MIT OpenCourseWare http://ocw.mit.edu 16.346 Astrodynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Exercises 06 The equations describing the hyperbolic locus of velocity vectors are not valid when the transfer angle is θ = 180◦ . However, by expressing the velocity vector components in polar coordinates rather than along skewed-axes, the singularity disappears. 1. Show that the velocity components along skewed axes are related to ordinary polar coordinate components as c v c = vθ1 csc φ1 = vθ1 csc θ r2 vρ = v r1 − vθ1 cot φ1 = v r1 − r2 cos θ − r1 vθ1r2 sin θ Hint: Equate the velocity vector in both sets of coordinates and calculate the scalar product with each of the unit vectors. 2. Derive the equation of the hyperbolic locus of velocity vectors in the form v sin θ − v 2 cos θ − r1 = µ (1 − cos θ)r1vθ1 θ1 r2 r1 and show that it exhibits no difficulties as θ approaches 180 degrees. Refer to Figure
View Full Document
Unlocking...