# Launch azimuth from orbital inclination WRT rotating earth?

The formula for launch azimuth given latitude and target orbital inclination goes like this:

$$azm = \sin^{-1} \left(\frac{\cos(inc)}{\cos(lat)}\right)$$

I assume the orbital inclination $$inc$$ and launch azimuth $$azm$$ are both defined in an inertial frame.

Trouble is... that in an inertial frame, a rocket sitting on the pad already has a sizable eastward velocity imparted by rotation of the earth---some 400 m/s at Kennedy.

That initial velocity vector adds to the delta-v imparted by the rocket engines along the launch azimuth... so that if you point at the $$azm$$ given by the equation above, it seems you'll hit orbit slightly off from your target orbital inclination?

Is this right? How do you account for earth's rotation in your calculation of the launch azimuth so that when you reach orbit you're at just the right orbital inclination?

• I don't know much more on this one, but please look at this link. In order to render the equations correctly, you can copy/paste them in a MathJax supported environment (like a StackExchange draft reply), remove the "math" tags, and replace with  respectively.orbiterwiki.org/wiki/Launch_Azimuth Jan 30 at 16:13
• A weaness of the above explanation is that it "assumes" that the launch azimuth is converted to Interial frame, but it does not show how. This is covered on section 6.4.1 in Vallado's "Fundamentals of Astrodynamics and Applications". In there, the author converts the latitude in the inertial frame, and the resulting azimuth is then expressed in the inertial frame. I will attempt to write a full reply later tonight, I hope those pointers are somewhat helpful, but once again I do not know the topic very well. Jan 30 at 16:20
• Thanks @Manny! This is just what I needed.
– user36480
Jan 31 at 6:43  