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 .

v_a = \omega_E r_a

v_p = r_a v_a / r_d = \omega_E { r_a }^2 / r_d

Orbital energy:

{v_a}^2 - 2 \mu / r_a = {v_p}^2 - 2 \mu / r_d

Solve for r_a :

( \omega_E r_a )^2 - 2 \mu /r_a = ( \omega_E {r_a}^2 / r_d )^2 - 2 \mu / r_d

{r_a}^3 - ( 2 \mu / {\omega_E}^2 ) = {r_a}^5 / {r_d}^2 - 2 \mu r_a / ( {\omega_E}^2 r_d )

Divide by {r_d}^3 :

( r_a / r_d )^3 - ( 2 \mu / {\omega_E}^2 {r_d}^3 ) = ( r_a / r_d )^5 - ( 2 \mu / {\omega_E}^2 {r_d}^3 ) ( r_a / r_d )

Normalize for R = r_a / r_d and M = 2 \mu / ( {\omega_E}^2 {r_d}^3 ) .

R^3 - M = R^5 - M * R

R^5 = R^3 + M * (R - 1) ... given M, solve iteratively for R