SatCom #10: Rate of Precesion of Orbit Plane (Nodal Precession) - MATLAB Cody - MATLAB Central

Problem 45812. SatCom #10: Rate of Precesion of Orbit Plane (Nodal Precession)

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Satellite and Space Engineering - Problem #10
This is part of a series of problems looking at topics in satellite and space communications and systems engineering.
Problem #5 has already looked at the 'Sun-Synchronous Orbit' which has the special feature that the plane of the orbit precesses (rotates) in inertial space at exactly the same rate as the earth rotates around the sun. Therefore, the orbit plane always maintains a fixed angle with respect to the sun, which means that the satellite always passes over the same point on the ground at the same local mean solar time.
A more general way of looking at what is going on for this type of orbit is to note that the rate of precession changes depending on the way that the orbit is configured. The unequal forces on the satellite caused by the equatorial bulge of the Earth tends will make an inclined orbit precess at a different angular rate depending on the orbit configuration. This precession is often called the 'nodal precession' of the orbit, because the points at which the orbit crosses the equator (the 'nodes') precess around the Earth as the orbit precesses.
You are given the satellite orbit's apogee and perigee altitudes (in km) and the inclination (in degrees). You should calculate the nodal precession rate (in degrees per day) for that orbit.
Hint : See https://formulasearchengine.com/wiki/Nodal_precession for a detailed explanation of how to derive the nodal precession rate of a satellite orbit.
You should take the radius of the Earth to be 6378137 m, the second zonal gravity harmonic of the Earth (J2 term) as 0.0010826269, and the Earth standard gravitational parameter as 3.986004418e14 (m^3/s^2).
Example: The CLOUDSAT satellite has an apogee of 710 km and a perigee of 709 km. It's orbit inclination is approximately 98.2 degrees. Its nodal precession rate is approximately 0.9825.
Some future problems in this series will build on work done in previous problems, so if you get a working solution I suggest you hang onto the code!
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26.96% Correct | 73.04% Incorrect
Last Solution submitted on Jan 19, 2025

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