Astro 210 Lecture 16 Feb 23, 2018

Transcription

1 Astro 210 Lecture 16 Feb 23, 2018 Announcements HW4 due online in PDF, Today 5:00 pm HW5 posted today, due online next Friday Night Observing next week...weather permitting Campus Observatory. Mon, Tue, Wed, Thur 7 9pm bring report form available on Moodle take and submit selfie while there 1

2 The Solar System Overview Clear trends & organization observed Q: examples? Q: basic solar system architecture? Q: expected trends in planet temperatures? how to test? 2

3 Measuring Planetary Temperatures key point: the visible light that lets you see the planet is not blackbody emission reflected sunlight but: there is black body emission at longer λ: infrared www: IR Moon, Mars General trends: indeed, T drops with distance d but less strongly than 1/d (not an inverse square law for T!) we would like to (and can!) understand in detail 3 Understanding Temperatures Q: what are sources of heating? of cooling? Q: what physical laws/conservation principles are important? Q: what sets planetary surface temperatures?

4 The Astrophysics of Planetary Temperatures surface heating dominated by radiation (i.e., energy flow) from the Sun s emission, peaked at optical λ geo thermal (usually) small contribution (but large for Jupiter) cooling also due radiation (blackbody emission, peaked at IR) 4 Note: Sun is steady source of light i.e., constant luminosity = power = Wattage = L each planet constantly receives this radiation and also emits its own, according to its T planet T constant value (in time), set by an equilibrium: incoming/outgoing energy flows exactly balance! Q: Why? What would happen if inflow > outflow? vice versa?

5 planetary energy flows (i.e., power or wattage W): W in = absorbed sunlight energy: constant power flow in W out = blackbody emission: strong increasing function of T What if W out > W in? energy conservation planet has net energy loss suffers cooling reduces W out if new W out > W in? still, then lather, rinse, repeat until W out = W in! equilibrium achieved! What if W out < W in? can convince yourself: planet warms until W out = W in! equilibrium achieved in this case too! 5 lesson: all roads lead to equilibrium! If Sun s emission steady, then planet T must go to steady value, set by energy-conserving balance: W out = W in

6 Planetary Temperatures Calculated Can get excellent estimate of planetary T from (fairly) simple first-principles calculation! d key is energy balance: R R absorption = emission sun Absorption recall: surface of area S surf emits flux F surf then radiated power = luminosity [energy/sec] is L = F surf S surf Sun: L = F S = 4πR 2 σt2 at planet, flux is F = L/4πd 2 = σt 4 (R /d) 2 [energy/area/sec] 6...but we know not all incoming sunlight is absorbed! Q: Why not? Q: What substance would absorb all incident sunlight? Q: What substance would absorb no incident sunlight? Q how could we simply quantify all of this?

7 Not all sunlight absorbed...or else wouldn t see Earth from space! some is reflected! recall: perfect absorber is blackbody perfect reflector: ideal mirror real substances/planets: somewhere between Define: albedo A = amount of light reflected incident light ideal mirror: A = 1 blackbody A = 0 Earth surface (average value) A Earth 0.4 (1) 7 In fact, A(λ) depends on λ, but roughly constant over visible band

8 iclicker Poll: Sun, Shadows, and the Earth Which of these is larger? A The surface area of sunlit portion of Earth B The area of Earth s shadow C (a) and (b) are equal 8

9 Planetary Energy Balance: Absorption albedo A is fraction of incident radiation reflected fraction absorbed is 1 A absorbed flux is F abs = (1 A)L /4πd 2 = (1 A)σT 4 (R /d) 2 over sunlit surface of planet, energy absorbed per sec: 9 W abs = area intercepting sunlight F abs (2) ( ) = πr 2 (1 A)σT 4 R 2 (3) d = (1 A)πR 2R2 d 2 σt4 (4) note: effective absorbing area is πr 2 planet s cross section i.e., area of shadow

10 Planetary Energy Balance: Emission emitted flux: F emit = σt 4 (avg surface T) what area emits? case I: slowly rotating, no atm: backside cool only dayside emits case II: both sides hot both sides emit slow rotate thin/no atm. fast rotate thick atm. Top View emitting area: S emit = { 2πR 2 slow rot 4πR 2 fast rot (5) energy emitted per sec: 10 W emit = S emit σt 4 (6)

11 Planetary Temperatures: The Mighty Formula In equilibrium: W abs = W emit (1 A) πr 2 (R /d) 2 σt 4 = ( 2πR 2 4πR 2 ) σt 4 { slow rot fast rot (7) and so planet surface temperature is 11 T = [ (1 A)/2 (1 A)/4 ] 1/4 ( R d ) 1/2 T { slow rot fast rot note: planet T set by Sun surface T independent of planet radius R! drops with distance from Sun, but as T 1/ d (8)

