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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
