大気と星の構造学 地球惑星大気物理学入門

2010 6 (7 5 )

0. 0–1 0–2 1 2 3 4 , - 5 6 7 8 1

1 1–1 1) 2) 3) 4) () ()H 2.79 × 10 10He 2.72 × 10 9C 1.01 × 10 7N 3.13 × 10 6O 2.38 × 10 7Ne 3.44 × 10 6Mg 1.076 × 10 6Si ≡ 1 × 10 6S 5.15 × 10 5Ar 1.01 × 10 5Fe 9.00 × 10 5 (O,Mg,Si,Fe,...) 2006 IAU 11) 2) 3) (1 ( 10 ) 2

1–2 or HHe HHe H 2 OCH 4 NH 3 () (2006 ) ( km) (1221.5) Fe (3480.0) Fe (6350) Mg,Si,O (6371) Si,Al,Ca,O (6371) H,O (>6371) N,O 1 1-1 1 1) 2) H 2 O (1g/cc) H 2 O H2O 3) ( 8g/cc) 1-2 () ( O 2− ) Si 4+ Mg 2+ Mg Si O 2:1:4 3

(a) 0.3871 AU 88.0 day 58.65 day

2. 2–1 ∼ H H 2 CO 2 N 2 O 2 H 2 O2–2 T = σT 4 σ σ = 5.67 × 10 −8J s −1 m −2 K −4 T eq 4πR 2 σT 4 eq = πR 2 (1 − A)FR A F 1AU F = 1370 J s −1 m −2 T e f f 4πR 2 σT 4 e f f = []T s >> T eq , T e f f CO 2 T e f f > T eq T s > T eq , T e f f H 2 OCO 2 6

(Natial Space Science Center, NASA, USA) P(bar) 0.13 () 1.013 92 0.004∼0.0087 - T s (K) 6430 () 288 737 210 165 (at 1bar) (K)T e f f 5780 254.3 231.7 210.1 110 (K)T eq — 254.3 231.7 210 124 (vol%) H (91.0) N 2 (78.08) CO 2 (96.5) CO 2 (95.32) H 2 (90)He (8.9) O 2 (20.95) N 2 (3.5) N 2 (2.7) He (10)O(78ppm) Ar (0.93) SO 2 (150 ppm) Ar (1.6) CH 4 (0.3)2–3 < 3 MmkT < G2 R k G M m k = 1.38 × 10 −23 J/K. v e 12 v2 e = G M R32 kT < 1 2 mv2 e 2 2-1 -2-2 0.3 254 K 2-3 2-4 7

3. 3–1 ( km) (1221.5) Fe (3480.0) Fe (6350) Mg,Si,O (6371) Si,Al,Ca,O (6371) H,O (>6371) N,O = 13.6 eV 1eV=1.602×10 −19 J=1.727×10 4 K3–2 P, T, () ()8

H 2 O • • 2 • • 3 3 • 3–3 n A A (g) (l)P, T G(P, T, n g , n l ) = min. (3.1)9

n g , n l G G = n g µ g (P, T) + n l µ l (P, T)= (n − n l )µ g (P, T) + n l µ l (P, T)= nµ g (P, T) − n l {µ g (P, T) − µ l (P, T)} (n g = n, n l = 0) µ g < µ l P, T (n g = 0, n l = 0) µ g > µ l µ g = µ l G ≡ U + PV − TS, µ i ≡ u + Pv − T s (TS )• • P, T µ g = µ l (3.2) (3.2) ∂µ∂PT∆s = ∆h ∂µ g∂P δP + ∂µ g∂T δT = ∂µ l∂P δP + ∂µ lδT (3.3)∂T= v,∂µ∂T= −s v g δP − s g δT = v c δP − s c δTdPdTdPdT∣ = ∆seq ∆v∣ = ∆heq T∆v-(3.4)(3.5) ∆v ≈ v g =RT/P dPdT ∣ = ∆heq RT P (3.6)2 ∆h () 10

