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.

Starting from the capture time T seconds after construction orbit apogee, we can compute the capture angle L radians to the west of apogee.

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 )

Let's define two new known parameters in terms of the known parameters r_{pl}, r_i, and v_i$ ,

\alpha \equiv { \Large { { v_i^2 r_pl } \over \mu } } ~~~~~~~~~~~~~ hence ( e_l \sin( L ) )^2 = \alpha ( 1 - e_l )

\beta \equiv { \Large { r_pl \over r_i } } ~~~~~~~~~~~~~ hence e_l \cos( L ) = \beta ( 1 - e_l ) - 1

This is enough to create a quadratic equation:

( e_l \sin( L ) )^2 + ( e_l \cos( L ) )^2 ~=~ e_l^2 ~=~ ( \beta - \beta e_l -1 )^2 + \alpha ( 1 - e_l )

0 = ( \beta^2 - 1 ) e_l^2 + ( 2 \beta - 2 \beta^2 - \alpha ) e_l + ( \beta^2 - 2 \beta + \alpha + 1 )

Define the coefficients of the quadratic polynomial:

A \equiv ( \beta^2 - 1 ) ...... B \equiv ( 2 \beta - 2 \beta^2 - \alpha ) ....... C \equiv ( \beta^2 - 2 \beta + \alpha + 1 )

And solve for e_l :

e_l = { { -B + \Large { \sqrt{ B^2 ~-~ 4 A C } } \over { 2 A } } }

The other root of the equation yields e_l = 1 which is nonphysical .

Here's the result of a spreadsheet