At a given instant t(0), a 1000kg earth orbiting satellite has the inertial position and velocity vectors, r(0) = 3207i + 5459j + 2714k (km) and v(0) = -6.532i + 0.7835j + 6.142k (km/s). solve eq 2.22 numerically to fin the maximum altitude reached by the satellite and the time at which it occurs.

equation 2.22: r'' = -(u/r^3)(r)

u = gravitational parameter = g(m1 + m2)

r = position vector

(ans: using matlab's ode45 function, distance = 456,500km, speed = 5km/s.)