3–4 () AB A B ABAB AB ⇋ A + B(R3.1)G(P, T, n AB , n A , n B ) = min. (3.7) ξ AB n mol A+B ξ = n G ∂G∂ξ = 0 (3.8) ∂n AB∂ξ= −1, ∂n A∂ξ= 1, ∂n B∂ξ= 1 (3.9) − ∂G∂n AB+ ∂G∂n A+ ∂G∂n B= 0 (3.10)−µ AB + µ A + µ B = 0 (3.11) µ i = µ ◦ i (P 0, T) + RT ln(P i /P 0 ) P A P B= P 0 exp ( )µ ◦ AB −µ◦ A −µ◦ BRTP AB (3.11)= K(T) T x P A : P AB = x : 1 − x (3.5) P = P AB + P A + P BxP = K(T) (3.12)1 − x2 T P () () 3 3-1 ∆h (3.6) T T = T 0 P = P 0 3-2 100 C 1 100 C 2260J g −1 0 C 350 C 11

4. 4–1 Pa=N/m 2 =J/m 3 (c.f. : 1 2 ρv2 + P + ϕ = const) ρ d g P P = ρgd (4.1) zdPdz= −ρg (4.2) z r 4–2 • ()• • ()• ()12

PV = NRT (4.3) N 1 mol µ N = Vρ/µ P = ρ RT (4.4)µ (4.4) (4.2) dPdz = − µgRT P (4.5) ( P = P 0 at z = z 0 )[ ]−µg(z − z0 )P = P 0 expRT(4.6)H = RTµg[ ] −(z − z0 )P = P 0 expH(4.7)(4.8) (4.7) H . 1/e (4.7) mgH = kT (4.8) m 1 R = N A kµ = N A m (N A ) (4.8) 1 dPP dz = − 1H(z)H(z) =RT(z)µ(z)g(z) H (4.9)(4.10)13

4–3 (4.6) z → ∞ P 0 P 1v 2 esc( ) 2 2kTv T =m m 1 • ()ϕ esc ∫ϕ esc = v z f (u)d 3 u (4.13)|u|≥v esc ,v z ≥0 f (m) 3/2f (u) = n exp⎡⎢⎣− m(v2 x + v 2 y + v 2 ⎤z)⎥⎦ (4.14)2πkT2kT14

λ ϕ esc = n v T2π 1/2 (1 + λ)e−λ (4.15)λ = mv2 esc2kT = v2 escv 2 T 4 4-1 a) b) 1 km 10 K 500 hPa 1000 hPa 300 K 30 g mol −1 10 m s −2 4-2 R (1) r(< R) ρ r (2)(1) r (r = R) 0 (3) ρ c ρ m (ρ c > ρ > ρ m )4-3 (1) (4.15) ()(2) 1. a σ = 1 × 10 −19 m 2 15

5. 5–1 dA dt FdAdtF () () F cond = −k dT(5.1)dlF cond l k J K −1 m −1 s −1 k ≈ 1 3 ρc pv T l. (5.2) ρc p (κ) () () 5-2 16

0 K (T > 0) ()() ν ∼ ν + dν = ×1 L n x , n y , n z ( nx πx) ( ny πy) ( nz πz)sin sin sin e i2πνt . (5.3)L L Ln x , n y , n z 1 2 ( 2 )ν = c √n2L2 x + n 2 y + n 2 z (5.4) c ν ∼ ν + dν (n x , n y , n z ) √2Lν/c ≤ n 2 x + n 2 y + n 2 z ≤ 2L(ν + dν)/c. (5.5) (n x , n y , n z ) 2Lν/c 2Ldν/c 1/8 4πL 3 ν 2 dν/c 3 L 8πν 2 dν/c 3 . (5.6)1 ν 1 0, hν, 2hν, 3hν, · · · nhν, · · · () nhν ∝ exp(− nhν )kT < ε ν > ∑ ∞n=0 nhν exp ( )− nhνkT< ε ν >= ∑ ∞n=0 exp ( ) . (5.7)− nhνkT17

