Astronomy 1 Full Solutions

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  • University of Glasgow

    Department of Physics and Astronomy

    Astronomy 1Problems Handbook

    20092010Instructors ManualValues of astrophysical constants

    speed of light c 2.998 108 m s1gravitational constant G 6.673 1011 N m2 kg2Planck constant h 6.626 1034 J sBoltzmann constant k 1.381 1023 J K1Stefan-Boltzmann constant 5.671 108 W m2 K4Rydberg constant R 1.097 107 m1Avogadro constant NA 6.022 1023 mol1gas constant R 8.315 J mol1 K1

    proton mass mp 1.673 1027 kgelectron mass me 9.109 1031 kgelementary charge e 1.602 1019 Celectronvolt eV 1.602 1019 Jastronomical unit AU 1.496 1011 mparsec pc 3.086 1016 mlight year ly 9.461 1015 msolar mass M 1.989 1030 kgsolar radius R 6.960 108 msolar luminosity L 3.826 1026 WEarth mass M 5.976 1024 kgEarth radius R 6.378 106 mobliquity of the ecliptic 23 26

  • Contents

    1 Solar System Physics 1

    2 Positional Astronomy 30

    3 Dynamical Astronomy 70

    4 Stellar Astrophysics part I 92

    5 Stellar Astrophysics part II 116

    6 Observational methods 133

    7 Cosmology 150

    NotationIn many cases the marks allocated to each question, or section of a question, are shown in squarebrackets in the right-hand margin. Question numbers followed by a small e are recent examquestions. Answers to some numerical problems are shown in curly brackets. Where available,solutions are shown immediately after their questions (which are given in smaller type), with ahorizontal line separating each problem.

    Dr M.A. HendrySeptember 2009

  • 1 Solar System Physics 1

    1 Solar System Physics

    1.1e List five main differences between the terrestrial and jovian planets. [5]

    Solution: no solution available

    1.2e State Newtons law of gravitation, defining all the symbols you use. Hence derive an expression for the surfacegravity of a spherically symmetrical uniform planet in terms of its mass, radius, and the gravitational constant,G. Calculate the value of the surface gravity of Jupiter, using planetary data from the lecture handout or coursetextbook. [5]

    Solution: Newtons law of gravitation: the magnitude of the gravitational force F between twomasses m1 and m2 separated by a distance R is

    F = Gm1m2R2

    where G is the gravitational constant, and where the force exerted by m2 on m1 is along the linebetween m1 and m2, and vice-versa. If m is the mass of an object near the surface of a body ofmass M , radius R, then the gravitational attraction of the mass M on m at the surface is given by

    F = mg = GMmR2

    where g is the surface gravity of the object of mass M . Rearranging, we have the surface gravityg of an object of mass M and radius R is given by

    g = GMR2

    Inserting the values given yields g = 24.9ms1 for Jupiter.

    1.3 State Newtons Law of Gravitation and show that the acceleration due to gravity at the surface of a planet ofmass M and radius R is given by

    g = G MR2


    stating clearly any assumptions that you make. [3]Given that the mass of Venus is 0.817 times that of the Earth and the radius of Venus is 0.97 times that of theEarth, calculate the acceleration due to gravity at the surface of Venus if its value at the surface of the Earth is9.8 m s2. {8.5 m s2} [3]

    Solution: Fgrav = GMm/r2 where m and M are masses of bodies separated by r . Force isattractive and acts along line joining bodies, mg = GMm/R2 g = GM/R2, assuming planetspherical with spherically symmetric density. Thus

    gVgE= MV


    = 0.817(0.97)2



    gV =0.817(0.97)2

    9.8 = 8.5 m s2.

  • 2 Astronomy 1 Problems Handbook, 20092010

    1.4e Calculate the ratio of the surface gravities of the Earth and Jupiter, given the following dataRJupiter = 11.2REarth and MJupiter = 317.8MEarth

    where R and M denote respectively the radius and mass of the planet. [2]

    Show that, for two planets of equal mean density, the ratio of their surface gravities is equal to the ratio oftheir radii. [3]

    Solution:g = GM

    R2We can write

    M = 43R3

    g = G 43


    R2 = 4


    Henceg1g2= R11

    R22= R1

    R2if 1 = 2

    [3]For Jupiter,

    gJ upitergEarth

    = MJME




    = 317.8 (


    )2= 2.53


    1.5e Starting from Newtons Law of Gravitation, derive an expression for the gravitational acceleration at the surfaceof a planet in terms of its radius and average density. Given that Mars and Earth are both terrestrial (rocky)planets, and using planetary data from the lecture handout or textbook, estimate the ratio of the gravitationalacceleration at the surface of Mars to that of Earth, stating any assumptions you make. [5]

    Solution: Newtons law of gravitationF = GMm


    F= gravitational force between bodies of mass M and m separated by distance r .

