# 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 ) \Delta v_t = v_0 ( 1 - \cos( \phi ) ) v_z = v_0 \sin( \phi ) v_r = e v_0

These simplify to:

 r = { \Large { { 2 r_a r_p } \over { r_a + r_p } } } v_0 = \sqrt{ \Large { \mu \over r } } \Delta v_t = v_0 ( 1 - \cos( \phi ) ) v_z = v_0 \sin( \phi ) v_r = e v_0

Note that v_0 is the circular orbit velocity at radius r .

Assume a launch loop at 5° south latitude with a altitude of 100 km and a perigee radius of 6478 km. What are the equator crossing velocities relative to circular orbits?

 Destination plane crossing \DeltaV compared to circular radius r_0 v_0 \Delta v_t v_z v_r km km km/s km/s km/s km/s ISS 6800 6635 7.7508 0.0295 0.6755 0.1880 M288 12770 8596 6.8097 0.0259 0.5935 2.2260 M320 13532 8762 6.7449 0.0257 0.5879 2.3777 O3B 14420 8940 6.6773 0.0254 0.5820 2.5376 GPS 26538 10414 6.1867 0.0235 0.5392 3.7590 GEO 42164 11231 5.9576 0.0227 0.5192 4.3707 Moon 384400 12741 5.5932 0.0213 0.4875 5.4078

Note that lunar r_0 comes pretty close to M288, the original server sky orbit. It may be better to change the server sky orbit to M320 (R=2.122 Re), which moves it farther from the proton belt, LAGEOS, and the lunar crossing, and provides more sunshine. That makes the orbital repeat cycle 9 overhead passes in two days, rather than 10.

Note also that the numbers differ from 12789 km for M288 and 14441 km for O3B/M360. Brain fail, but this may be the J₂ oblateness effect. If not, I've got a lot of pages to correct.

PlaneCrossV (last edited 2017-03-25 06:55:45 by KeithLofstrom)