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_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 } } |
v_t = v_0 \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 \Delta V compared to circular |
|||||
|
radius |
r_0 |
v_0 |
v_t |
v_z |
v_r |
|
km |
km |
km/s |
km/s |
km/s |
km/s |
ISS |
6800 |
6635 |
7.7508 |
7.7213 |
0.6755 |
0.1880 |
M320 |
13532 |
8596 |
6.8097 |
6.7838 |
0.5935 |
2.2260 |
O3B |
14420 |
8762 |
6.7449 |
6.7192 |
0.5879 |
2.3777 |
GPS |
26538 |
8940 |
6.6773 |
6.6519 |
0.5820 |
2.5376 |
GEO |
42164 |
10414 |
6.1867 |
6.1632 |
0.5392 |
3.7590 |
Moon |
384400 |
12741 | | 5.5932 |
5.5719 |
0.4875 |
5.4078 |
|