Question

(1s equal to pt) A s spacecraft of mass m is in a circular orbit around the Sun with the radius R average Earth-Sun distance. At point A, the spacecraft is firing its thrusters Sun, to point E the tangentially to the closer to the s e trajectory such that it will fall on an elliptical orbit, which will bring it un, to point B. At point B, it once again fires its thrusters tangentially to its ry to go into a new circular orbit with the radius R, -0.4R,.The Sun adhere to the following notation as you address the upcoming Take v as the speed on the large circle as the speed at A after the first at B prior to the second "boost" and v, as the speed on the final mass is M. Please boost", v, as circular orbit. (a) In terms of v, what are v, V, and v,? EXplai (b) In terms of the energy E, of the initial circular orbit, what is tihe change in energy of swers AE, and after the second man the spacecraft's orbit after the first maneuver, ? Explain your answers.

Answer 1

Radius of initial circular orbit = R_A Radius of final circular orbit, R_B = 0.4 R_A therefore semi-major axis of the transfer orbit a = R_A + R_B/2 = R_A + 0.4 R_A/2 = 0.7 R_A v_1 = speed on the large circle v_1 = squareroot G M/R_A where M = Mass of earth Energy of the satellite in transfer orbit, E = -G M m/2a = -G M m/1.4 R_A where m = mass of satellite therefore (Kinetic energy + potential Energy) at point A in transfer orbit = E 1/2 m v^2_A - G M m/R_A = -G M m/1.4 R_A v^2_A = 2 G M (1/r_A - 1/1.4 R_A) = 2 G M .4/1.4 R_A V_A = squareroot 4 G M/7 R_A = > V_A = 2 squareroot G M/7 R_A similarly at point B 1/2 m v^2_B - G M m/0.4 R_A = -G M m/1.4 R_A V^2_B = 2 G M m (1/0.4 R_A - 1/1.4 R_A) = 25