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differential equation applications 229Here r is the planet-CM distance, G is the universal gravitational constant, and theattractive force lies along the line connecting the planet and the sun (Figure 9.10left). The hard part for Newton was solving the resulting differential equationsbecause he had to invent calculus to do it and then had go through numerousanalytic manipulations. The numerical solution is straightforward since even forplanets the equation of motion is stillwith the force (9.78) having components (Figure 9.10)f = ma = m d2 xdt 2 , (9.79)f x = F (g) cos θ = F (g) x r , (9.80)f y = F (g) sin θ = F (g) y r , (9.81)r = √ x 2 + y 2 . (9.82)The equation of motion yields two simultaneous second-order ODEs:d 2 xdt 2 = −GM x r 3 ,d 2 ydt 2 = −GM y r 3 . (9.83)9.17.1 Implementation: Planetary Motion1. Assume units such that GM =1and use the initial conditionsx(0)=0.5, y(0)=0, v x (0)=0.0, v y (0)=1.63.2. Modify your ODE solver program to solve (9.83).3. You may need to make the time step small enough so that the elliptical orbitcloses upon itself, as it should, and the number of steps large enough suchthat the orbits just repeat.4. Experiment with the initial conditions until you find the ones that produce acircular orbit (a special case of an ellipse).5. Once you have obtained good precision, note the effect of progressivelyincreasing the initial velocity until the orbits open up and become hyperbolic.6. Using the same initial conditions that produced the ellipse, investigate theeffect of the power in (9.78) being 1/r 4 rather than 1/r 2 . You should find thatthe orbital ellipse now rotates or precesses (Figure 9.10). In fact, as you shouldverify, even a slight variation from an inverse square power law (as arisesfrom general relativity) causes the orbit to precess.−101COPYRIGHT 2008, PRINCET O N UNIVE R S I T Y P R E S SEVALUATION COPY ONLY. NOT FOR USE IN COURSES.ALLpup_06.04 — 2008/2/15 — Page 229