B ν (T) = 2hν3 c 3 1. e hν/kt 1. (4) For the solar radiation λ = 20µm photons are in the Rayleigh-Jean region, e hν/kt 1+hν/kT.

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1 Name: Astronomy 18 - Problem Set 8 1. Fundamental Planetary Science problem 14.4 a) Calculate the ratio of the light reflected by Earth at 0.5 µm to that emitted by the Sun at the same wavelength. The solar radiation peaks at 0.5µm. The flux of solar radiation at 1AU is F = L /4πa 2. (1) The amount of light reflected by the Earth is L = Aπr 2 F = AL (r /2a) 2 (2) where A 0.3 is the albedo of the Earth, L is the luminosity of the Sun, and r is the radius of the Earth. Therefore, at λ = 0.5µm, L /L (λ = 0.5µm) A(r /2a) (3) b) Calculate the ratio of the thermal radiation emitted by Earth at 20 µm to that emitted by the Sun at the same wavelength. Sun s black body temperature T 6000K and the equilibrium temperature at Earth T 300K (the equilibrium temperature is actually 260K, but it is raised to 300K due to a mild Greenhouse effect). For λ = 20µm, hc/λ = m 2 kgs ms 1 / m, kt = m 2 kgs 2 K K so that hν/kt = For the Planck s radiation law B ν (T) = 2hν3 c 3 1 e hν/kt 1. (4) For the solar radiation λ = 20µm photons are in the Rayleigh-Jean region, At λ = 20µm, hν/kt = 0.12 << 1 so that The ratio of luminosity L /L (λ = 20µm) e hν/kt 1+hν/kT. (5) B ν (T ) = 2ν2 kt c 3. (6) ( r r ) 2 ( ) hν/kt (7) e hν/kt 1 c) Repeat the above calculations for Jupiter. For Jupiter, the equilibriumtemperatureist J = T / 5.2 = 114Kanditsradiusisr J = 0.1r. Thus, the ratio of luminosity ( ) 2 ( ) rj hν/kt L J /L (λ = 20µm) (8) e hν/ktj 1 r 1

2 2. Fundamental Planetary Science problem 14.5 What types of extra solar planets are most easily detected by the following methods and why a) radial velocity (Doppler) surveys High-mass (m p ) planets with relatively small periods (P). This method measures the reflex motion of the host star V. Short-period planets have large Keplerian velocity V p. The host stars of relatively massive planets have relatively large reflex motion with velocity V = (m p /M )V p. There are more conspicuous. b) astrometry High-mass (m p ) planets with period of a few years. The displacement of the host star is D (m p /M )a so more distant planets can cause bigger wobbles in the position of their host stars. But their periods P a 3/2. If we wish to see objects within a few years, it is better to select those with a reasonably modest period. c) transit photometry Planets with large radii R p and close-in in orbits. The transit depth is proportional to (R p /R ) 2. The transit probability is proportional R /a. d) coronagraphy Long-period young gas giant planets. We use the coronagraphic technique to directly measure the light from the planets while blocking out that of the stars. Long period planets are further away from their host stars and their images are more easily separated from that of the host star. In order to overcome the effect of defraction (due to the limited size of the telescopes), it is desirable to image distant planets. But these planets are so far away from their host stars that they do not reprocess much of their host stars light. Young stars release their own intrinsic radiation and therefore are better targets. 3. Fundamental Planetary Science problem Radii and masses measured for five hypothetical transiting planets are listed below. State the ranges in composition consistent with these measurements. In some cases, the ranges allowed by the error bars include some unlikely or unphysical compositions. Which cases are these and why? a) R p = 3±1R, M p = 3±1M The density is one tenth that of the Earth. It must have an extensive hydrogen dominated atmosphere. One possibility is that it s a water world with a significant amount of water molecules dissociated into hydrogen and oxygen. In order for the hydrogen not to escape, it is located either near a low mass and low luminosity host star or beyond 1AU around a solar type star. b) R p = 1±0.5R, M p = 3±1M With a density times that of the Earth. it is likely a planet made primarily of iron. 2

