Convert latitude of lowest altitude to argument of perigee?

Question

I am designing an orbit around Mercury. I know the values I want for the semi-major axis, eccentricity, inclination, and RAAN. I want the altitude of closest approach (periapse) to occur at $60.0^{\circ}$ N latitude above Mercury's surface.

My Question: How do I convert this constraint for the latitude of closest approach into an argument of periapsis, $\omega$?

Specifically, this is modeled after the orbit below:

Answer 1

The answer is:
$ \sin(\omega) = \frac{\sin(lat)}{\sin(i)} $

this expression has has two solutions:
$ \omega = \arcsin \left( \frac{\sin(lat)}{\sin(i)} \right), \ \ \text{and} \ \ \pi - \arcsin \left( \frac{\sin(lat)}{\sin(i)} \right) $

The second solution matches the diagram.

So if the latitude is 60° and the inclination is 82.5°, then the argument of periapsis (ω) is 119.13225456935°.

(The term ''perigee'' applies only to orbits around the Earth; the correct term for Mercury is ''perihermion''. The general term is ''periapsis''.)

Derivation:
The perihelion direction vector is: $ \vec{q} = [1 \ 0 \ 0] \cdot \operatorname{rot}_z(\omega) \cdot \operatorname{rot}_x(i) $.
where rot_z is the rotation matrix about the z axis, and rot_x is the rotation matrix about the x axis.

Convert the perihelion direction vector to spherical coordinates, set the latitude equal to 60°, then solve for ω.