Construction2

The older rendezvous page, Construction1, started from a launch time and an exact associated longitude; calculations were complicated and the radial velocity at capture was large.

A better approach is to start with the capture angle at the construction orbit, which defines an exact radius and radial velocity. From there, find a launch orbit with an apogee, launch angle, and launch time that arrives at the construction orbit with the same radius and arrival velocity. The tangential velocity will be different, on the order of 100 m/s, but we can accommodate that with a "passive net capture system".

So, given the construction orbit, choose an angle from apogee \theta . Construction orbits have low perigees and very high apogees, far beyond GEO, so the apogee velocity is small and the angular velocity very small near apogee. Given a desired arrival time t_{ac} referenced from t_{ac} = 0 at apogee, we can estimate \theta given the apogee radius r_{ac} and apogee velocity v_{ac}

The construction orbit perigee should be well above LEO; assume an altitude of 2000 km above the equatorial radius 6378 km, thus r_{pc} = 8378 km. The construction orbit semimajor axis is defined from the orbit period, which should be a multiple N of an 86164.0905 second sidereal day. From that, we can compute the apogee radius, velocity, and angular velocity \omega_{ac} , then iterate from the desired arrival time to the exact capture angle.

Note that the estimate is pretty good, off by 0.000154 radians for one hour delay and the 1 sidereal day construction orbit; that corresponds to an earth rotation of 2.11 seconds, or about 1 kilometer of distance. Good for capability estimation, too much error for accurate aiming from the loop, which will need precise (millisecond) timing.

Other tables can be constructed for other vehicle capture times.

Given the launch orbit perigee ( r_{pl} = 6378 km + 80 km loop altitude), the intercept radius r_i, and the intercept radial velocity v_i, we can algebraically compute the launch orbit perigee r_{al} and the angle between intercept and launch perigee L . From those parameters, we can compute the semimajor axis a_l, the eccentricity e_l, the period P_l, then the launch time and launch angle.

The radius and the radial velocity at angle L are given by:

r_i = { \Large { { ( 1 - e_l^2 ) a_l } \over { 1 + e_l \cos( L ) } } } ~~~~~~~~~~~~~~~~~~~~~~~~~ v_i = { \Large \sqrt{ \mu \over { ( 1 - e_l^2 ) a_l } } } \sin( L )

Define D = ( 1 - e_l^2 ) a_l , invert the first equation, and square the second equation. The equations become:

{ \Large { 1 \over r_i } } = { \large { { 1 + e_l \cos( L ) } \over D } } ~~~~~~~~~~~~~~~~~~~~~~~~~ v_i^2 = { \Large { \mu \over D } } \sin( L )^2

Move D to the numerators and e_l , and \mu to the denominators:

{ \Large { D \over { r_i ~ e_l } } } = { \Large { 1 \over e_l } } + \cos( L ) ~~~~~~~~~~~~~~~~~~~~~~~~~ { \Large { { D ~ v_i^2 } \over \mu } } = \sin( L )^2

Move the { \Large { 1 \over e_l } } term in front, and square the first equation:

\left( \Large { { D \over { r_i ~ e_l } } ~-~ { 1 \over e_l } } \right) ^2 = \cos( L )^2 ~~~~~ And now add them:

\left( \Large { { D \over { r_i ~ e_l } } ~-~ { 1 \over e_l } } \right) ^2 + { \Large { { D ~ v_i^2 } \over \mu } } = \cos( L )^2 + \sin( L )^2 ~=~ 1 ~~~~~~~ so: ~ \left( \Large { { D \over { r_i ~ e_l } } ~-~ { 1 \over e_l } } \right) ^2 + \left( { \Large { { D ~ v_i^2 } \over \mu } } ~-~ 1 \right) ~=~ 0

Move e_l up (because it could become zero): ~ \left( { \Large { D \over r_i } } ~-~ 1 \right) ^2 ~+~ \left( { \Large { { D ~ v_i^2 } \over \mu } } ~-~ 1 \right) e_l^2 ~=~ 0

Hooray! We are left with an equation with unknowns a_i and e_i , which are both functions of r_{al} , our only remaining unknown. Lets see if we can clean up this dog's breakfast of terms. Start with the definitions for a_l , e_l , then refactoring D :

a_l ~=~ 0.5 ( r_{pl} + r_{al} ) ~~~~~~~~~~~~ e_l ~=~ { \Large { { r_{pl} - r_{al} } \over { r_{pl} + r_{al} } } } ~~~~~~~~~ D ~=~ { \Large { { ( 1 - e_i ) a_l ( 1 + e_i ) a_l } \over a_l } } ~=~ { \Large { { 2 ~ r_{pl} ~ r_{al} } \over { r_{pl} + r_{al} } } }

Since both D and e_l have r_{pl} +r_{al} in the denominators, if we multiplied out all the denominators we might end up with a fifth order polynomial. At this point I will punt and use a spreadsheet with trial and error to search for zero, then automate the trial and error process with a computer program.

To be continued ...

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