From the triangle formed by go, g, and (22Rcos L, we have g sin &=02 Rcos Lsin L=(32R/2)sin2L.We expect 8 to be small, and, therefore sin d A 0, and g A go. Thus ArE which is maximum when L =+45. In this case, we have Q2=7. 29(10-)rad /s, Re=6370 km, and 6max=17(10-3)rad≈0.1° We now consider a couple of three dimensional examples. In the first example, the motion is known, and we are asked to determine the forces required to obtain that motion. In the second example, the motion is unknown and the trajectory needs to be obtained by integrating the equation of motion. Example Aircraft flying at constant velocity We now consider an aircraft A, flying with constant velocity v= UNeN Uwew uvey relative to the surface of the earth. We assume our inertial observer to be at the center of the earth. and our accelerated observer to be at the aircraft(e. g. A= B) y The angular velocity of the Earth, n, can be expressed as n=S cos LeN +Ssin Leu. The aerodynam force, R, that an aircraft is required to generate in order to maintain its course is R+mgo-maB-2mn XUNWU =0 Since aB=92×(!xTB),andg=go-×(g×rB)≈-geU, we have, R=-2mQww sin LeN +2mQ(UN sin L-vU cos L)ew+(mg+ 2mnuw cos L)er For instance, for an aircraft to fly horizontally (i.e. vu =0), it will require a horizontal force, RH 2mQsin L(-WW eN +UN ew). We see that for L>0(northern hemisphere), this force is always to the"left of the aircraft, and needs to be generated aerodynamically in order to maintain a straight path. The reverse true in the southern hemisphere.From the triangle formed by g0, g, and Ω2R cosL, we have g sin δ = Ω2R cosL sinL = (Ω2R/2) sin 2L. We expect δ to be small, and, therefore, sin δ ≈ δ, and g ≈ g0. Thus, δ ≈ Ω 2Re 2g sin 2L , which is maximum when L = ±45o . In this case, we have Ω = 7.29(10−5 ) rad/s, Re = 6370 km, and δmax = 1.7(10−3 ) rad ≈ 0.1 o . We now consider a couple of three dimensional examples. In the first example, the motion is known, and we are asked to determine the forces required to obtain that motion. In the second example, the motion is unknown and the trajectory needs to be obtained by integrating the equation of motion. Example Aircraft flying at constant velocity We now consider an aircraft A, flying with constant velocity v = vN eN + vW eW + vU eU relative to the surface of the Earth. We assume our inertial observer to be at the center of the Earth, and our accelerated observer to be at the aircraft (e.g. A ≡ B). The angular velocity of the Earth, Ω, can be expressed as Ω = Ω cosL eN + Ω sinL eU . The aerodynamical force, R, that an aircraft is required to generate in order to maintain its course is R + mg0 − maB − 2mΩ × (v)NW U = 0 . Since aB = Ω × (Ω × rB), and g = g0 − Ω × (Ω × rB) ≈ −geU , we have, R = −2mΩvW sinL eN + 2mΩ(vN sinL − vU cosL) eW + (mg + 2mΩvW cosL) eU . For instance, for an aircraft to fly horizontally (i.e. vU = 0), it will require a horizontal force, RH = 2mΩ sinL(−vW eN +vN eW ). We see that for L > 0 (northern hemisphere), this force is always to the “left” of the aircraft, and needs to be generated aerodynamically in order to maintain a straight path. The reverse is true in the southern hemisphere. 6