Orbital Insertion

Kourou is located approximately 500 kilometres (310 mi) north of the equator, at a latitude of 5°10'. Ariane 5 GTO launch azimuth 80 degrees from north. Is the launch azimuth designed to clear the coast (running NW-SE) for safety reasons? The inclination of the GTO orbit is 0.5°

Launch loops will be located below the equator, to avoid storms and hurricanes associated with the Inter Tropical Convergence Zone (ITCZ). Launching due east from a launch loop below the equator places the apogee well north of the equatorial plane; this orbit will cross the equatorial plane at an orbital angle ±90° from perigee.

A previous attempt (OLD_LaunchAzimuth) assumed that launching with an azimuth to the north would help orbital insertion by accomplishing this with one rocket firing. While this is possible for lower target orbits, it adds too much \delta V to the firing - the azimuth must be removed, and the transfer orbit is farther from the ideal Hohmann.

Instead, the loop should launch payloads at zero azimuth, due east. The orbit will still be inclined by the launch latitude, which must be corrected. However, angle changes are easier at lower velocities, that is, high in the orbit.

Since launch loop will throw payloads directly into transfer orbits, launch azimuth should be arranged to put the equatorial plane crossing near apogee as far from earth as possible. For logistical convenience and safety, the launch loop west station should be south of the intertropical convergence zone (ITC) in January, and the whole system should be erected when the rain and wind is minimized. A loop with a 2000 kilometer track and 300 kilometer inclines will be 23 degrees west to east.

What is the optimum azimuth for injection into a geostationary circular equatorial orbit (42164 km radius), or into a server sky M288 circular equatorial orbit (12789 km radius)?

Equatorial Launch, Two Impulse Hohmann

If we launch from the equator, latitude \phi = 0 , the best trajectory is a two impulse Hohmann, with the Launch Loop providing launch velocity v_L at an altitude a_L , radius from the center of the Earth r_L = R_E + a_L .

Start with an azimuth angle ( launch angle deflection to the north) of \alpha = 0 and zero elevation angle, parallel to the earth's surface (or close to parallel).

Ignoring second order effects, the launch vehicle will follow an elliptical transfer orbit to an apogee at the destination orbit, and require one velocity change \Delta v_d to enter the destination orbit. We want to minimize \Delta v_d , because that scales the size of the rocket engine needed at apogee.

A Newtonian orbit with a perigee of r_L and an apogee of r_P has a semimajor axis r_S = { \Large { { r_P + r_L } \over 2 } } and an eccentricity e = { \Large { { r_P - r_L } \over { r_P + r_L } } } . Note that r_P may differ from r_D if the transfer orbit is not equatorial and coplanar.

Let \theta be the angle from perigee; the elliptical equatorial transfer orbit will intersect the destination orbit where \theta = \pi .

The orbit radius is r = { \Large { { ( 1 + e ) r_L } \over { 1 + e \cos( \theta ) } } } .

The orbit's characteristic velocity is \large v_0 = { \Large { \sqrt{ { \mu \over 2 } \left( { 1 \over r_L } + { 1 \over r_P } \right) } } } .

Earth's standard gravitational parameter is \mu = 3.9860044e14 m³/s².

The velocity as a function of angle is v = v_0 ( 1 + e \cos( \theta ) , so perigee velocity (or absolute launch velocity) v_L = v_0 ( 1 + e ) and apogee velocity v_A = v_0 ( 1 - e ) .

The perigee velocity is the exit velocity relative to the loop vector-summed with the Earth rotation velocity at the exit latitude, 7.292115e-5 radians/sec \times r_L \cos( \phi ) . For a loop exit altitude of 120 km, and an exit latitude of 5 degrees south, r_L = 6498 km and the rotation velocity at that altitude is 472 m/s.

This will get a drawing later

• x is the direction of the perigee at the equator, pointing out of the page, in the equatorial plane.

• y is the direction 90 degrees east of perigee orthogonal to x, pointing right on the page, in the equatorial plane.

• z is north, pointing up on the page, orthogonal to x and y,and the equatorial plane.

