11

Your concerns are all perfectly valid - giving just a number doesn't tell a lot. So, in all proper publications, the reference system has to be mentioned. Typically, the reference is the body the spacecraft is mainly influenced by. I.e. in a Earth orbit it is the center of Earth, in lunar orbit the Moon. For interplanetary probes in transit it's usually the ...


9

Yes. This is a classical astrodynamics problem of orbit determination. The technique you would use is called Gauss' method. It allows you to determine an approximate orbit from three timed observations of azimuth and elevation. The details are well-described in the link, but are too lengthy to reasonably list here.


9

There are many different coordinate systems (X, Y, Z axes as you refer to them). The center of the Earth is often used as the origin "center" of the coordinate system, at least for calculations concerning the space in the vicinity of Earth. There are many such Earth-centered coordinate systems as well. In the Space Shuttle Program we used the "Mean of 50" ...


8

The standard for any tidally locked body, of which Europa is a member, is to have the 0 longitude be the point at the center of the planet-facing side. That being the case, the middle of the map should be the portion facing Jupiter, the edges the part that never faces Jupiter. See Wikipedia for the referenced quote below: Tidally-locked bodies have a ...


5

An LVLH frame is easy to construct. One construction: Let $\boldsymbol r$ and $\boldsymbol v$ denote the spacecraft's position and velocity with respect to the center of the planet as expressed in an inertial frame. Construct $\hat {\boldsymbol x}$ as the unit vector directed along the spacecraft position vector: $\hat {\boldsymbol x} = \boldsymbol r/||\...


5

For the H-1 engines of Saturn I/IB, it appears that at some point in development, the origin was ...almost... at the bottom of the engines, according to this drawing I found linked from AlternateWars: That drawing shows "approximation" for the station 0/station 1 reference for the bottom of the engine bell, while another Rocketdyne document has the bottom ...


5

Galactic centre: -29.01° declination and 17.76 hours right ascension, Voyager 1 12.44° declination and 17.163 hours right ascension, missing by ~40° Voyager 2 −55.29° declination and 19.888 hours right ascension, missing by ~50° Pioneer 11 -8.80° declination and 18.83 hours right ascension, missing by ~30° Pioneer 10 25.99° declination and 5.2 hours ...


4

While this answer was written for a different question/project, it points out that the Two Line Element sets and the SGP4 propagator work together and each is designed specifically to work only with the other. The elements of a TLE are not exactly Keplerian orbital elements, even though the parameter names overlap with Keplerian element names. This is ...


4

Here is information on an open source Java package that has code for TLE propagation: https://www.orekit.org/forge/projects/orekit/wiki/Tle It also contains classes for all the components you need to get your range/angles to the satellite. Look at the topocentric frame interfaces in the javadoc; it's got interfaces to get the azimuth/elevation/range: ...


4

Are you modeling drag? If you are not, you don't even need to model the Earth's rotation because the J2 effect depends on latitude only. This is a low fidelity simulation (there are lots of effects other than J2; e.g., drag, third body effects, higher order gravity terms, solid body and ocean tides, solar radiation pressure, relativistic gravity, ...) But ...


4

The Explanatory Supplement to the Astronomical Almanac has all the equations you need. Take a look at chapters 3 and 4. Keep in mind you need some clear definitions of what you mean by ECEF and ECI. Most people utilize WGS84 for ECEF, but that is not a requirement. Similarly ECI could be J2000 or ICRF In general you will need 4 steps to convert from ECI ...


3

Given either geocentric latitude, longitude, and radius or geodetic latitude, longitude, and altitude, the computation of Earth-centered, Earth-fixed cartesian coordinates is fairly simple. For geocentric coordinates $R,\theta,\lambda$, one uses $$ \begin{aligned} R\ &\text{is the radial distance from the center of the Earth} \\ \theta\ &\text{is ...


3

SOFA The IAU’s Standards of Fundamental Astronomy (SOFA) library has a ANSI C and FORTRAN version. See the documentation on the C2T06A function for converting from ECI to ECEF. note: You must look up what are called Earth Orientation Parameters (EOP) if you wish to convert ECI coordinates to ECEF. You can find them here.


3

No, you would need to know the range to the satellite as well. Think of it this way - if you draw a diagram of what you describe above, the vector representing the line of sight from your ground position to the satellite (az, el) would cross a swath of longitudes, except for the trivial case where el=90


3

In section 3 of your reference, it states The EROS-A legacy pass-file contains all metadata in one file, but requires a lot of exceptional processing associated with [sic] customized coordinate system (so-called Q-frame). Satellites usually have many frames associated with different aspects of the body of the satellite. For instance, there will be at ...


