midterm_fall2023_solutions

Name: Physics 341: Midterm Exam Wednesday, October 25, 2023 2:00 - 3:20 pm The exam consists of 25 multiple choice questions. Each question is worth 2 points, for a total of 50 points. Note some problems are conceptual while others require calculations (and will thus take longer to answer). Try to pace yourself accordingly during the exam. There is also 1 bonus question worth 3 points at the end. A sheet of constants and nature and useful quantities is provided at the end of the exam, and you may use a non-programmable calculator. You may consult 1 page of hand-written notes. You may NOT consult any books, neighbors, neighbors’ answers, or smartphones during the exam; the penalty for violating this rule will be a failing grade for the course. You have 80 minutes for this exam. 1

Multiple choice: Be sure to write the answers on the bubble sheet. [2 pts each] 1. You wish to use dimensional analysis to estimate the “event horizon” of a black hole, inside of which nothing can escape the gravity of the black hole (not even light). You can figure out two potential estimates: I. R ∼ GM c 2 = [ M − 1 L 3 T − 2 ][ M ] / [ L 2 /T 2 ] II. R ∼ ℏ Mc = [ ML 2 T − 1 ] / [ M ][ L/T ] Which of the following is true? A. Neither of these is correct, because the dimensions are not length, [ L ] B. (I) is correct because this problem depends on gravity and the point at which not even light can escape the black hole’s gravity. C. (II) is correct because this problem depends on quantum mechanics and the point at which not even light can escape the black hole’s gravity. D. (II) is correct because less massive black holes have larger event horizons. E. These two arrangements are equally valid. Answer: B. 2. Stars form by the gravitational collapse of interstellar gas clouds. Noting that the average cloud has a mass density of ρ = 1 M ⊙ pc − 3 , use dimensional analysis, the estimated time it takes for a star to collapse is: A. 15 × 10 3 yr B. 15 × 10 4 yr C. 15 × 10 5 yr D. 15 × 10 6 yr Answer: D. The quantities and dimensions we have to work with are gravity G [ M − 1 L 3 T − 2 ] density ρ [ M L − 3 ] We are looking for a timescale. Notice that the combination Gρ eliminates both the mass and length and has dimensions [ T − 2 ]. So then we can derive t ∼ 1 Gρ 1 / 2 = [ T ] We just need to convert the density 2

Figure 2: Set up for question 4. D. total energy E. none of the above Answer: E 5. An object is in uniform circular motion at a constant speed. Which of the following could be a plot of its tangential velocity in polar coordinates, v Θ ? Figure 3: Options for question 5. 6. Imagine a planet is orbiting a massive star. If all of the star’s mass suddenly collapses down to form a black hole, what will happen to the planet? A. It will be inexorably sucked into the black hole due to its tremendous gravity B. It will orbit the black hole, but at a smaller semi-major axis than it had before C. It will orbit the black hole, but at a larger semi-major axis than it had before D. It will orbit the black hole at its original semi-major axis as if nothing had happened Answer: D 4

7. For a planet in a bound elliptical orbit around a star, which of the following is true: A. The planet’s kinetic energy is greatest at apocenter. B. The planet’s kinetic energy is greatest at pericenter. C. The planet’s kinetic energy is the same at all points in the orbit. D. The planet’s kinetic energy is zero; it only has potential energy. Answer: B. 8. If you lived on an asteroid with an average distance to the Sun of 3 AU , how long would it take to go once around the Sun? A. 2.6 years B. 3 years C. 5.2 years D. 9 years Answer: C 9. Which of the following statements describe a characteristic of the solar system that is explained by Kepler’s first law? I. Pluto moves faster when it is closer to the Sun than when it is farther from the Sun. II. Venus orbits the Sun faster than Earth orbits the Sun. III. Earth is slightly closer to the Sun on one side of its orbit than on the other side. V. Inner planets orbit the Sun at higher speed than outer planets. VI. The Sun is located slightly off-center from the middle of each planet’s orbit. A. I B. I, II, and IV C. III and VI D. All of them can be described by Kepler’s 1st law. Answer: C. 10. You are using the Doppler shift of water masers to derive the observed velocities of stars at the center of NGC 4258. Figure 4 below is the result of your observations. You recognize the results as a Keplerian rotation curve (a characteristic decline in velocity as a function of radius indicative of the 1-Body problem). What is the mass of the central object that the stars at ∼ 4 pc are orbiting? Assume a circular orbit. A. 2 × 10 6 M ⊙ 5