The planar equatorial orbit has a direction vector of:

\left[\matrix{ r_x \\ r_y \\ r_z }\right] ~ = ~ \left[\matrix{ \cos( \theta ) \\ \sin( \theta ) \\ 0 }\right] { \Large { { (1 + e) r_L } \over { 1 + e \cos( \theta ) } } } ~ = ~ { \Large \left[\matrix{ { { (1 + e) r_L \cos( \theta ) } \over { 1 + e \cos( \theta ) } } \\ { { (1 + e) r_L \sin( \theta ) } \over { 1 + e \cos( \theta ) } } \\ 0 }\right] }

And a velocity vector of:

\left[\matrix{ v_x \\ v_y \\ v_z }\right] ~ = ~ \left[\matrix{ - \sin( \theta ) \\ \cos( \theta ) \\ 0 }\right] { v_0 ( 1 + e \cos( \theta ) ) } ~ = ~ \left[\matrix{ - v_0 ( 1 + e \cos( \theta ) ) \sin( \theta ) \\ v_0 ( 1 + e \cos( \theta ) ) \cos( \theta ) \\ 0 } \right]

Rotation by Launch Latitude

The Launch Loop exit will be in the southern hemisphere at latitude \phi , inclined towards the north, to avoid hurricanes and the inter tropical convergence zone (ITCZ). The eastern exit end of the loop can tolerate more weather than the western elevator end. Wind loads determine the area-to-mass scaling and the operating schedule of the loop, so less wind is better.

Since the loop is to the south, the equatorial orbit is inclined by the latitude \phi . That will put the perigee well north of the destination equatorial orbital plane. To compute the new position and velocity vectors, we rotate both components around the y axis by an angle of \phi . Another matrix multiplication in front of the planar rotation:

\left[\matrix{ r_x \\ r_y \\ r_z }\right] ~=~ \left[\matrix{ \cos( \phi ) & 0 & - \sin( \phi ) \\ 0 & 1 & 0 \\ \sin( \phi ) & 0 & \cos( \phi ) }\right] \left[\matrix{ \cos( \theta ) \\ \sin( \theta ) \\ 0 }\right] { \Large { { (1 + e) r_L } \over { 1 + e \cos( \theta ) } } }

\left[\matrix{ v_x \\ v_y \\ v_z }\right] ~=~ \left[\matrix{ \cos( \phi ) & 0 & - \sin( \phi ) \\ 0 & 1 & 0 \\ \sin( \phi ) & 0 & \cos( \phi ) }\right] \left[\matrix{ - \sin( \theta ) \\ \cos( \theta ) \\ 0 }\right] { v_0 ( 1 + e \cos( \theta ) ) }

Reducing the Matrices to a Unit Vector

\left[\matrix{ r_x \\ r_y \\ r_z }\right] ~=~ \left[\matrix{ \cos( \theta )\\ \cos( \alpha ) \sin( \theta ) \\ \sin( \phi ) \cos( \theta ) + \cos( \phi ) \sin( \alpha ) \sin( \theta ) }\right] { \Large { { (1 + e) r_L } \over { 1 + e \cos( \theta ) } } }

\left[\matrix{ v_x \\ v_y \\ v_z }\right] ~=~ \left[\matrix{ -\cos( \phi ) \sin( \theta ) - \sin( \phi ) \sin( \alpha ) \cos( \theta ) \\ \cos( \theta ) \\ -\sin( \phi ) \sin( \theta ) }\right] { v_0 ( 1 + e \cos( \theta ) ) }

One delta V to the Destination Orbit

The destination orbit has radius r_D equal to the vector sum of r_x and r_y , with r_z = 0 . That tells us a lot; we know that the bottom term of the distance unit vector is zero:

0 ~=~ \sin( \phi ) \cos( \theta ) + \cos( \phi ) \sin( \alpha ) \sin( \theta )

and

r_D ~=~ { \Large { { (1 + e) r_L } \over { 1 + e \cos( \theta ) } } }

Solving the penultimate equation for \theta_D :

\cos( \phi ) \sin( \alpha ) \sin( \theta_D ) ~=~ - \sin( \phi ) \cos( \theta_D )

\Large { { \sin( \theta_D ) } \over { \cos( \theta_D ) } } ~=~ - { { \sin( \phi ) } \over { \cos( \phi ) } } { 1 \over \sin( \alpha ) }

\theta_D ~=~ \arctan\left( \Large { { -\tan( \phi ) } \over { \sin( \alpha ) } } \right) ... this has two solutions, choose the solution closest to \pi such that \cos( \theta_D ) < 0 .

