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| || $ 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 || | || $ 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 || |
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| || $ r = a ( 1 - e² ) $ || $ v_t = v_0 \cos( \phi ) $ || $ $ v_r = e v_0 $ || $ v_z = v_0 \sin( \phi ) $ || | || $ 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? ||<-2> Destination ||<-5> 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 || |
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 |
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radius |
r_0 |
v_0 |
v_t |
v_z |
v_r |
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km |
km |
km/s |
km/s |
km/s |
km/s |
ISS |
6800 |
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