Astronomy 111 Midterm #1

Transcription

1 Astronomy 111 Midterm #1 Prof. Douglass 11 October 2018 Name: You may consult only one page of formulas and constants and a calculator while taking this test. You may not consult any books, digital resources, or each other. All of your work must be written on the attached pages, using the reverse sides if necessary. The final answers, and any formulas you use or derive, must be indicated clearly (answers must be circled or boxed). You will have one hour and fifteen minutes to complete the exam. Good luck! First, work on the problems you find the easiest. Come back later to the more difficult or less familiar material. Do not get stuck. The amount of space left for each problem is not necessarily an indication of the amount of writing it takes to solve it. You must show your work to receive full credit. Numerical answers are incomplete without units and should not be written with more significant figures than they deserve. If you need a physical or astronomical constant that is not on your equation sheet, do not give up: estimate its value. If your estimate is reasonable, you will not lose any credit. Remember, you can earn partial credit for being on the right track. Be sure to show enough of your reasoning that we can figure out what you are thinking. Honor Pledge Please write out and sign the Honor Pledge shown below. I affirm that I will not give or receive any unauthorized help on this exam, and that all work will be my own. Signature: 1

2 R = cm R = cm M = g M = g L = erg/s G = dyn cm 2 g 2 T = 5772 K c = cm/s 1 AU = cm k = erg/k 1 pc = 206, 625 AU σ = erg s 1 cm 2 K 4 Question Points Score Total: 105 Page 2

3 1. Short answers. Please write in complete sentences, and feel free to use equations and/or sketches to help explain your thoughts. (a) (5 points) Explain what is incorrect with this statement: According to Kepler s second law, a planet s radius vector sweeps out equal areas in equal times. Since the major and minor axes of an ellipse divide the area enclosed by the ellipse into four equal-area quadrants, it takes a planet the same amount of time to travel each quarter-orbit between the ends of the major and minor axes. Solution: The radius vector originates at one of the foci of the ellipse, not at the intersection of the major and minor axes. Thus, the area of the four quadrants is not the area swept out by the planet s radius vector, and therefore we cannot apply Kepler s second law to it. (b) (5 points) A spinning, uniform-density planet heats up from radioactivity, melts, and differentiates: its higher-density contents sink toward the center without changing the planet s radius. Does this make the planet spin faster or slower? Why? Solution: The moment of inertia of a uniform-density sphere is 2 5 MR2. The moment of inertia of a differentiated planet is 4 15 MR2. The moment of inertia decreases as a planet becomes differentiated. Since angular momentum is conserved and L = Iω, the differentiated planet will spin faster. Page 3

4 (c) (5 points) Give two independent reasons why we think the Moon s maria are younger than the Moon s highlands. Solution: The maria are less cratered than the highlands. Radioisotope dating results in younger values for the maria than the highlands. (d) (5 points) How far apart are the brightest stars in the Northern Cross (Deneb, α Cygni) and the Sourthern Cross (Acrus, α Crucis)? Express your answer in degrees. α Cyg: α = h δ = α Cru: α = h δ = Solution: Remember, we need to convert the R.A. angles from hours to degrees by multiplying by 360 /24 h. Ψ = cos 1 [cos δ 1 cos δ 2 cos(α 1 α 2 ) + sin δ 1 sin δ 2 ] ( ( )) 360 = cos [cos( ) cos( ) cos ( h h ) 24 h Ψ = sin(45.48 ) sin( )] Page 4

5 2. (10 points) A planet with mass M and radius R is in orbit around a star with luminosity L. The planet is heated by both starlight and radioactive decay, with a current decay rate of Λ; it is cooled by blackbody radiation. Derive a formula for the planet s distance from the star, r, as a function of its surface temperature T. Solution: P in = P out L 4πr 2 πr2 + MΛ = σt 4 4πR 2 LR r = 2 16πR 2 σt 4 4MΛ Page 5

6 3. Samarium-neodymium dating. One radioactive isotope of samarium, 147 Sm, decays by α-radiation into a relatively rare, stable isotope of neodymium, 143 Nd, with a half-life of yr (decay constant λ = yr 1 ). Another stable neodymium isotope not involved in a decay chain, 144 Nd, provides a good relative-abundance reference. (a) (10 points) A rock sample is analyzed, and the relative abundances of the included minerals are determined to be: Mineral 147 Sm/ 144 Nd 143 Nd/ 144 Nd A B How old is the rock (in years), and what was the original 143 Nd/ 144 Nd relative abundance? Solution: t = 1 ( ) ( ) λ ln DA D B = N A N B ln yr t = yr ( ) ( ) DA D B D 0 = D A N A = N A N B D 0 = Page 6

7 (b) (5 points) Explain the origin of this rock (terrestrial, lunar, meteoritic,... ). Solution: This rock s age is older than any lunar or terrestrial rock that we have seen, but it is similar in age to meteorites. Therefore, this rock is probably from a meteorite. (c) (5 points) Another radioisotope of samarium, 146 Sm, also decays by α-radiation into yet another relatively rare, stable isotope of neodymium, 142 Nd, with a half-life of yr. Why would analysis of this isotope not be as useful as Rb Sr or 147 Sm 143 Nd for measuring the ages of the oldest rocks in the Solar System? Solution: Its half-life is so much shorter than the ages of the oldest rocks in the Solar System (3 4.5 billion years) that most of the original radioactive nuclides will have decayed by now. There is still plenty of 87 Rb and 147 Sm, though, because their half-lives are much longer and will therefore provide much more accurate abundance measurements. Page 7

