This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Due: Friday, March 12, 2010 Name: Solutions Embry-Riddle Aeronautical University AE 313 – Space Mechanics Homework 5 Transfers II - interplanetary 1. In the Heliocentric Ecliptic Inertial coordinate system, Earth and Jupiter have the following coordinates at a certain date: AU R AU R Jupiter Earth ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ = 5.2 1 Assuming Earth and Jupiter are in the ecliptic plane on circular orbits and the parking orbit altitude for the Earth is 300 km. Also using an interplanetary GENERAL transfer with PeriHELION at 1 AU and Apohelion at 6 AU. Given: μ sun = 1.3271E+11 km^3/s^2, μ Jupiter = 1.267E+08 km^3/s^2. a) Draw a diagram of the initial geometry. b) Calculate the semi major axis of the transfer orbit. (ANS: 3.5 AU) a t = (r a + r p )/2 = (6+1)/2 AU = 3.5 AU c) Calculate the time of flight for the interplanetary part (Kepler’s equation) Since we start at Perihelion the initial true, eccentric and mean anomalies are 0 rad. Given the perihelion and apohelion we can calculate the transfer orbit values for a, e, and p. Then we can calculate the anomalies at the given radius of Jupiter of 5.2. This gives: a t = 3.5 AU, e t = 5/7=0.714, p t = 1.714 AU, ν 2 =2.789 rad (159.8°), E 2 = 2.319 rad (132.8°), M 2 = 1.795 rad. TOF = M 2 /n t = 1.795 rad/0.1527 rad/TU = 11.752 TU = 683.2 days , where n t = sqrt(mu sun /a t ^3) = 0.1527 rad/TU d) What is the synodic period for the Earth-Jupiter constellation? Tsyn = 2 π /abs(n 1-n 2 ) = 2 π /abs(1 rad/TU – 0.0842 rad/TU) = 6.861 TU = 399 days e) How many TUs and Days until the next launch possibility? (Hint: 1 TU = 365 days/2 π ) Note that the planets are lined up with one another. Note that the planets are lined up with one another....
View Full Document
- Fall '10
- Velocity, Planet, ΔT, Solutions Embry-Riddle Aeronautical University AE