Question

80.A spacecraft is approaching Mars after a long trip from the Earth. Its velocity is such that it is traveling along a parabolic trajectory under the influence of the gravitational force from Mars. The distance of closest approach will be 300 km above the Martian surface. At this point of closest approach, the engines will be fired to slow down the spacecraft and place it in a circular orbit 300 km above the surface. (a) By what percentage must the speed of the spacecraft be reduced to achieve the desired orbit? (b) How would the answer to part (a) change if the distance of closest approach and the desired circular orbit altitude were 600 km instead of 300 km? (Note: The energy of the spacecraft-Mars sys- tem for a parabolic orbit is E = 0)

Answer 1

From conservation of energy, 0 + 0 = 1/2 mv_r^2 - G Mmars^m/r V_r = squareroot 2G Mmars/r Sigma = ma: G Mmars^m/r^2 = mv_0^2/r V_o = squareroot G Mmars/r V_r - V_0/V_r = squareroot 2 V_o - V_o/squareroot 2V_o = squareroot 2 - 1/squareroot 2 times 100 % = 29.3 % The answer to part (a) applies with no changes, as the solution to part (a) shows