Astroparticle Physics

356 17 SolutionsSect. 6.51. E = GMm γREE= GMc 2for H ≪ R.2. a) P ≈ G c 2 ω6 m 2 r 4 .− GMm γR + H = GMm HγR(H+ R) ,HR(H+ R) = GMR 2 c 2HRH + R = g ⊙c 2For ω = 1/year, m = 5.97 × 10 24 kg, r = 1 AU one getsP ≈ 1.34 × 10 22 W .HRH + R ≈ g ⊙c 2 H ≈ 3 × 10−12Even though this power appears rather large, it is only a fraction of about 3 × 10 −5of the solar emission in the optical range.b) Assume ω = 10 3 s −1 , m = 10 kg, r = 1 m which leads toP ≈ 7.4 × 10 −8 W .17.7 Chapter 71. p = F A = mgA = 1.013 × 105 N m 2 ,mA = p g=1.013 × 1059.81kgm 2 ≈ 10 326 kg/m2 ≈ 1.03 kg/cm 2 .2. p = p 0 e −20/7.99 m 8.29 × 103 kg kg≈ 82.9hPa, ≈ ≈ 845A 9.81 m2 m 2 = 84.5g/cm2 .3. The differential sea-level muon spectrum can be parameterized by∫N(E)dE ∼ E −γ dE with γ = 3 ⇒ I(E) = N(E)dE ∼ E −2 ,where I(E) → 0forE →∞. The thickness of the atmosphere varies with zenith angleliked(θ) = d(0)cos θ .For ‘low’ energies (E