Plane Crossing Velocity
Launch loops will be located slightly south of the equator, so they will launch into latitude-inclined transfer orbits, crossing other circular equatorial orbits. What is the relative velocity compared to an object in a circular equatorial orbit?
The launch orbit has a perigee of r_p and an apogee of r_a . Earth's standard gravitational parameter \mu = 398600.44 km³/s². The orbital radius and velocity are:
r = { \Large { { a ( 1 - e² ) } \over { 1 + e \cos( \theta ) } } } |
a = { \Large { r_a + r_p } \over 2 } ~ ~ ~ ~ e = { \Large { { r_a - r_p } \over { r_a + r_p } } } ~ ~ ~ ~ ~ ~ \phi is inclination ~ ~ ~ ~ ~ ~ \theta is orbital angle |
tangential v_t = v_0 ( 1 + e \cos( \theta ) ) \cos( \phi ) |
v_0 = \Large { \sqrt{ { \mu \over 2 } \left( { 1 \over r_a } + { 1 \over r_p } \right) } } |
|| radial v_r = e v_0 \sin( \theta ) || north/south v_z = v_0 ( 1 + e \cos( \theta ) ) \sin( \phi ) $
The transfer orbit crosses the equatorial plane at \theta = \pi/2 = 90 degrees, the semi latus rectum, so the equations simplify to:
r = a ( 1 - e² ) |
v_t = v_0 \cos( \phi ) |
v_r = e v_0 || v_z = v_0 \sin( \phi ) $ |