Assuming a circular orbit ( e = 0), then | ⃗v | = r ω = ℓ r = √ GMa a = GM a 1 / 2 Alternatively, you could calculate the Period using Kepler’s third law and use v = 2 πr/P . At r = 30 pc from a black hole of mass M = 9 × 10 8 M ⊙ , the orbital speed is v = 6 . 67 × 10 − 11m 3 kg − 1 s − 2 ) × (9 × 10 8 × 1 . 99 × 10 30 kg) 30 × 3 . 09 × 10 16 m 1 / 2 = 3 . 6 × 10 5 m / s which is about 0.11% of the speed of light. 13. What is the angular resolution (in arcseconds) of your eye, at a typical optical wave- length of 500 nm ? Assume the pupil of the eye is 5 mm across. A. 0.6 arcmin B. 0.2 arcmin C. 0.8 arcmin D. 0.4 arcmin Answer: D At a typical optical wavelength of λ = 500 nm = 5 × 10 − 5 cm, the resolution of your eye (5 mm = 0.5 cm) is θ ≈ 1 . 22 λ D = 1 . 22 5 × 10 − 5 cm 0 . 5 cm = 1 . 2 × 10 − 4 rad × 180 deg π rad × 3600 ′′ 1 deg = 25 ′′ This is about half an arcminute. Note the conversion from radians. 14. Which rotation curve in the plot below describes Keplarian motion (e.g., planetary orbits in the solar system)? A B 7

A. The left figure A B. The right figure B Answer: B Left: For the merry-go-round, there is solid body rotation, hence v ∝ r , as all of the must must rotate once in a period. Right: For the solar system, essentially all of the mass is sitting at the center in the Sun, so the rotation curve is Keplerian, v ∝ 1 / √ r . 15. A typical quasar luminosity is about 10 12 L ⊙ , where L ⊙ = 3 . 83 × 10 33 erg s − 1 is the luminosity of the Sun. If the energy is released by mass falling into a SMBH, estimate the mass accretion rate in solar masses per year. You can assumed ε = 0 . 1 . A. 0 . 67 M ⊙ yr − 1 B. 0 . 81 M ⊙ yr − 1 C. 1 . 10 M ⊙ yr − 1 D. 1 . 23 M ⊙ yr − 1 Answer: A. The energy emitted as matter falls into a black hole has the form L ≈ ε ˙ Mc 2 , where ˙ M is the mass accretion rate, and ε is a dimensionless parameter representing the efficiency of energy release. If we take ε = 0 . 1, then setting L = 10 12 L ⊙ and solving for ˙ M , we get: ˙ M = L ε c 2 = 10 12 × 3 . 83 × 10 33 ergs s − 1 0 . 1 × (3 × 10 10 cm s − 1 ) 2 = 4 . 3 × 10 25 g s − 1 = 0 . 67 M ⊙ yr − 1 If we assumed a different efficiency ε , it would change the derived accretion rate a little bit, but would not substantially alter our conclusions. 16. Assuming a roughly constant mass accretion rate of matter falling into a black hole of 0.4 M ⊙ yr − 1 , how long would it take to build a mass of 10 9 M ⊙ ? A. 2.5 × 10 9 yr B. 5 × 10 9 yr C. 25 × 10 9 yr D. 50 × 10 9 yr 8