8 4. A spherical planet with radius R has density ρ(r) given by ( r ) 2 ρ(r) = ρ 0 (1) R where ρ 0 is a constant. (a) (10 points) What is the mass M of the planet, in terms of ρ 0 and R? Solution: ρ = M V dm = ρ(r) dv = M = 4π ρ 0 R 2 R 0 = 4π ρ 0 R 2 1 M = 4π 5 ρ 0R 3 5 r5 0 r 4 dr R ρ(r)r 2 sin θ dr dθ dφ Page 8

9 (b) (15 points) What is the moment of inertia of the planet, in terms of M and R? Solution: di = (r sin θ) 2 dm = r 2 sin 2 θρ(r)r 2 sin θ dr dθ dφ di = ρ 0 R 2 r 6 sin 3 θ dr dθ dφ I = 8π 3 = 8π 3 ρ 0 R 2 R 0 ρ 0 1 R 2 I = 8π 21 ρ 0R 5 7 r7 0 r 6 dr R We can derive an expression for the density ρ 0 in terms of M and R from our solution to the last part. Therefore, M = 4π 5 ρ 0R 3 ρ 0 = 5 M 4π R 3 I = 8π 5 M 21 4π R 3 R5 I = MR2 = 0.476MR 2 Page 9

10 (c) (5 points) Compare this result to the moments of inertia of the inner planets in our Solar System. Would you expect this sort of planet to be common among other planetary systems? Why or why not? Solution: This moment of inertia is larger than the moments of inertia of the terrestrial planets in our Solar System (all with I 0.33MR 2 ). In fact, it is larger than the moment of inertia of any planet in our Solar System. This moment of inertia describes an interior density somewhere between uniform and a hollow shell the planet is anti-differentiated. This is a very unlikely composition for any planet based on our theories of planet formation. Page 10

11 5. The NASA Dawn satellite was in orbit around 4 Vesta (M = g) from At first, Dawn conducted a mapping survey in a circular orbit about Vesta with a radius r 2 = cm and after a few weeks was moved to another circular orbit with radius r 1 = cm to make a more detailed images of the surface. (a) (10 points) Calculate the semimajor axis length (in cm) and eccentricity of a Hohmann transfer orbit which Dawn could have used to switch between its circular mapping orbits. How long would it take (in hours) to travel along the transfer orbit between that orbit s apoapse and periapse? Solution: The radius at periapse is r p = cm, and the radius at the apoapse is r a = cm. The semimajor axis is the average of the periapse and apoapse: a = r p + r a 2 a = cm = cm cm 2 Knowing the semimajor axis, we can solve for the ellipticity of the orbit: r p = (1 ε)a ε = 1 r p a = cm cm ε = 0.52 The travel time from periapse to apoapse is equal to half of the period of the elliptical Hohmann transfer orbit. 4π2 a 3 t = P 2 = 1 2 = GM π 2 ( cm) 3 ( dyn cm 2 g 2 )( g) t = s = 18.4 hr Page 11

12 (b) (10 points) Dawn s mass at launch was m = g. Suppose the spacecraft included a standard hydrazine engine with thrust F = dyne. Calculate the durations (in seconds) of the two burns necessary to inject Dawn into the transfer orbit and to circularize it in the smaller orbit. For each of these burns, what was the direction of thrust relative to the spacecraft s velocity in orbit? Solution: We can use Newton s second law to solve for the thrust time. F = p t t = p F = m v F The thrust time at the apoapse (to place the space craft in the elliptical Hohmann transfer orbit) is t a = m F (v i v t,p ) ( = m ( GM 2 GM 1 ) ) F r 2 r 2 a = m ( ) GM r1 1 F r 2 a = g ( dyn cm 2 g 2 )( g) dyn cm ( ) cm cm t a = 57.7 s The thrust time at the periapse (to circularize the orbit) is t p = m F (v t, a v f ) ( = m ( 2 GM 1 F r 1 a = m ( ) GM r2 F a 1 r 1 = g dyn t p = 77.8 s ) ) GM r 1 ( dyn cm 2 g 2 )( g) cm ( cm cm 1 The spacecraft needs to be slowed down at each burn, so the thrusters should be pointed in the direction of motion so that the direction of the thrust is opposite the direction of motion. ) Page 12

13 (c) (5 points) Extra credit Dawn does not have an ordinary rocket engine for maneuvering. Instead, it has a xenon ion drive, capable of only F = 10 4 dyne thrust; this engine makes up for its puny output by being capable of producing thrust for a very long time. Describe how this engine could be used to transfer the satellite between its two circular orbits, and the path that the spacecraft takes between the orbits. (Do not try to do the calculation.) Solution: The ion drive should be pointed into the direction of motion of the spacecraft for a long time. This will gradually slow down the spacecraft as it falls towards Vesta along a spiral path. The drive should be shut off when the correct distance and speed are achieved. Page 13