According to my textbook, the square of orbital period $P_{orb}$ is given by $$P_{\mathrm{orb}}^{2}=\frac{4\pi^{2}a^{3}}{GM},$$ where $a$ is the semi-major axis.
My question is, does the inclination $i$ affect orbital period?
I read that the launch of a rocket often gets help from Earth's angular momentum, for example, if you launch at Earth's equator, the velocity you need to obtain is less than, say, launching near the poles. However, I don't see inclination $i$ being part of the equation for orbital period.
Does the inclination have any effect on orbital period? For example, a planet orbiting around a star at $0^{\circ}$ inclination, would its period change when the inclination rises to $60^{\circ}$?
Best Answer
No, inclination does not affect the period (but see below).
The reason is that gravity is a central force, so for something orbiting a spherically symmetric mass there is no dependency on what reference plane or directions you use. You could always select another one and get a different inclination measure, but the period does not change if somebody tilts their coordinate system.
Around Earth a satellite is affected by the oblateness of Earth (and even more by the Moon), that does introduce a dependency on which plane it rotates in. These deviations mean that technically there is no given period since the orbit does not perfectly close (it is no longer the standard 2-body problem, and the plane precesses around Earth), although in practice this is a minor effect. One can approximate the effect, and then there is a slight effect of inclination on period.