Astroparticle Physics

338 17 Solutionsv 2 = r 2 ω 2 = G M ♁⇒ r 3 = GM ♁rω 2 .Taking into account that for the geostationary satellite the revolution period T ♁ =1day = 86 400 s and ω ♁ = 2π/T ♁ , one obtains√r = 3 GM ♁ T♁24π 2 ≈ 42 241 km .The altitude above ground level therefore is H = r − R ♁ = 35 871 km.Geostationary satellites can only be positioned above the equator because only therethe direction of the centrifugal force can be balanced by the direction of the gravitationalforce. In other words, the centrifugal force for a geostationary object pointsoutward perpendicular to the Earth’s axis and only in the equator plane the gravitationalforce is collinear to the former.2. The centrifugal force is balanced by the Lorentz force, somv 2ϱ= evB ⇒ p = eBϱ ⇒ ϱ =peB .Since the kinetic energy of the proton (1 MeV) is small compared to its mass (≈938 MeV), a classical treatment for the energy–momentum relation, E = p 2 /2m 0 ,isappropriate from which one obtains p = √ 2m 0 E kin .Then the bending radius is given byϱ = p √eB = 2m0 E kineBDimensional analysis:{ } kg mp = eBϱ, psor ϱ ≈ 2888 m .{ m }c = (pc) {J} =e {As} B {T} ϱ {m} c {ms −1 } ,s(pc) [J]×1.6 × 10 −19 J/eV = 1.6 × 10 −19 × B [T]×ϱ [m]×3 × 10 8 ,(pc)[eV] =3 × 10 8 B[T] ϱ[m] =300 B [Gauss] ϱ [cm](pc) [eV]ϱ =300 B [Gauss] cm = 2.888 × 105 cm .3. From Fig. 4.2 one can read the average energy loss of muons in air-like materials to be≈ 2MeV/(g/cm 2 ). This number can also be obtained from (4.6). The column densityof the atmosphere from the production altitude can be read from Fig. 7.3. It is approximately940 g/cm 2 . Finally, the average energy loss in the atmosphere is− dEdx x ≈ 2MeV/(g/cm2 ) × 940 g/cm 2 = 1.88 GeV .