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Revision 5 as of 2017-03-23 22:37:32
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Revision 6 as of 2017-03-23 22:41:02
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|| 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 ||
|| M288 || 12770 || 8596 || 6.8097 || 6.7838 || 0.5935 || 2.2260 ||
|| M320 || 13532 || 8762 || 6.7449 || 6.7192 || 0.5879 || 2.3777 ||
|| O3B || 14420 || 8940 || 6.6773 || 6.6519 || 0.5820 || 2.5376 ||
|| GPS || 26538 || 10414 || 6.1867 || 6.1632 || 0.5392 || 3.7590 ||
|| GEO || 42164 || 11231 || 5.9576 || 5.9349 || 0.5192 || 4.3707 ||
|| Moon || 384400 || 12741 || 5.5932 || 5.5719 || 0.4875 || 5.4078 ||

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

M288

12770

8596

6.8097

6.7838

0.5935

2.2260

M320

13532

8762

6.7449

6.7192

0.5879

2.3777

O3B

14420

8940

6.6773

6.6519

0.5820

2.5376

GPS

26538

10414

6.1867

6.1632

0.5392

3.7590

GEO

42164

11231

5.9576

5.9349

0.5192

4.3707

Moon

384400

12741

5.5932

5.5719

0.4875

5.4078

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