# Calculating Apollo launch azimuth

The launch azimuth for Apollo missions to the general area of the lunar equator was approximately 70 degrees. Could someone explain in simple terms how this figure is arrived at?

I presume that if the moon and earth equators were both aligned with the ecliptic plane and your trip was from equator to equator you would use a 90 degree azimuth. If so, is the 70 degree Apollo figure just this 90 degrees adjusted for the actual alignment of the two bodies?

I know this answer is a couple of years late, but I hope you will find it helpful.... The stuff in the 2nd paragraph of your question (beginning with 'I presume....') is not correct.

The moon is (or was, at the time of Apollo) in an orbit around the earth that had an inclination of approximately 28.5 degrees. Therefore, in a month's time (actually about 28 days) the moon would move in a 'great circle' in a path between the 28 deg North and 28 deg South latitude.

At the same time, the earth rotates on its axis at approximately 15 degrees per hour West-to-East. Therefore, over short time periods, the moon appears to move East-to-West over the surface of the earth at an essentially fixed latitude.

For the Apollo lunar flights, a minimum-energy-type trajectory was employed that required injecting the spacecraft into translunar trajectory at a point very near 180 degrees around the earth from the moon's position. This point is called the 'lunar antipode'. It forms the 'perigee' of the translunar trajectory, with 'apogee' out near the moon.

The idea was for the Translunar Injection to occur after about 1 3/4 orbits of the Earth. Therefore, at this time (the time of Translunar Injection) the spacecraft's orbital trajectory had to intercept the lunar antipode. The planned time of liftoff was such that on a 72 degree launch azimuth, the orbital trajectory of the spacecraft would intercept the antipode at the right time. 72 degrees was the lowest launch azimuth allowed by the Air Force Range Safety people at the time.

If launch was delayed, the lunar antipode moved further to the West on the Earth's surface. In order to intercept the antipode at a 'new' Translunar Injection time (i.e., the planned time plus the time of launch delay) a new launch azimuth was necessary. As launch azimuth was increased, the orbital track of the spacecraft would intercept the lunar antipode at progressively later times, at a point progressively further west on the surface of the earth.

The above is somewhat simplified, but it contains the essence of the answer to your question.

Thank you very much.

This document says (in rather general terms, as you requested)

This flight azimuth, dependent on the launch time, launch day and month, is calculated using polynomial coefficients taken from the guidance presettings in order to achieve the desired translunar targeting parameters. The translunar targeting parameters are functions of the moon position, earth parking orbit inclination, earth-moon distance, and moon travel rate.

One important contributing parameter is the inclination of the Earth, around 23 degrees (this causes seasons). Without that inclination, the launch azimuth would be around 90. Other contributing factors are of course the plane in which the Moon moves, and more.