# Why do rockets launching to the ISS from Florida point at 45 degrees to reach a 51.6 degree inclination?

In the Everyday Astronaut video about roll programs, it was mentioned in passing that in order to launch to a 51.6 degree inclination (to rendezvous with the ISS), the rocket must point at about 45 degrees.

He mentioned the answer had to do with the rotation of the Earth and spherical trigonometry. So, why is this?

When launching from the Kennedy Space Center at a latitude of about 28.6 degrees, The Everyday Astronaut Tim Dodd says:

If you want to go to 51.6 degrees and rendez-vous with the International Space Station, you don’t actually point at 51.6 degrees, you actually point at about 45 degrees.

cued at 07:56:

• Interesting question! I've added a link and cued at the quote to make it easier for others to find the relevant bit, and added a block quote for those don't want to.
– uhoh
Apr 23, 2020 at 20:58
• orbiterwiki.org/wiki/Launch_Azimuth & orbiter-forum.com/showpost.php?p=51004&postcount=5 explains it all, if someone wants to make an answer of it. I'm going to bed right now, I won't do it before tomorrow. Apr 23, 2020 at 21:22
• there are also some potentially helpful equations here and here
– uhoh
Apr 23, 2020 at 22:08
• "You had me at 'spherical trigonometry' " :-) . Apr 24, 2020 at 11:15

# The $$51.6^\circ$$ inclination of the ISS's orbit is measured from the equator, not from the latitude of the launch site (for KSC, $$28.6^\circ$$).

Intuitively, as the orbit moves toward a higher latitude, the angle it makes with those lines of latitude decreases, until it is parallel with them, then tilts negative, heading to lower latitudes, crossing the equator, and then doing the same in reverse. This is just what happens when you project a circular(ish) inclined orbit onto the latitude/longitude coordinate system.

This is easiest to see on a picture. Annotating the diagram from this answer should make the situation clear:

As the trajectory moves northward, the angle it subtends with lines of latitude decreases.

# How do we math this?

The launch angle is given by the following simple formula, with $$i$$ the inclination of the orbit, $$\phi$$ the launch site latitude, and $$\beta$$ the orbital inclination: $$\beta = \text{arccos}\!\left(\frac{ \text{cos}(i) }{ \text{cos}(\phi) }\right)$$ It is of course straightforward to plug in the numbers for this situation: $$\beta = \text{arccos}\!\left(\frac{ \text{cos}(\approx\!51.6^\circ) }{ \text{cos}(\approx\!28.6^\circ) }\right) \approx 45.0^\circ$$

Note that you will sometimes incorrectly see arcsine instead of arccosine. This comes from an easy-to-make error in the derivation, which I will highlight below. Arcsine will (coincidentally) produce almost the correct result for KSC, making it plausible, but running a sanity check at $$0^\circ$$ launch latitude will prove it is the wrong formula.

# Where does that math come from?

I was unable to find a diagram, so I drew one:

The north pole is at $$A$$, the ascending node of the orbit is at $$B$$, and the launch site is at $$C$$. We apply the spherical law of sines: $$\frac{\text{sin}(A)}{\text{sin}(a)} = \frac{\text{sin}(B)}{\text{sin}(b)} = \frac{\text{sin}(C)}{\text{sin}(c)}$$ (which is where the "spherical trigonometry" quip comes from). Since $$c$$ is $$90^\circ$$ by construction and $$B=\pi/2-i$$ and $$b=\pi/2-\phi$$ by inspection, simple algebra and basic properties of sine/cosine (every step shown in the diagram) gives: $$\text{sin}(C) = \frac{\text{cos}(i)}{\text{cos}(\phi)}$$ Now if you weren't paying attention, you might take the arcsine of both sides and think you were done. However, we don't want $$C$$; we want $$\beta$$. By vertical angles, we can see that $$C=\pi/2+\beta$$. Therefore, we have: $$\text{sin}(C) = \text{sin}(\pi/2+\beta) = \text{cos}(\beta)$$ And our final answer is: $$\beta = \text{arccos}\!\left(\frac{\text{cos}(i)}{\text{cos}(\phi)}\right)$$

• Great explanation and diagram. Apr 25, 2020 at 1:33