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Astronomy 111 Recitation #10 Astronomy 111 Recitation #10 10-11 November 2011 Formulas to remember Tidal acceleration The difference in tidal acceleration across a satellite with dimensions in the radial and transverse directions ∆r and ∆ , lying a distance r away from a parent body with mass M: 2GM M grtr = ∆ r3 GM ∆ gtφ =−∆ r3 ∆r if it has no angular momentum, and r 3GM grtr = ∆ r3 if it is in orbit. Roche limit The Roche limit is the smallest radius at which a satellite (density ρ ) can orbit a parent body (density ρplanet , radius Rplanet ) without being torn apart: 13 ρplanet aRRoche≅ 2.46 planet ρ Hill radius The Hill radius is the range of orbital radii over which a satellite’s (mass m, in orbit with semimajor axis length a around planet with mass M) tidal forces dominate the motions of smaller bodies; thus approximately the half-width of the gap it would induce in a ring: 13 m raH = . 3M Perturbations Transverse acceleration (in orbital plane, perpendicular to radius vector): 2011 University of Rochester 1 All rights reserved Astronomy 111, Fall 2011 Increases a, small change in ε, slows down orbit At periapse: increases a and ε. At apoapse: Increases a, small increases a, decreases ε. change in ε, speeds up orbit Radial acceleration (in orbital plane, parallel to radius vector): Applied at apoapse: no Applied here, a and ε change in a and ε; increase. satellite slows down in orbit. Applied at periapse: no change in a or ε; satellite speeds up in orbit. Applied here, a and ε decrease. Normal acceleration (perpendicular to orbital plane): Applied at apoapse, Applied here, decreases orbital precesses orbital plane inclination counterclockwise. Applied at periapse, Applied here, increases orbital precesses orbital plane inclination clockwise. Torque and angular momentum Lrp=×=ω I N = rmvsinθnˆ F ddLω θ N=×= rF =I dt dt r =rFsinθnnˆˆ = r⊥ F 2011 University of Rochester 2 All rights reserved Astronomy 111, Fall 2011 N is perpendicular to the plane of r and F, in the direction given by the right-hand rule. Trigonometry reminder γ 2 22 a Law of cosines: a=+− b c2 bc cosα b sinαβγ sin sin β Law of sines: = = abc α c Found on the Unit Circle: sin(πθ−=) sin θ cos( πθ−=−) cos θ Workshop problems (do after discussing Homeworks #7 and #8): Warning! The workshop problems you will do in groups in Recitation are a crucial part of the process of building up your command of the concepts important in AST 111 and subsequent courses. Do not, therefore, do your work on scratch paper and discard it. Better for each of you to keep your own account of each problem, in some sort of bound notebook. 1. If Earth had an icy ring, where would it be? 2. Consider the properties of Io and Europa, and without doing any calculations, tell me whether their tidal heating comes predominantly from Jupiter, or from each other, and why. 3. Now prove it. a. Calculate the difference between the maximum and minimum radial tidal accelerations by Io and Europa on each other. b. Calculate the difference between the maximum and minimum radial tidal accelerations exerted by Jupiter on Io and Europa. c. Revisit problem 1. 4. “Tidal” torque on a dumbbell. (And a little instruction in how physicists make approximations.) In this problem you will retrace some of the steps in the tidal-torque calculations from lecture this week, in a better-defined system in which the “tidal bulge” is a built-in property of the planet. Note that it is not intended to take long. Refer freely to your lecture notes, and consider dividing the parts of the problems among your group. Each part will break into at least two sub-parts, for the near and far distances. There will be a large number of calculations, but all the calculations will be simple. A moon with mass m executes a circular orbit with radius r about a planet that consists, oddly, of two spheres with mass M 2 held with their centers a distance 2br apart by a rigid, massless bar. The moon revolves, and the dumbbell rotates about its center of mass, currently with the same angular frequency, but the near side of the dumbbell is currently ahead of the moon by an angle φ, as shown below. 2011 University of Rochester 3 All rights reserved Astronomy 111, Fall 2011 2b M/2 φ M/2 r 2 2 a. What are the squares of the distances, rn and rf , between the moon and the near and far weight of the dumbbell? b. We are told that the moon is very far away from the dumbbell compared to the distance between the weights (that is, br 1 ). Use the approximation n (1+x) ≅+ 1 nx if x 1 , 2 2 and show that the reciprocals of rn and rf are given by 11 2bb11 2 ≅+1 cosφφ and ≅−1 cos . 