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a) The initial ratio of deuterium (D) to hydrogen (H) in a planets atmosphere was 175000; however, the present ratio is 1/15

Units 1-3 Equations Fundamental Constants Tidal Forces: G gravit. constant 6.67 x 10-1 Nmkg -2 Astronomical Unit 1.49 x 101

Molecular Mass Planetary Surface Temperatures: H hydrogen = MH H2 molecular hydrogen = 2MH D = deuterium = 2MH belium = 4MH H

a) The initial ratio of deuterium (D) to hydrogen (H) in a planet's atmosphere was 175000; however, the present ratio is 1/1500 and the initial and final abundances of D are 5 x 10° atoms per m3 and 9 x 106 atoms per m2, respectively. What fraction of deuterium has been lost, and what fraction of hydrogen has been lost in this planet's atmosphere, assuming the only loss mechanism for either atom is escape to space? b) Assuming that atmospheric gases will begin to escape from a planet's atmosphere if the average gas velocity is one-sixth the escape velocity of the planet, at what temperature might N2 be able to of Venus is about 750 K, would N2 be able to escape? begin to escape from Venus? If the surface temperature
Units 1-3 Equations Fundamental Constants Tidal Forces: G gravit. constant 6.67 x 10-1 Nm'kg -2 Astronomical Unit 1.49 x 101 AU k Boltzmann's constant= 1.38 x 10- J/K Angular size of an object: dFe oxrdr ma dFe xr"dr dFt x mr-3dr oxr-* Force on Earth (Moon): Mass of Tidal Bulge: a= Stef.-Bocht3.00 x 10 m JmK-s-1 MH= mass of hydrogen 1.67 x 10-27 kg Mn=mass of neutron or proton 1.67 x 10-7 kg Net Force on Moon (Bulges): Stefan-Boltzmann law: Roche's Limit: F = aT 1/3 Solar System Data R Luminosity Equations: TR2.5PM Two touching particles: L= (4mR2)- F Lref A Lin Leb, = (1 - A) Lin 1/3 Luminosity (sphere): Reflected luminosity: Absorbed luminosity: Mo= mass of Sun= 1.99 x 1030 kg Ro= radius of Sun= 6.96 x 10° m 026 w TR2,44PM Pm Zero-strength sphere: Lo= luminosity of Sun3.8 Fo=lux of Sun = 1.36 x 103 w/m2 1/3 PM R Spherical bodies > 40 km: TAO1.38 Kepler's Laws: radius of Earth = 6.378 x 10° m kg 0 017 RE ME 1/3 PM r= 1.19 mass of Earth 5.98 x 1st Law Body about to impact: eccentricity (Earth) 2nd law constant RM=radius of Moon= 1.738 x 10° m MM=mass of Moon = 7. 3.844 x 10 m aMesemi-major Radiation Pressure: 2 (Mm) 3rd law: Lra2Q P= eccentricity (Moon)= 6.055 Gravitation Law eMe C RMe= radius of Mercury 2.439 x 10 m MMe= mass of Mercuryerury) 0.387 AU Me etricity (Mercury) = 0.206 GMm F= Poynting-Robertson Effect: tPR 7.0(100)apr/Q Velocity Equations: eMe Virial Theorem: Ry radius of Venus 052 GM Circular velocity Ucire 15T pTGRHM avsemi-major axis (Venus)= 0.723 AU eveccentricity (Venus) = 0.007 For collapse to start: Escape velocity: Velocity Equation: v2= GM (2-1) Wien's Law: RM radius of Mars642 x 104 m GMasemi-major axis (Mars) 1.524 AU eMaeccentricity (Mars) =0.093 0.00290 T Doppler Effert Radioactive Decay Equation: radius of Jupiter 7.140 x 107 m ass of Jupiter = 1.90 x 1027 kg aj semi-major axis (Jupiter) 5.203 AU ej=eccentricity (Jupiter) = 0.048 RJ N(t)= Noexp- Relevant Physics Equations: F= ma Force: Collision Velocities mu FT vimp Centripetal Force: Rş= radius of Saturn= 6.033 x 10 m Ms mass of Saturn5.69 x 10 kg assemi-major axis (Saturn) 9.539 AU es= eccentricity (Saturn)= 0.056 GM Maximum Rotation Rates: Acceleration due to Gravity: 3m Pe= 1 KE Ry=radius of Uranus= 2.556 x 107 m Mu= mass of Uranus= 8.68 x 10*kg A1t ay = semi-major ) 0.047 eu eccentr Kinetic Energy: PE= mgh Potential Energy: Maximum Sizes of Irregular Bodies GMm U = Gravitational Potential Energy: 2Gp (R2 - RN= radius of Neptune= 2.476 x 107 m MN mass of Neptune 1.02 AU Pressure of Overlying Material: 2Tr T 2 Period of Revolution: Strength: S1.34(10-10)pR. sen eN aN ty(Neptune) = 0.009
Molecular Mass Planetary Surface Temperatures: H hydrogen = MH H2 molecular hydrogen = 2MH D = deuterium = 2MH belium = 4MH H20 water = 18MH N2 molecular nitrogen = 28MH O2 molecular oxygen = 32MH CO carbon monoxide = 28MH CO2 carbon dioxide = 44MH CH4= methane = 16MH (4) () Perfect blackbody: Taur/ = То 1 - A He= Including e and ¢: = eoT cos d 1-А For a latitude of 30° T 285 K ea2 Ideal Gas Law PkT P = иМн Atmospheric Structure: Ne neon = 20MH dP MHgdz kT Pressure in atmosphere: More Planetary Data MH KT P= Po exp mean density of Earth = 5520 kg/m3 PE PM mean density of Moon = 3340 kg/m PMe= mean density of Mercury = 5420 kg/m3 pv mean density of Venus = 5250 kg/m3 P = Poexp KT Н%. иМн9 Scale Height: PMa mean density of Mars = 3940 kg/m3 PJmean density of Jupiter = 1314 kg/m3 ps=mean density of Saturn = 690 kg/m3 pu= mean density of Uranus = 1290 kg/m3 PN= mean density of Neptune = 1640 kg/m3 Velocity needed to escape atmosphere: V2 fvec 3kT ГиМн f= 0.25 Thermal escape of atmosphere Hydrostatic Equation: t= 3g Y dP = pg dz z increasing downwards: z increasing upwards: P increases, r decreases: GMm dP= -pg dz dP= -pg dr Y = m uMH Pressure Inside Planets: 2TGP(R-r) 1.4(1010)F(R2 - -2) P, Volcanic Eruptions: P= PMg(z Depth z in a magma chamber: h) = pRgz h=PR PM PM Height h of volcano:
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initial COnd di on 5000 Final Condition IV1500 lnitial 4 D 5X1Daton & 3 m Funal Present qx106autDMA Fin al fraction deuterim9X106 atomg /m 1500 X Ratio ehydro gen lost 500 X 9X106 atsmIm3 5000x5 X108 aton/m3 Fraction 21 hy.drogen lost 500d 1 Stan daV 3KT 2 t Bx 1380 X10-23x 750 468Xlo- 26 L413m/A Nidregen &14 3m have 10.36 Km velocy Gnd Venuw escape vehnity required upto.2 3 KT V 0360) 3 x t. 3807 (0-23 xT 4.68 x10-26 60 05 K So at be T2 12 /267 6605 K able to esCape might venu N2 Ics Scanned w

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