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$ r_p = \huge { r_a \over { ( 2 \mu / ( {r_a}^3 {\omega_E}^2 ) ) - 1 } } $ $ r_p = \Large { r_a \over { \Large { { \huge { 2 \mu } \over { {r_a}^3 {\omega_E}^2 } } } - 1 } } $

Testing empirically with a spreadsheet:

|| $ r_a $ || $ r_p $ ||
|| 29 000 || 5 634 || smash! ||
|| 29 790 || 6 378 || earth surface ||
|| 30 000 || 6 678 || 300km altitude ||
|| 34 383 || 12 789 || M288 ||
|| 39 267 || 26 600 || GPS ||
|| 42 164 || 42 164 || GEO ||
|| 50 964 || 384 400 || Moon ||
|| 53 120 || inf || Escape ||

So, a 20 iteration binary search of $ r_a $ between 29700 km and 53000 km will converge $ r_p $ on our target $ r_d $

Orbit Circularization

What is the ΔV needed for apogee insertion into a circular equatorial orbit from a launch loop transfer orbit, and for perigee insertion from a space elevator transfer orbit?

The destination orbit has a radius of r_d and a velocity of v_d = \sqrt{ \mu / r_d } were \mu = 398600.4418 km3 / s2.


The launch loop calculation is fairly simple.

An 80 kilometer breech altitude launch loop defines a transfer orbit with a perigee r_p = 6378 + 80 km = 6458 km . The semimajor axis is a = 0.5 * ( r_p + r_d ) , the eccentricity e = ( r_d - r_p ) / ( r_d + r_p ) , the characteristic velocity is v_0 = \sqrt{ \mu / ( a * ( 1 - e^2 ) ) } , and the apogee velocity is v_a = ( 1 - e ) v_0 . Combining and simplifying:

{v_a}^2 = ( 1 - e )^2 {v_0}^2 = ( \mu / a ) ( 1 - e )^2 / ( 1 - e^2 )

{v_a}^2 = ( 2 \mu r_p ) / ( r_d ( r_d + r_p ) )

v_a = \sqrt{ ( 2 \mu r_p ) / ( r_d ( r_d + r_p ) ) }

\Delta V = v_d - v_a


The space elevator calculation is more complicated.

For transfer orbits below geosynchronous, the apogee of the orbit is the release radius r_a , and the angular velocity is \omega_E = 2 \pi / P_{sidereal} , where P_{sidereal} is the sidereal day, 86164.0905 seconds. We want to find r_a and the perigee velocity v_p . We will have to work backwards.

v_a = \omega_E r_a

{v_a}^2 / 2 \mu = 1 / r_a - 1 / ( r_a + r_p )

Solve for r_p , perigee radius, the candidate r_d

r_p = \Large { r_a \over { \Large { { \huge { 2 \mu } \over { {r_a}^3 {\omega_E}^2 } } } - 1 } }

Testing empirically with a spreadsheet:

r_a

r_p

29 000

5 634

smash!

29 790

6 378

earth surface

30 000

6 678

300km altitude

34 383

12 789

M288

39 267

26 600

GPS

42 164

42 164

GEO

50 964

384 400

Moon

53 120

inf

Escape

So, a 20 iteration binary search of r_a between 29700 km and 53000 km will converge r_p on our target r_d

OrbitCirc (last edited 2017-03-13 16:20:29 by KeithLofstrom)