Astronomy 111 Practice Midterm #1

Size: px

Start display at page:

Download "Astronomy 111 Practice Midterm #1"

Alannah Watkins
5 years ago
Views:

1 Astronomy 111 Practice Midterm #1 Prof. Douglass Fall 218 Name: If this were a real exam, you would be reminded of the Exam rules here: 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! Your results will improve if you take this practice test under realistic test-like conditions: in one sitting, with your already-prepared cheat sheet at hand, and with the will to resist peeking at the solutions until you are finished. Also, as usual: 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. Numerical answers are incomplete without units and should not be written with more significant figures than they deserve. 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 = 26, 625 AU σ = erg s 1 cm 2 K 4 Page 2

3 1. Short answers. Please write in complete sentences and show equations whenever it is useful. (a) (5 points) Give one reason why we think that the main asteroid belt contains both differentiated and undifferentiated bodies. Solution: There are some main-belt asteroids whose average densities are larger than the density of silicate rock, but whose surfaces are covered in silicate rock. There are other mainbelt asteroids which have densities much lower than their silicate-rock crust. Meteorites come from the main belt, and they are made of iron, stony-iron, and achondrites with densities and element abundances indicative of the fragments of heavily differentiated bodies. However, they also include objects like ordinary and carbonaceous chondrites, which have never completely melted. (b) (5 points) What kinds of minerals are most common in the rocks of the Moon s highlands? Of the Moon s maria? Solution: Highlands: light-colored rocks like anorthosite, rich in felsic minerals such as plagioclase Maria: mostly basalt, made of the mafic minerals, pyroxenes, and olivines (c) (5 points) Give two independent reasons why we think Mercury, Venus, and Earth are differentiated. Solution: Their moments of inertia are smaller than that of a uniform-density sphere (implying that the mass is more concentrated in the center). The bulk density of these planets is much larger than the densities of the surface rock. Page 3

4 (d) (5 points) The bulk density of the asteroid 16 Psyche is 6.7 g cm 3. Describe its composition and structure, and give its classification. Solution: The density is much higher than that of silicate rock, so it must be solid and made mostly of iron. This would give it an M-type asteroid classification. (e) (5 points) What is the sidereal time at midnight (standard time) tonight? Give your answer in hours to an accuracy of two significant figures. Solution: ST was approximately h at midnight on Sept. 22, the autumnal equinox. Since 17 days have passed since then, the sidereal clock has advanced by ST = h = 1.12 h (1) (f) (5 points) There are essentially no main-belt asteroids with an orbital period of 4 years, though there are plenty with larger and smaller periods. Explain. Solution: Those that would have an orbital period equal to 4 years would find themselves in a mean-motion orbital resonance of about 3:1 with Jupiter (in one of the Kirkwood Gaps). Any asteroid placed in the main belt with this period would be quickly moved to some other orbit as a result of perturbations from Jupiter s orbit. Page 4

5 2. A main-belt asteroid with mass M A = 1 18 g and radius R A = cm initially lies in a circular orbit with radius r = 3 AU. It suffers a collision that suddenly and drastically changes its orbit; afterwards, its elliptical orbit is completely enclosed by its old one, and its perihelion distance is 1 AU. (a) (1 points) What is the semimajor axis length (in AU) and the eccentricity of the new orbit? Solution: Its perihelion is 1 AU and its aphelion is 3 AU, so the semimajor axis is 3 AU + 1 AU a = 2 a = 2 AU We can solve for the orbit s eccentricity with the equation for the perihelion (the same could be done with the aphelion). f p a = 1 ε ε = 1 f p a = 1 1 AU 2 AU ε =.5 (b) (5 points) Lying in wait for the asteroid at r = 1 AU is the Earth. Suppose Earth were scheduled to be at the asteroid s perihelion at the same time as the asteroid. How fast will the asteroid be going relative to Earth when it reaches us? Will it hit Earth from behind, or will the Earth overtake the asteroid? Solution: The speed of the asteroid at perihelion is ( 2 v A = GM 1 ) f p a ( = 8 dyn cm 2 ( g 2 )( g) cm/au 1 AU 1 ) 2 AU v A = cm/s The speed of the Earth as it moves around the Sun is GM ( v = = 8 dyn cm 2 g 2 )( g) a (1 AU)( cm/au) v = cm/s Therefore, the speed of the asteroid relative to Earth is v A v = cm/s (2) The direction is the same as for both bodies, so the asteroid would hit the Earth from behind. Page 5