2222 rrnfrrrr From here on, use this same first-order approximation: that is, neglect terms proportional to 2 (br) or higher powers of br, in sums with terms proportional just to br, or to terms independent of br. c. What is the magnitude of the gravitational force that the moon exerts on each weight? d. What is the orbital angular frequency? (No, that’s not a trivial question. Show that, to first order in br, the total force on the satellite has magnitude GMm r2 and is directed toward the dumbbell’s center of mass, and thus that the angular frequency of its orbit is ω = GM r3 . Hint: consider using a Cartesian coordinate system with origin at the dumbbell’s center and one axis through the satellite.) e. What is the torque the moon exerts on the dumbbell (magnitude and direction)? f. What, therefore, is the total torque the dumbbell exerts on the moon (magnitude and direction)? g. What is the magnitude and direction of the moon’s orbital angular momentum? The dumbbell’s rotational angular momentum? 2011 University of Rochester 4 All rights reserved Astronomy 111, Fall 2011 h. Does the torque the dumbbell exerts on the moon increase or decrease the moon’s orbital angular momentum? How will its orbit change as a result? Learn your way around the sky, lesson 10. (An exclusive feature of AST 111 recitations.) Use the lab’s celestial globes, TheSky running on the lab computers, and any other resources you would like to use, to answer these questions about the celestial sphere, the constellations, and the orbits of the planets. 5. Summer is gone and winter almost upon us, so as the Summer Triangle (Vega, Deneb and Altair; see Recitation 2) sets early in the evening, the Winter Triangle rises. The Winter Triangle is composed of three of the brightest stars in the winter sky – Betelgeuse (α Orionis), Sirius (α Canis Majoris), and Procyon (α Canis Minoris) – and is a useful reference point for finding other constellations. a. Calculate the dimensions of the Winter Triangle – that is, the angular width of each side. b. Suppose I know where to look, to find the Winter Triangle. Describe to me how I can find the constellations Orion and Gemini. c. The constellation Monoceros (the Unicorn) is notable for the star-formation regions it contains. Neither these regions, nor any of the stars in the constellation, are very bright to the naked eye: none of its stars are brighter than fourth magnitude. Nevertheless, few constellations are so easy to find. Tell me how to find Monoceros. 2011 University of Rochester 5 All rights reserved Astronomy 111, Fall 2011 Problem solutions −3 −3 1. The Roche limit for ice (ρ = 0.94 gm cm ) in orbit around Earth (ρ = 5.5 gm cm ) is 4.43R⊕ , so the outer edge would be 3.43R⊕ = 21900 km above the surface. The inner edge would be limited by the atmosphere, since wind resistance would cause the particles to fall, eventually. The top of the atmosphere, for these purposes, can be taken to be about 400 km. (Two ways to tell: this is the elevation of the orbit of the Hubble Space Telescope, which NASA did not want to be subject to air drag. And the Space Shuttle orbit, 280 km up, is not quite high enough; satellites in this orbit, prominently the International Space Station, need occasional boosting to keep from falling in.) So: between about 400 and 21900 km elevation, right above the equator. 2. Io and Europa are not terribly different in size, mass and composition, so if their tidal heating were dominated by each other, their internal states would be similar. They’re not; Io is much hotter inside than Europa, as is evident in its extreme volcanic activity. In fact, any small difference with this cause would be in the wrong direction: since Io is more massive than Europa, one would expect Europa to be subject to the greater tidal heating. The other important difference between Io and Europa is distance from Jupiter: Io is a good deal closer to the planet. Since tidal forces are much larger at smaller distances, this makes it more likely that Jupiter’s tides heat Io and Europa. 3. a. Io and Europa are in very nearly circular orbits; the difference between their perijove and apojove distances is small compared to their orbital semimajor axes. So their minimum distance is the difference between their orbital radii, and their maximum distance is the sum of their orbital radii: 33Gm∆∆ r Gm r ∆=g − , tr 33 (aEuropa−+ aa Io) ( Europa a Io ) where m is the mass of one moon, and ∆r the length of the other along the line between them (i.e.
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