3 c) R p = 12 ± 2R, M p = 300 ± 100M Gas giant with a density times that of Jupiter. It is probably mostly made of hydrogen. d) R p = 3±1R, M p = 30±10M Super earth with a density 1±0.5 that of the Earth. It is probably composed mostly of silicon. e) R p = 2±0.2R, M p = 10±1M Super earth with a density 1±0.1 that of the Earth. Its composition is likely to be rocky and mostly made of silicon. 4. Fundamental Planetary Science problem 15.5 a) Calculate the rate of growth, dr/dt, of planetary embryos of radius R p =4,000km and mass M p = kg in a planetesimal disk of surface density σ ρ = 100 kg m 2, temperature T = 300K and velocity dispersion v=1 km s 1 at a distance of a =2AU from a star of 3M. You may use the two-body approximation for planetesimal/protoplanetary embryo encounters. Accretion rate can be found in Equation (13.22) dm p /dt = ρvπr 2 pf g (9) wherev isthevelocitydispersionandv e = (GM p /R p ) 1/2 = 4kms 1 isthe escape velocity from the surface of the planet and F g = 1+(v e /v) 2 = 17. For a planet with a density ρ p = 3M p /4πR 3 p = kg m 3 and a mean motion n = (GM /a 3 ) 1/2 2.e 7s 1, the rate of radius growth can be found in Equation (13.26) such that dr p /dt = ( ) 1/2 3 σ ρ n F g m s 1. (10) π ρ p The radius doubling time scale τ = R p /(dr p /dt) yr. b) What will halt (or at least severely slow down the accretion of such a protoplanetary embryos? What will is mass be at this point? When it acquires an isolation mass (Equation 13.28) such that ( a ) ( ) 3 3/2 ( ) 1/2 M σρ M g 1AU 1gcm g 0.3M. M (11) 3

4 5. Fundamental Planetary Science problem 15.9 A planet of mass M p and radius R p initially spins in the prograde direction with zero obliquity and rotation period P rot. It is impacted nearly tangentially at its north pole by a body of mass m, whose velocity V before the encounter was small compared with the escape speed from the planet s surface. a) Derive an expression for the planet s spin period and obliquity after the impact. You may assume that the projectile was entirely absorbed. Since the impact is tangential at the planet s north pole, its spin rate does not change. But its obliquity does change after the impact. The angular momentum gained in the direction normal to the original spin axis is L = mvr p. (12) We assume the impact is on a parabolic trajectory which barely grazed the planet s surface near the north pole so that V = (2Gm/R p ) 1/2 and L = m(2gm p R p ) 1/2. The obliquity change (tilt angle) is where γ is the moment of inertia. θ = tan 1 L/γM p R p (2π/P rot ) (13) b) Numerically evaluate your result for M p = 1M, R p = 1R, P rot = 10 5 seconds, and m = 0.02M. (Of course, a truly tangential impactor is likely to skip off rather than being absorbed, but even for a trajectory only 10 o from the horizontal, most ejecta can be captured at velocities considered here. The case of a polar impactor is also a singular extremum, but both of these effects, together only add a factor of a few to the angular momentum provided by a given mass impacting with a random geometry, and they make the algebra much easier.) From the above equation, we find θ = tan 1 L/γM p R p (2π/P rot = tan 1 m(2gm p R p ) 1/2 /γm p R p (2π/P rot (14) We can apply the above numbers into this equation and obtain the actual value. 4

5 6. Fundamental Planetary Science problem Calculate the rise in temperature if the Earth differentiated from an initially homogeneous density distribution to a configuration in which one-third of the planet s mass was constrained in a core whose density was twice that of the surrounding mantle. You may assume infinite conductivity. Gravitational energy GM(R) E p = dm(r). (15) R In a planet with a homogeneous density ρ i, M = 4πρ i R 3 and E p = 1 3 Ri 0 G4πρ i R 2 4πρ i R 2 dr = 3 GM 2. (16) 5 R i For the final state with ρ fc in the core and ρ fm in the mantle and ρ fc = 2ρ fm. The radius R c where the core and mantle are separated can be found from 4πρ fc R 3 c/3 = M/3 = 8πρ fm R 3 c/3 (17) 4πρ fm (R 3 p R 3 c)/3 = 2M/3. (18) Assume the planet s total radius R p does not change from R i after the differentiation, we find from the above equations, The core mass R c = 5 1/3 R i (19) ρ fm = (5/6)ρ i, ρ fc = (5/3)ρ i. (20) M c = 4πρ fc R 3 c/3 = 2/5M (21) We then work out the gravitational potential ( E p = 16π2 Gρ 2 Rc i 25 Ri R4 dr+ R c 36 R4 dr ) +GM c Ri R c 5 6 ρ irdr [ (22) 25 Rc 5 = 9 Ri ( ) 1 R5 c 36 Ri ( )] 1 R2 c 3 GM 2 12 Ri 2. (23) 5 R i The third term takes into account the mass in the core when we calculate the gravitational potential in the mantle. Assuming all the released energy goes into heating the interior of mantle and core (due to a perfect heat conductivity) and without any heat loss, E = E p E p = M p C p T (24) where C p = erg K 1 g 1 is specific heat capacity. From these equations we find T 10 3 K. (25) 5

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