1. Approximating ODES [40 points] Imagine you dug a tunnel straight through the center of the earth to the other side, and jumped in. You will experience acceleration due to gravity (g), but this will decrease as you approach the center of the planet (ignore the molten core) and then start increasing again as you start to move to the surface on the other side. You will also experience a drag force proportional to your velocity in the direction opposite your motion. The governing equation of motion is expressed by the following ODE for velocity: dv dt m- msign(r)g(r) - sign(v)cav², where vis velocity (m/s); t = time (s); g(r) is the changing gravitational constant (m/s²) as a function of height above the Earth's center (r, in m); sign(x) is a function that returns -1, 0, and 1 for negative, zero, and positive values of x, respectively (you can use the Matlab function sign() for this); ca is the drag coefficient (kg/m); and m is mass (kg). Acceleration due to gravity is given by g(r) = 4 Gpr 4π 3 where G = 6.674 x 10-11 m³kgs-2 is the gravitational constant and p = average density of the Earth. 5515 kg/m³ is the In both cases, calculate the position and velocity of the jumper for t = 0 to 3 x 10's (a bit over 80 hours) given the following parameters: m = 65.8 kg, and cd = 0.25 kg/m. Assume that the initial height is the radius of the Earth, 6.371 x 100 m, and zero initial velocity. Hint: Start by writing the equation in terms of position. Could we maybe turn that into..a system? a) Try using the midpoint method (i.e., modified Euler method) with a step size of 0.5 s. (Perhaps just solve to an end time of 100 s.) Can you obtain a reasonable solution? What does this tell you? b) Now solve this using ode45. Plot the resulting position and velocity as a function of time. Interpret the behavior-does this seem reasonable? Explain. Also, what is the range of time-step sizes that the integrator chose?

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