3

You must be using a computer from the 1960s to have even the most precise IAU earth orientation computation take "5-10 seconds". That said, computing the Earth's orientation using an extremely accurate algorithm at some epoch time and then rotating the Earth by $2\pi$ radians per sidereal day ($7.29211585275553\times10^{-5}$ radians per second) about the ...


2

I am not aware of any particular standard for the primary body, however for the Pluto system, in this image the 0,0 point is on the opposite side of the hearth: And in this image, the hearth is shown to also be at the opposite side from Charon. So I assume you are right.


2

I find the accepted answer unclear, so I'll try: The two maps in discussion use coordinate systems based on the convention that 0 longitude is the point directly facing Jupiter, but they have chosen to put that point on the far right-hand edge of the map. Dyfed Regio is on Europa's Trailing and Anti-Jovian hemispheres (not the Sub-Jovian). Note that the ...


2

At the bottom of your linked web page is the sentence: If you need really accurate data for a planet or asteroid, try the JPL HORIZONS System. (These guys were navigating spacecraft through the solar system when I was still playing with GI Joe!) ...which is indeed an excellent answer to your question! A step-by-step example of how to obtain "rectangular ...


2

$lon_{Sat}=f(lon, lat, ht, el, az)$ Where: $lon$ Receiver longitude $lat$ Receiver latitude $ht$ Receiver height in metres (does not have a major effect but adding for completeness) $el$ Satellite elevation in degrees $az$ Satellite azimuth in degrees $a = 6377.301243$ (Semi-Major Axis of Earth in Kilometres) $f = \frac{1}{ 298.257223563}$ (Flattening of ...


2

Moon position The Moon position is given with respect to the Earth Moon barycenter, whose NAIF ID is 3. Note that the barycenter of the Earth-Moon system is very close to the center of the Earth (otherwise it wouldn't have taken humans that long to figure out where tides came from). You should be able to list all the bodies of a given BSP file by looking at ...


2

I don't think there's precisely one selenographic coordinate system, but section 2.3 of the Lunar Constants and Models Document (JPL D-32296) seems to recommend spheres. 2.3 Moon Shape Parameters The general shape of the Moon is very nearly a perfect sphere, excluding local topography variations. In fact, the magnitude of the local topography ...


2

The main ways I'd go about this, depending on the application and precision you need: Approximations you can code up simply: Plain-text keplerian elements are available on the NASA JPL website: https://ssd.jpl.nasa.gov/?planet_pos These give good approximations up to the year 2050 If you need something more precise then the Planetary Ephemerides DE431 is a ...


2

Okay this isn't a complete answer but it may be of some help until you can find a better site for your question. See also this answer for more information. Use a simple pinhole camera approximation where a ray that passes from the object through the center of the camera's pupil travels a straight line from object to the focal plane. Assuming that the ...


2

Hill coordinates represent a local reference frame widely used in relative motion when only the main body gravity (in this case the Sun) is considered. The Hill coordinates origin and axes are defined as follows $Ox_hy_hz_h$: centred in the target center of gravity (in this case Ryugu gravity center) $x_H$: the line joining the target and the Sun, positive ...


2

It looks like you're working from the JSpOC Spaceflight Safety Handbook for Operators (https://www.space-track.org/documents/JSpOC_Spaceflight_Safety_Handbook_For_Operators.pdf). In this case they define the RIC frame as identical to what is often called the UVW frame (https://www.space-track.org/documents/JSpOC_Pc_4Aug16.pdf pg 3). This frame is defined ...


2

There are multiple reference ellipsoids for the Earth. Given a latitude, longitude, and altitude, the corresponding ECEF cartesian coordinates will differ with different ellipsoids. The reverse is also true. There is no clear cut answer to this question because you didn't specify which reference ellipsoid Justin Mooney's script uses. That said, the ...


2

Such a rotation matrix does not exist. In Vallado's book I found the following: "Because these stations [used to derive ITRF] are affected by plate tectonic motion ... conversions between ITRF solutions involves translation, scale, and rotation"


1

--Formulas-- radius = sqrt(x^2 + y^2 + z^2) latitude = cotan(z/(sqrt(x^2+y^2))) Actually, the link you have shows $$ \Phi'(t) = \tan^{-1} \frac{z(t)}{\sqrt{x^2(t) + y^2(t)}}$$ That's the inverse tangent function, not the cotangent. For the figures you show, the inverse tangent would be $-0.25598\, \text{rad}$ or $-14.666\,\text{deg}$ That said, there's ...


1

Does a solar geometry library exist for satellites in LEO? Almost certainly, but it's also almost certainly hidden. Your question can be broadened: Does a generic, freely available geometry library exist for satellites in any orbit about any planet (or even in transit between planets)? Here the answer is YES, not just a boring yes in the standard font ...


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