12 Calculate for T = 5800 K, and d in AU: T = 332 K 279 K ( ( ) 1/4 1 A slow rot d 2 AU 1 A d 2 AU for solar system objects, with d AU in AU ) 1/4 fast rot (9) Example: the Earth inputs: d AU = 1 AU, try A = 0 atm: day and night temp roughly same fast rot (case II) 12 T average = 279 K 6 C 43 F (10) pretty close!... but a little low but using A 0.4 gives T = 246 K = 27 C: yikes! but: haven t accounted for greenhouse effect, small deviations from perfect blackbody emission,... Q: so what do we expect for the Moon?

13 Atmospheres: Gas Properties gases on microscopic scale a swarm of particles, for example atoms or molecules gas particles have empty space between them not packed together as in liquid or solid gas particles are in constant random motion collide with each other, container walls (if any) exchange energy & momentum distribution of speeds 13 on macroscopic scales (i.e., how we see things) particle motions perceived as temperature

14 iclicker Poll: Gas Particle Speeds consider a parcel of gas: macroscopically, gas is at rest (not moving/blowing) at room temperature T in this gas: the average particle velocity v and speed v = v are: A v = 0 and v = 0 B v = 0 and v > 0 C v 0 and v = 0 14 D v 0 and v > 0

15 average particle velocity vector vanishes: v = 0 why? not because particles are still rather: equal numbers with v x > 0 vs v x < 0 averages to zero otherwise: gas would have net v x, wouldn t be at rest! note microscopic macroscopic (particle bulk) correspondence: micro: equal probabilities for particle v > 0 and v < 0 macro: corresponds to bulk gas speed u gas = 0 since particles are moving, speeds v 2 = v 2 x + v2 y + v2 z > 0 avg KE of one gas particle is nonzero Q: what does this mean for bulk, macroscopic gas? 15

16 particle motion particle kinetic energy proportional to bulk temperature KE per particle = 1 2 µ v2 = 3 kt (11) 2 example of general rule of thumb: in thermal system, typical particle energy E particle kt with k = Joules/Kelvin: Boltzmann s constant for thermal gas: average particle speed ( root mean square ) is v rms = 3kT where µ = mass of 1 gas particle: molecular weight if a gas particles has µ (12) 16 A = tot # of n, p = sum atomic weights (13) then µ = Am p, with m p = proton mass = kg

17 also: peak speed (most probable) v p = 2kT/µ < v rms Q: how can this be less than average? Note: v rms T: hotter faster on avg T measures avg particle energy & speed v rms 1/ µ: more massive particles slower on avg Example: air is mostly N 2 and O 2 molecules Q: in this room, which faster:? 17 but: if peel orange, smell does not propagate at v rms 500 m/s: don t smell it within 10 ms! Q: why not?

18 18 Director s Cut Extras

19 More on Equilibrium Temperatures we argued that a planet s temperature T is set by and equilibrium in which the energy flow into the planet is exactly balanced by the energy flow out Here we look at this in more detail The energy flow onto a planet is set by the flux of sunlight ( ) ( ) de = W abs = (1 A)F sunlight S shadow = πr 2 R 2 (1 A) σt 4 dt in d (14) which is essentially constant and most importantly is independent of the planet s temperature 19 The energy flow out of a fast rotator is ( ) de = W emit = S emit F bb = 4πR 2 σt 4 (15) dt out

20 by energy conservation, the a planet s heat energy content changes due to the energy flows in and out, i.e. ( ) ( ) ( ) de de de = + (16) dt heat dt out dt in and since E heat T, we have 20 dt dt W abs W emit (17) = 4πR 2 σt 4 + πr 2 (1 A) we can rewrite this last expression as dt 4πR 2 σ dt [ T 4 1 A 4 = 4πR 2 σ ( T 4 T 4 eq ) ( R d ( ) R 2 σt ] 4 d ) 2 σt 4 (18) (19) (20) where T eq is the fast-rotator equilibrium temperature found above

21 Let s look at the three possibilities for the temperature change dt dt 4πR2 σ ( T 4 Teq 4 (21) if T = T eq planet has exactly equilibrium temperature then dt/dt = 0: temperature does not change with time if T > T eq planet hotter than equilibrium temperate then dt/dt < 0: temperature decreases: the planet cools cooling continues until T = T eq if T < T eq planet cooler than equilibrium temperature then dt/dt > 0: temperature increases: the planet warms warming continues until T = T eq ) 21 So: no matter what temperature a planet starts with it will always be driven to the equilibrium temperature T = T eq and then temperature remains constant We have already guessed this, but now we have proven it