Now, given \theta_D and r_D and r_L , solve for e :

e ~=~ { \Large { { r_D - r_L } \over { r_L - \cos( \theta_D ) r_D } } }

Target orbit velocity at insertion

If the circular equatorial target orbit has a radius of r_D , then we can easily compute the orbital velocity v_D :

v_D ~=~ \sqrt{ \Large{ \mu \over r_D } }

Define \delta as the angle of the r_D vector at insertion; remember that z = 0 :

\left[\matrix{ r_{Dx} \\ r_{Dy} }\right] ~=~ \left[\matrix{ \cos( \delta ) \\ \sin( \delta ) }\right] r_D

\left[\matrix{ v_{Dx} \\ v_{Dy} }\right] ~=~ \left[\matrix{ -\sin( \delta ) \\ \cos( \delta ) }\right] v_D

What is \delta ? We know that

\left[\matrix{ r_{Dx} \\ r_{Dy} }\right] ~=~ \left[\matrix{ r_x \\ r_y }\right]

\left[\matrix{ \cos( \delta ) \\ \sin( \delta ) }\right] r_D ~=~ \left[\matrix{ \cos( \phi ) \cos( \theta_D ) - \sin( \phi ) \sin( \alpha ) \sin( \theta_D ) \\ \cos( \alpha ) \sin( \theta_D ) }\right] { \Large { { (1 + e) r_L } \over { 1 + e \cos( \theta_D ) } } }

But we already know that: r_D ~=~ { \Large { { (1 + e) r_L } \over { 1 + e \cos( \theta_D ) } } }

So we can divide both sides to get:

\left[\matrix{ \cos( \delta_D ) \\ \sin( \delta_D ) }\right] ~=~ \left[\matrix{ \cos( \phi ) \cos( \theta_D ) - \sin( \phi ) \sin( \alpha ) \sin( \theta_D ) \\ \cos( \alpha ) \sin( \theta_D ) }\right]

Ah! Two unit vectors, so the components of each must match! Equate the y terms:

\sin( \delta ) ~=~ \cos( \alpha ) \sin( \theta_D )

so

\delta ~=~ \arcsin( ~ \cos( \alpha ) \sin( \theta_D ) ~ ) ~~~~~~ Again, there are two solutions for \delta , so choose the solution closest to \pi , so that \cos( \delta ) < 0 .

And so the velocity change vector \Delta V is:

\Delta V ~=~ \left[\matrix{ v_{Dx} \\ v_{Dy} \\ 0 }\right] - \left[\matrix{ v_x \\ v_y \\ v_z }\right] ~=~ \left[\matrix{ -\sin( \delta ) \\ \cos( \delta ) \\ 0 }\right] v_D ~-~ \left[\matrix{ -\cos( \phi ) \sin( \theta_D ) - \sin( \phi ) \sin( \alpha ) \cos( \theta_D ) \\ \cos( \alpha ) \cos( \theta_D ) \\ -\sin( \phi ) \sin( \theta_D) + \cos( \phi ) \sin( \alpha ) \cos( \theta_D ) }\right] { v_0 ( 1 + e \cos( \theta_D ) ) }

What a complicated mess! However, starting with gravitational parameter \mu, launch loop exit radius r_L, destination radius r_D, latitude \phi, and a guess for launch azimuth \alpha, we have already calculated all the terms in the equation: velocities v_0 and v_D, transfer orbit angle \theta_D, transfer orbit eccentricity e, and and destination orbit angle \delta, so we can compute thrust vector \Delta V and its magnitude (which we hope to minimize). At this point, I punt, and shove all the parameters and calculations into a spreadsheet, then manually tweak the launch azimuth \alpha until I minimize the magnitude of \Delta V. I hope a better mathematician can give me a closed form equation for the optimum!

MoreLater