6 (c) (5 points) Suppose that we observe the interaction that changed the asteroid s orbit and therefore know that the asteroid is headed our way. How much warning do we have (in years)? Solution: The asteroid would be traveling from aphelion to perihelion, which takes exactly half of its period. t = P 2 = 1 4π 2 a 2 GM 3 (2 AU = π 13 cm/au) 3 ( dyn cm 2 g 2 )( g) t = s = 1.42 yr (d) (5 points) Estimate the level of danger that this poses. How much energy would be dissipated in the impact? Express your answer in megatons: 1 megaton = erg. (The total yield of the world s nuclear weapons is about 2, megatons, down from the Reagon-ear peak of 4, megatons.) Solution: The Earth s speed would hardly change as a result of this impact, since the asteroid is so much less massive than Earth. The dissipated energy is then just equal to the asteroid s kinetic energy (from the Earth s perspective) immediately before impact. K = 1 2 M A(v A v ) 2 = 1 2 (118 g)( cm/s) 2 K = erg = megatons Page 6

7 3. An achondrite meteorite is dissected into minerals, and its minerals are analyzed for Rb-Sr radioisotope dating. The results are as follows: Mineral 87 Rb/ 86 Sr 87 Sr/ 86 Sr Plagioclase Silica (a) (1 points) How old is the meteorite, and what was the original 87 Sr/ 86 Sr relative abundance? Solution: The decay rate for 87 Rb is λ = yr 1. Labeling the silica as mineral A and the plagioclase as mineral B, and remembering that N refers to the ratio of radioactive nuclei ( 87 Rb) to reference nuclei ( 86 Sr), and that D refers to the ratio of daughter nuclei ( 87 Sr) to reference nuclei ( 86 Sr), we can solve for the age of the meteorite. t = 1 ( ) λ ln DA D B + 1 N A N B = ln yr 1 t = yr ( ) The original ratio of daughter to reference nuclei is equal to the difference between the current ratio of daughter nuclei and the number of radioactive nuclei which have decayed. ( ) DA D B D = D A N A N A N B ( ) =.7147 (.17512) D =.6997 Page 7

8 (b) (5 points) Compare the age of this meteorite to other Solar System bodies we have discussed as well as to the age of the Solar System itself. How long after the formation of the Solar System did this achondrite solidify? Solution: We estimate the age of the Solar System to be Gyr. This meteorite solidified about yr after the Solar System formed; it is much older than our Moon. (c) (1 points) This achondrite happens to be one of the 63% of all achondrites which are fragments of 4 Vesta. What does the meteorite s type, composition, and age suggest about the interior structure of Vesta? Solution: Achondrites are poor in iron and other siderophile elements and are rich in elements common to silicate minerals. As a result, they are thought to come from the outer parts of differentiated bodies. Since the meteorite solidified so early in the Solar System s lifetime, it is likely that Vesta formed early in the Solar System s history, was molten about 3 Myr after the Solar System formed, and differentiated while molten. Page 8

9 4. A spherical planet with radius R has density ρ(r) given by ( r ) s ] ρ(r) = ρ [1, R s > (3) where ρ is a constant. The density decreases from a value of ρ at the center to zero at the surface. (a) (1 points) Show that ρ is given by ρ = 3M s + 3 4πR 3 s (4) Solution: ρ(r) = dm dv dm = ρ(r) dv = ρ(r)r 2 sin θ dr dθ dφ 2π M = ρ dφ ρ = π sin θ dθ R ( ( r ) s ) r 2 1 dr R R ) = ρ (2π) ( cos θ) π (r 2 r2+s R s dr ( ) 1 1 R = 2πρ (1 + 1) 3 r3 (s + 3)R s rs+3 ( 1 = 4πρ 3 R3 1 ) s + 3 R3 = 4πρ R 3 s s + 3 3M s + 3 4πR 3 s Page 9

10 (b) (1 points) What is the moment of inertia of the sphere, in terms of M and R? Solution: di = r 2 dm = r 2 sin 2 θρ(r) dv 2π I = ρ dφ = 8π 3 ρ R π sin 3 θ dθ R (r 4 1R ) s rs+4 dr = 8π ( ) 1 3 ρ 1 R 5 r5 (s + 5)R s rs+5 = 8π ( 1 3 ρ 5 R5 1 ) s + 5 R5 = 8π 15 ρ R 5 s = 8π 15 s + 5 3M s + 3 4πR 3 s I = 2 5 MR2 s + 3 s + 5 R 5 s s + 5 ( ( r ) s ) r 4 1 dr R Page 1

Astronomy 111 Practice Final Exam

Astronomy 111 Practice Final Exam Prof. Douglass Fall 2018 Name: If this were a real exam, you would be reminded of the Exam rules here: You may consult only one page of formulas and constants and a calculator