Simple Nature - Light and Matter

figure t/2, and we assume that the craft is in a nearly circular orbitaround the sun, hence the 90-degree angle between the directionof motion and the incoming sunlight. We assume that the sail is100% reflective. The orientation of the sail is specified using theangle θ between the incoming rays of sunlight and the perpendicularto the sail. In other words, θ=0 if the sail is catching thesunlight full-on, while θ=90 ◦ means that the sail is edge-on to thesun.Conservation of momentum givesp light,i = p light,f + ∆p sail ,where ∆p sail is the change in momentum picked up by the sail.Breaking this down into components, we have0 = p light,f ,x + ∆p sail,x andp light,i,y = p light,f ,y + ∆p sail,y .As in example 53 on page 189, the component of the force thatis directly away from the sun (up in figure t/2) doesn’t change theenergy of the craft, so we only care about ∆p sail,x , which equals−p light,f ,x . The outgoing light ray forms an angle of 2θ with thenegative y axis, or 270 ◦ −2θ measured counterclockwise from thex axis, so the useful thrust depends on −cos(270 ◦ − 2θ) = sin 2θ.However, this is all assuming a given amount of light strikes thesail. During a certain time period, the amount of sunlight strikingthe sail depends on the cross-sectional area the sail presents tothe sun, which is proportional to cos θ. For θ=90 ◦ , cos θ equalszero, since the sail is edge-on to the sun.Putting together these two factors, the useful thrust is proportionalto sin 2θ cosθ, and this quantity is maximized for θ ≈ 35 ◦ . A counterintuitivefact about this maneuver is that as the spacecraft spiralsoutward, its total energy (kinetic plus gravitational) increases,but its kinetic energy actually decreases!A layback example 71The figure shows a rock climber using a technique called a layback.He can make the normal forces F N1 and F N2 large, whichhas the side-effect of increasing the frictional forces F F 1 and F F 2 ,so that he doesn’t slip down due to the gravitational (weight) forceF W . The purpose of the problem is not to analyze all of this in detail,but simply to practice finding the components of the forcesbased on their magnitudes. To keep the notation simple, let’swrite F N1 for |F N1 |, etc. The crack overhangs by a small, positiveangle θ ≈ 9 ◦ .In this example, we determine the x component of F N1 . The othernine components are left as an exercise to the reader (problem81, p. 236).Section 3.4 Motion In Three Dimensions 205