Fig. 3.— Radial velocity observations for HIP 50796 and fitted orbit. The dotted line represents the center-of-mass velocity of the binary. Errors are smaller than the size of the points. Figure 5: Radial velocities of the star HIP 50796, from Torres et al. (2006, AJ, 131, 1022). I. The unseen companion is likely a planet II. The center of mass of the system is moving away from us III. The orbital eccentricity is measurably different than zero A. I only B. II only C. III only D. I and III E. II and III Answer: E 19. Based on the data in Figure 5, at which of these orbital phases is the star the closest to its companion? A. 0.0 B. 0.3 C. 0.5 D. the orbits are circular, so the separation is constant Answer: A 10

star 1 star 2 Figure 6: Radial velocities of the double-lined binary system V1061 Cygni, from Torres et al. (2006, ApJ, 640, 1018). 20. The hydrogen Hα line has a rest wavelength λ rest = 656.28 nm. If we see Hα from both stars in V1061 Cygni at orbital phase = 0.25 (see Figure 6), at what wavelengths would we observe the lines ( λ 1 for star 1, and λ 2 for star 2)? Hint: Think about the possibilities; you won’t need to calculate anything. A. λ 1 = 656.02 nm, λ 2 = 656.06 nm B. λ 1 = 656.54 nm, λ 2 = 656.50 nm C. λ 1 = 656.54 nm, λ 2 = 656.06 nm D. λ 1 = 656.02 nm, λ 2 = 656.50 nm E. impossible to tell without knowing the inclination angle i Answer: D 21. We observe a double-lined spectroscopic binary and measure the radial velocity ampli- tudes, K 1 and K 2 , and the orbital period P but not the orbital inclination i . Which of the following can we determine? I. The mass ratio, m 1 /m 2 II. The total mass, M = m 1 + m 2 III. The individual masses, m 1 and m 2 IV. The reduced mass, µ = m 1 m 2 / ( m 1 + m 2 ) 11

The semimajor axis for the Sun’s motion due to Mercury is then a Sun = µ M ⊙ a = 3 . 3 × 10 23 kg 2 × 10 30 kg (5 . 79 × 10 10 m) = 9 . 55 × 10 3 m The resulting speed is then given by v Sun = 2 πa 1 P = 2 π (9 . 55 × 10 3 m) 7 . 60 × 10 6 s = 0 . 0079 m/s . 24. White dwarf stars have a radius like the Earth, R WD = R ⊕ , and so their transit light curve would look very similar to a planet, but the typical mass of a white dwarf is much higher, M WD = 0 . 6 M ⊙ . Calculate the radial velocity amplitude we would measure for a white dwarf orbiting around a star similar to our sun with an orbital period of 1yr if we observed the system edge-on to its orbital plane. A. 5.0 km/s B. 8.0 km/s C. 13 km/s D. 21 km/s Answer: C For a circular orbit the observed radial velocity of the star satisfies ( m 2 sin i ) 3 ( m 1 + m 2 ) 2 = P 2 πG K 1 = ⇒ K 1 = 2 πG P ( m 1 + m 2 ) 2 1 / 3 m 2 sin i For a white dwarf and sun-like star system, m 1 + m 2 = 1 . 6 M ⊙ and the same orbital period: 1 K 1 = 2 π × 6 . 67 × 10 − 8 cm 3 g − 1 s − 2 3 . 16 × 10 7 s × (1 . 6 × 1 . 99 × 10 33 g) 2 1 / 3 × 0 . 6 × 1 . 99 × 10 33 g × sin 90 ◦ = 1 . 3 × 10 6 cm s − 1 = 13 km s − 1 So the reflex motion of the star is much larger if the companion is a white dwarf rather than the Earth and we could easily discern whether the detected transit is due to a planet or a white dwarf. 1 Note that in the white dwarf case, because the total mass of the system is larger, the semimajor axis will also increase, for the same orbital period. 13

25. A hypothetical star has a transiting exoplanet “a” orbiting with a a = 1 AU and radius R a = 1 R ⊕ . It causes a 0.1% decrease in the brightness of the star during its tran- sit. If this star had another planet “b” orbiting at a distance a b = 2 AU with radius R b = 2 R ⊕ , how much would the star’s brightness decrease during a transit of planet b? A. 0.05% B. 0.1 % C. 0.2% D. 0.4% Answer: D. 26. Bonus Question: What is the theme of my class playlist? [3 pts] A. Soft rock B. Bad elevator music C. 1980’s disco (my personal fav) D. Astronomy! Answer: D. 14