of the companion: a. 7.5*1026 kg b. 1.2*1026 kg c. 6.9*1027 kg d. 5.1*1015 kg e. 3.3*1027 kg 23. Calculate the planetary phase angle (counterclockwise (counterclockwise from Earth, Earth, a = 1.0 AU) that a probe may correctly complete a Hohmann transfer orbit to Venus (a = 0.7 AU) a. 141 141 degr degreees b. 17.5 degrees c. 121 de degrees d. 241 degr degree eess e. 343 degr egrees ees 24.Calculate 24. Calculate the blackbody blackbody equilibrium equilibrium temperature temperature of Mars. Take Mars’s albedo albedo to be 0.25 and semimajor 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 semimajor axis of a satellite satellite orbiting the Earth 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 km d. 7900 km 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 Aldebaran (RA 4h 37m, 37m, Dec 16o 31’)
b. Pollux (RA 7h 45m, Dec 28o 2’) c. Regulus Regulus (RA (RA 10h 8m, Dec 11o 58’) d. Spica (RA 13h 25m, Dec -11o 14’) e. Antares Antares (RA 16h 29, Dec -26o 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 23h 56m b. 24h 15m c. 24h 24h 37m d. 24h 24h 40m e. 24h 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 0o b. 0h 90o c. 6h 0o d. 12h 12h 0o e. 12h 12h 90o 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” 0.3” b. 0.8” c. 1.6” 1.6” d. 2.5” 2.5” e. 1.2” 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 0.13 m b. 0.52 m c. 1.05 1.05 m d. 3.10 3.10 m e. 2.04 2.04 m
USAAAO First Round 2015 Answers 1. C 2. D 3. C 4. B 5. C 6. B 7. D 8. A 9. B 10. 10. C 11. B 12. 12. B 13. D 14. 14. C 15. 15. E 16. 16. B 17. C 18. 18. A 19. A 20.B 21. 21. C 22.A 23.D 24.B 25.D 26.A 27. 27. C 28.C 29.C 30.B
