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4 Pages

### P5_sol

Course: AST 111, Fall 2009
School: Rochester
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Word Count: 1055

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Set Problem #5, AST111 Solutions 1) Calculate the equilibrium temperature of the Moon as a function of latitude. Assume that the moon is a rapid rotator with zero obliquity, has a bond albedo of Ab=0.07 and an emissivity of 1. For a small ribbon of latitude, the sun only sees a projected area: Area = 2r 2 cos 2 d So flux absorbed for the ribbon is: L Fabs= (1 Ab ) (2r 2 cos2 d ) 2 4 d The surface area of the...

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Set Problem #5, AST111 Solutions 1) Calculate the equilibrium temperature of the Moon as a function of latitude. Assume that the moon is a rapid rotator with zero obliquity, has a bond albedo of Ab=0.07 and an emissivity of 1. For a small ribbon of latitude, the sun only sees a projected area: Area = 2r 2 cos 2 d So flux absorbed for the ribbon is: L Fabs= (1 Ab ) (2r 2 cos2 d ) 2 4 d The surface area of the ribbon is: 2r cos r d d r r cos d SA = 2 r cos d 2 The emitted flux for the ribbon is: Femit = 2 r 2cos dT 4 Setting the emitted flux equal to that absorbed and solving for the temperature we find: rd 2 r cos SA = 2 r 2 cos d (1 Ab ) L cos T ( ) = 4 2d 2 1 4 (1 0.07 ) ( 4 1033 erg s-1 ) cos = 4 2 (1.5 1013cm )2 ( 5.67 10-5 erg cm -2K 4s 1 ) T ( ) = 293.2K ( cos ) 1 4 1 4 2) According to a cookbook it takes about 10 minutes to cook a 1 inch (2.5cm) slab of steak to well done. Using a conductive diffusion approximation, estimate to order of magnitude the heat conductivity of meat. Note: the specific heat and density of meat are similar to those of water; c ~ 1 cal g -1 C-1, ~1g cm-3 . The heat conduction can be modeled with the diffusion equation T KT 2T = . So t cP z 2 KT z 2 ~ . Here z = 1.25cm if we flip the steak and cook it on both sides. The cP t time t = 10 60sec = 600sec . Solving for the heat conductivity we find z 2 (1.25) 2 cm 2 1g cm -3 1cal g -1C 1 KT ~ cP ~ = 2.6 103 cal s -1cm -1 C1 600s t 3) At a telescope, a star is observed during a minute long integration and measured to have a mag of m=16. The observation has a signal to noise S/N=10. a) What is the uncertainty in the mag? M=16X? Signal = S. From the definition of magnitude we know 2.5 log S = 16.0 . So S = 3.98 10 7 . Since S/N=10 we know that N is a tenth of this. Lets now consider the signal plus the typical uncertainty (the Noise, N). 1 2.5log(S + N ) = 2.5log(3.98 107 (1 + 10 ) = 15.9 2.5log(S ) 2.5log(S + N ) = 0.1 16 0.1 mag Another way to think about this: We would like to know S+N 1 2.5log(S ) 2.5log(S + N ) = 2.5log = 2.5log(1 + ) S 10 = 0.1 mag This problem can also be done with a first order expansion. d log10 ( x) 2.5dx 2.5dx 1.085dx dM = 2.5 dx = = = dx x log e (10) 2.3x x dx N 1 For us so that the difference in magnitude is dM = 0.1 and we can write = = x S 10 M = 16 0.1 . b) If you double the integration time what would be the S/N of the observation? The signal is twice as big but the noise is only 2 times as large. So the new S/N is 2 / 2 = 2 times the old one or 2 10 = 14 . c) If you double the integration time what would the uncertainty in the mag be? dM = 1.085 1 = 0.08 . 14 4) The solar system was formed from a disk composed mostly of molecular hydrogen ( mH2 = 3.35 1024 gm ) and some helium ( mHe = 6.64 1024 gm ) ; the planets might all plausibly be considered to have had an atmosphere dominated by hydrogen, as the giant planets like Saturn still do. Earth ( M = 5.97 10 27 gm ) has always had an average temperature of about 280 K, and Saturn ( MV = 95.184 M = 5.69 1029 gm ) an average temperature about 90 K. The Earth and Saturn have similar equatorial gravities surface g ~ 1000cm/s 2 . a) When the planets were newly formed, what was the thermal escape timescale of Earths hydrogen? What was the corresponding timescale for Saturn? First I estimate mean thermal velocities. For Hydrogen on Earth v0, H = 3kT / mH = 3 1.4 1016 280 =1.9 105 cm/s . 24 3.3 10 g Saturn has a temperature that is 90/280=0.32 times lower than that of the Earth. So on Saturn the mean thermal velocity is 0.32 = 0.57 times lower than that of Earth, and Saturn has a mean thermal velocity of v0, H = 1.1105 cm/s . Now I estimate scale heights. For both cases we assume an atmosphere dominated kT by hydrogen and the scale height h ~ . For the Earth h ~ 107 cm . For Saturn mH g with a 0.32 lower temperature h ~ 4 106cm . Both of these values are larger than their current ones because their atmospheres now also contain heavier molecules. By dividing h / v0 I get a characteristic timescale. For Earth and hydrogen this timescale is 77s. For Saturn and hydrogen this timescale is 36seconds. Now I estimate escape velocities. For Earth the escape velocity is ve = 2 gr = 1.1106 cm/s. Saturn has an equatorial radius that is 10 times larger and a similar acceleration so it has an escape velocity that is 10 times larger than that of Earth or ve ~ 3.6 106 cm/s . Now I need to compare mean thermal velocities to escape velocities. Y = ve 2 = 25 v0 2 for Earth and Y = 1000 for Saturn. eY / Y = 3 109 for Hydrogen on Earth and Saturn eY / Y ~ 10 400 on Saturn. Thermal escape timescales for Hydrogen from the Earth are of order 77 sec 3 109 ~ 3 1011 sec ~ 1000 years . For Saturn this timescale is way greater than a Hubble time (age of Universe) which ihs 1010years. b) Similarly, what was the timescale of escape for Earths nitrogen molecules ( mN 2 = 4.65 1023 gm ) ? Molecular nitrogen is about 14 times more massive than molecular hydrogen. ve 2 = 350 ...

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