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Two Station Ranging

Given the position of 2 stations and the distance between both and the satellite (in km, say) and the fact that the orbit is geostationary and positioned at 45.1(degrees) west, satellite position is to be determined.

What I've done so far is calculation of distance between stations (just to understand how things work), but I could not come up with a solution for the satellite position. We are also given a specific date (I assume that should help but how?)

------- EDIT ------- enter image description here What I already know from this illustration are: r1, r2, R1, R2, and that T is perfectly geostationary (let's assume an ideal world), and positioned at 45.1 deg West. I don't see how to easily obtain azimuth angles phi1 and phi2, and also the distances R1(y) and T(y). Based on the fact that the satellite is at 45.1 West, how do we calculate T(y)?. Any suggestions on that?

------- EDIT -------

Based on How to calculate Azimuth/Elevation of a satellite? I calculated azimuth angle and obtained Y coordinate of satellite. Formula for elevation is provided too but I can see that these all apply for single station positioning and it is not as precise as 2 station positioning. Paper https://iopscience.iop.org/article/10.1088/1742-6596/1168/5/052056/pdf describes the method for calculating the errors but not the actual values. Any suggestions on that?

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    $\begingroup$ The satellite's position is to be determined relative to what? The center of the Earth? (It is over 45.1 degrees W longitude and at geostationary height.) In azimuth and altitude relative to one of the stations? (You need the latitude and longitude of the station.) In Right Ascension and Declination of one of the stations? (You need latitude, longitude, and date.) $\endgroup$
    – JohnHoltz
    Oct 3, 2022 at 12:36
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    $\begingroup$ I think the knowns and unknowns you're trying to work with need clarification. You state you "already know" R1 and R2, then later state you need to obtain R1(y). $\endgroup$ Oct 4, 2022 at 14:07

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If the orbit is exactly geostationary, meaning inclination and eccentricity are both zero, then $z$ is zero, so you're done. The orbital period of a geosynchronous satellite is one sidereal day of 86,164 seconds, so its semimajor axis $a$ is 42,164 km, because $4 \pi^2 a^3 = \mu T^2$ and Earth's gravitational parameter $\mu=3.986\times 10^{14}\ m^3/s^2$. Thus your target satellite's position in the $x$-$y$ plane is this radius times the sine and cosine of the given longitude, and its vertical component vanishes.

More interesting things happen when you consider how to find the satellite's position if you only know that it is somewhere in a GEO orbit, but not at what longitude. Then you need to write the formulas for the distances $r_1$ and $r_2$ in terms of the known components of $\vec{R}_1$ and $\vec{R}_2$, and the unknown components of $\vec{T}$. If you continue assuming zero inclination, then you can write the equation in terms of the longitude as the only unknown, and thus find which location best fits your measured ranges. You don't actually need to know the azimuth and elevation angles to do this, but you can easily compute them from the dot and cross products of the site locations (capital $R_i$) with the lines of sight (lowercase $r_j$).

One warning about the source of your image: that paper is talking about measuring TDOA, which means time difference of arrival. In many practical situations of this kind of measurement, you can't measure $r_1$ and $r_2$ separately. Instead, you can only measure their sum, or only their difference. The sum case is what happens if you send a signal from ground site $R_1$ up to the satellite, which then transponds it back down to $R_2$. The difference case results from sending a signal from the satellite which is received at both stations at slightly different times.

Real satellites, on the other hand, are never precisely geostationary. In order to solve this problem with actual data, you need to know all the other orbital elements: eccentricity ($e$), inclination ($i$), right ascension of the ascending node (RAAN, $\Omega$), argument of perigee ($\omega$), and time, usually in the guise of mean anomaly ($M$). Time is important because in real life, everything is moving, so the best match between your measurements and you assumptions about the orbit properties is constantly changing. For more on this point, see How to calculate the ECEF coordinates of a satellite given its elevation and azimuth angle, plus the coordinates of a reference object on Earth

Nonzero eccentricity makes the orbit's ground track extend east and west, and nonzero inclination makes the orbit's ground track extend north and south. The one with the strangest-looking effect is argument of perigee. The familiar figure-8 (analemma) ground track only occurs when $\omega$ = 0 or $\pi$. At $\omega$ = $\pi$/2 and 3$\pi$/2, you get a sort of egg shape, an oval rounder at one end and pointier at the other. Now, instead of the ascending node being in the center of the orbit track, it is now on one edge or the other, with almost the whole orbit to either its east or west, as seen here:

ground tracks for w = 0, 90, 180, 270

In this image, which I made with STK, the blue ground track is $\omega$ = 0, the magenta is $\omega$ = $\pi$, $\pi$/2 is yellow and 3$\pi$/2 is green. If we add $\pi$/4 to each of those, we get this:

enter image description here

These orbits are greatly exaggerated with respect to usual geosynchronous orbits, in order to provide Brazil in the background for scale. If we switch from $e$ = 0.03 and $i$ = 10$^\circ$ to $e$ = 0.0003 and $i$ = 0.1$^\circ$, the ground track shapes are nearly the same; you just have to zoom the map farther in, and don't see land anymore.

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  • $\begingroup$ Thanks for an informative answer! Unfortunately it does not fully answer it, or I simply couldnt follow. I've edited the question with an attached illustration. Thanks $\endgroup$
    – pncln
    Oct 4, 2022 at 6:14

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