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.)

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

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