    Surface gravity (gravitational acceleration) = force per unit mass

    g = GMr2

    where r = Rplanet

    Express in terms of g = Gr2


    = 43Gr

  • 1 Solar System Physics 3

    So ratio of surface gravities of Mars and Earth

    ratio = MarsrMarsEarthrEarthAs both are terrestrial planets assume they have same






    1.6e Give a sketch of the Earths interior, clearly indicating the different regions, and their approximate dimensions. [5]

    Solution: no solution available

    1.7e Explain the terms

    (a) plate tectonics(b) volcanism(c) outgassing. [5]

    Solution: no solution available

    1.8e Give a sketch of the atmosphere of the Sun, indicating the approximate dimensions of the various regions andtheir approximate temperatures. [5]

    Solution: no solution available

    1.9e The Earth and Jupiters satellite Io both exhibit volcanism. Briefly describe the causes of this volcanism in eachcase. [5]

    Solution: Earth: volcanism is result of radioactive decay within interior heating rocks (radiogenicheating). Rocks melt lava comes to surface via cracks in crust.

    Volcanism found at subduction zones (boundary between two plates where oceanic crust issliding under continental crust), and also at weak points in plates hot spots such as at Hawaiianislands.

    Io: volcanic activity due to heating by tidal forces exerted primarily by Jupiter, and also byEuropa which is in orbital resonance with Io ( Europa orbital period = 2 Io orbital period). Io isclosest of Galilean satellites to Jupiter, Europa next.

    1.10e Sketch the internal structure of the Sun, indicating the approximate dimensions and temperatures of the variousregions. [5]

    Solution: no solution available

    1.11e Explain the mechanism of the greenhouse effect. Give the names of two main greenhouse gases in the Earthsatmosphere. [5]

  • 4 Astronomy 1 Problems Handbook, 20092010

    Solution: no solution available

    1.12e In what way do the surface features of the Moon and Earth radically differ and why? [5]

    Solution: Surface features of Moon Extensive circular impact craters. Smooth dark areas calledmaria. No water. Old rocks.

    Surface features of Earth 3/4 covered in oceans plus ice. Volcanically active. Younger rocks(range of ages). Surface (and atmosphere) modified by life.

    Basic differences due to lack of present day volcanic activity and no erosion processes on Moon(no atmosphere and no water), while these activities plus life have modified Earth surface. Moonprobably lost its volatile compounds (H2O, CO2 etc.) due to heating on impact with Earth of largeplanetesimal from which it is thought to have formed.

    1.13e The surface temperature of Venus is 700 K and that of the Earth is 300 K. Calculate the wavelengths at whichthe radiative flux from each planet is maximum. You may use Wiens law, maxT = 2.9 103 m K. [2]In what region of the spectrum do these wavelengths lie? [1]Explain carefully what is meant by the greenhouse effect in planetary atmospheres. Discuss this phenomenonwith reference to the atmospheres of the terrestrial planets. [10]How does the runaway greenhouse effect explain the near absence of water in Venuss atmosphere? [4]

    Solution: Using Wiens displacement law, with T = 700 K for Venus max = 41.4 nm and forEarth (T = 300 K) max = 97 nm. Both these wavelengths lie in the infrared region. (Radiation inthe infrared is largely absorbed by CO2 and H2O, hence greenhouse effect.) Atmosphere transmitslight in visible band, but absorbs or reflects light in infrared. Thus light from the Sun is transmittedthrough atmosphere since it is largely in visible because of high effective temperature (5 800 K) ofthe Sun. Light from the planets surface is however emitted at lower temperature (Venus 700 K, Earth300 K) and so is mainly in infrared, and so is absorbed by atmosphere and reradiated. This effectcomes about largely through the presence of CO2 (H2O and CO2 in the case of the Earth), althoughother molecules will contribute. The effect is particularly strong on Venus, whose atmosphere isdense and made largely of CO2. On the Earth it is also important, although not so evident. Onplanets with only thin atmospheres it is fairly unimportant. Venuss surface temperature mightbe expected to be around 350 K from simple modeling, i.e., assuming surface of planet emits aflux equal to the flux from the Sun, although in reality the surface temperature of Venus is around700 K. The runaway greenhouse effect is due to positive feedback. Water is evaporated by highsurface temperatures on Venus produced by the greenhouse effect due to CO2. The increase inwater vapour increases the greenhouse effect, which in turn increases the surface temperature andwater evaporation. Temperature continues to rise, and scale height of H2O increases until water isvulnerable to dissociation from UV and X-ray rad