ε ν >=hνexp ( ) . (5.8)hνkT − 1 ν ∼ ν + dν U(ν, T)dν = 8π hν 3c 3 exp ( ) dν. (5.9)hν − 1kT5–3 dA dΩ dt ν ∼ ν + dνI ν cos θdAdΩdtdν (5.10) θ dA J m −2 sr −1s −1 Hz −1 = J m −2 sr −1 I ν (radiation intensity) dΩθdAdA ∫ ∫∫F = I ν cos θdΩdν =F ν dν (5.11) F ν = ∫ I cos θdΩ dΩ u ν dA cdtu ν dAcdt. (5.12) dt dA I ν dtdA = u ν dAcdt. (5.13)18

I ν = u ν c (5.14)u ν ∫ I ν dΩ = 4πI ν I ν B ν (T) I ν = c U(ν, T) (5.15)4πB ν (T) = 2 hν 3c 2 exp ( ) . (5.16)hνkT − 1 ρ ds dI ν dI ν = −κ ν ρI ν ds + ρ j ν ds (5.17)κ ν j ν ν () j ν = κ ν B ν (T) (5.18) T dI ν = −κ ν ρI ν ds + ρ j ν ds = 0 (5.19) I ν = B ν (T) (5.18) - 100km 19

5 5-1 - ∫ ∞π 415 0x 3e x −1 dx =5-2 dA dΩ dt λ ∼ λ + dλI λ cos θdAdΩdtdλ (5.20)I λ = B λ (T) =2hc 2λ [ 5 exp ( ) ]. (5.21)hcλkT − 15-3 T (5.21) 3000/T(K) µm 5-4 T = 250 K 6000 K 20

6. 6-1 ( V, A) ∫ ∫dedV = − F · d A (6.1)dt VAe F F · d A > 0 F · d A < 0 c.f. ∫ ∫F = dΩ dν cos θI ν (θ, ϕ)6-2 • ( 4 )• ()• ()• dI ν = −κ ν ρI ν ds + ρ j ν ds (6.2)j ν = κ ν B ν (T) (6.3) I = ∫ I ν dν dI = −κρIds + κρB(T)ds (6.4)21

B(T) B(T) = σT 4π(6.5) z τ(z) =∫ ∞zρκdz. (6.6) (6.4) τ θ ds = dz/ cos θ (6.7) dτ = −κρdz (6.4) κρds = −dτ/ cos θ (6.8)cos θ dIdτ= I − B(T) (6.9)2 (6.9) I τ 2 • I • (0 < θ < π/2) (π/2 < θ < π) I + , I − ∫F up =∫F down = − ∫ dΩ(6.9) × cos θ dF updΩ cos θI = πI + (6.10)dΩ cos θI = πI − (6.11)23 dτ = F up − πB (6.12)− 2 dF down= F down − πB (6.13)3 dτ22

τ ∼ τ + dτ 0 = [] − [] = [F up (τ + dτ) + F down (τ)] − [F up (τ) + F down (τ + dτ)] (6.14)F up (τ + dτ) − F down (τ + dτ) = F up (τ) − F down (τ) (6.15)F up (τ) − F down (τ) = (6.16) (τ = 0) F down = 0F up = F up (τ) − F down (τ) = σT 4 e f f (6.12)+(6.13) (6.16) (6.12)-(6.13) 0 = F up + F down − 2πB(T). (6.17)2 d(F up + F down )3 dτ= σT 4 e f f . (6.18)F up + F down = 3 2 σT 4 e f f(τ + 2 )3(6.19) τ = 0 F up = σT 4 e f f , F down = 0 (6.17) πB(T) = σT(τ) 4 = 1 2 σT e f f( 32 τ + 1 )(6.20) = + σTs 4 = σT 4 e f f + F down(τ total ) (6.21) T s > T(τ total )σT 4 s = 1 2 σT 4 e f f( 32 τ + 2 )(6.22)23

6-3 dQ = 0() 1mol µ V = µ/ρP = ρRT/µ C v dT = −PdV (6.23)C v dT = RTdρ/ρ (6.24) (5.24) dPdz= −ρg (6.25)RdT + RTdρ/ρ = −µgdz (6.26)(C v + R)dT = −µgdz (6.27)dTdz = −µg/C p (6.28)T = T s − µgC pz 6 6-1 (1) 4/3 κ ν 1 10 m s −2 (2) 255 K (3) 24