# Tag Info

## Hot answers tagged orbital-elements

18

You are correct to a point that the RA of the ascending node and argument of perigee won't change over time without some external force acting upon the satellite. In a simplified gravitational field, an object's orbital plane remains fixed. Unfortunately, reality is a lot more messy. Earth's gravitational field differs significantly from that created by a ...

16

Why is the sidereal period of the Earth 362.392667 days? It's not. You are doing three things wrong: You are using the solar system barycenter and assuming that is an object (it isn't). You are using the Earth-Moon barycenter and assuming that is an object (it isn't, either). You are asking Horizons to compute the osculating Keplerian elements of these ...

13

Your assumption is a good starting place and its good to be cautious about it. Many of us are guilty of abbreviation or outright misuse of the terms for convenience. Here's my rough guide, not meant to be precise but rather to address the overlapping usage. I've spent a bit more time on the basic definitions as this is probably where variations in usage ...

11

It depends on a lot of factors, including: Time since epoch Altitude Atmospheric density variations Spacecraft maneuvers Accuracy, number and distribution of observations used to fit the TLE Fit span used for differential corrections Typical errors for a TLE for a non-maneuvering spacecraft at an altitude higher than 400km and with good observation data ...

10

Due to range safety requirements, which preclude launch trajectories that fly over populated areas, the maximum inclination by a standard launch from CCAFS/KSC is approximately 57 degrees. There was one mission, however, that exceeded that. STS-36, a classified shuttle mission, was launched to an inclination of 62 degrees, through the use of a "dog-leg" ...

10

The TLE gives mean motion ($n$) in $\frac{rev}{day}$. This needs to be converted to $\frac{rad}{s}$ which can be accomplished by multiplying the $n$ TLE value by $\frac{2\pi}{86400}$. Therefore, to go directly from $n$ in TLE to the semi-major axis $a$. We can use the following formula: $a=\frac{u^{1/3}}{\frac{2n\pi}{86400}^{2/3}}$. For $n=15.5918272 \,\,... 10 There is an international network of observers of classified satellites, organized around the Seesat-L mailing list: http://www.satobs.org/seesat/seesatindex.html They typically look for satellite passes using binoculars and a stopwatch, or using a camera. Then they fit TLEs to those observation, to be able to find the satellite on a later pass. You can get ... 10 Here's the heart of your problem: Center : Solar System Barycenter (SSB) [500@0] The ecliptic is defined in terms of the Earth's mean (or average) orbit about the Sun at the epoch rather than about the solar system barycenter. Had you instead chosen the Sun as the center you would have found an inclination of 0.000266 degrees (about one arc second) ... 10 Ok, that's embarrassing: You just have to add earth's radius (traditionally the equatorial radius) of about 6378 or 6378.137 km to apogee or perigee heights to get distances to the center. 9 Not surprisingly, one needs to use hyperbolic functions as opposed to trigonometric functions with regard to hyperbolic trajectories. The motivation is simple. Let's start with Kepler's equation,$M = E - e\sin E$. We're going to run into issues (but not impossibilities) with negative square roots. Kepler's equation works quite fine, as is, with hyperbolic ... 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. 8 The best guess available about the time scale used in TLE files is from the epic Revisiting Spacetrack Report #3 report from Celestrak, which they have put online here: http://www.celestrak.com/publications/AIAA/2006-6753/ They took the old SGP4 software, collected every single patch and improvement they could find in the dozens of versions online and ... 8 What's going on? You are learning: what osculating orbital elements are and are not, that real orbits are not Keplerian! @DavidHammen's answer is of course spot-on correct, but I understand why you would have thought that this might be the right period. It is true that the Earth-Moon barycenter might move along a more representative Keplerian orbit than ... 8 The first formula gives you the altitude at a particular point in the orbit, assuming that the position vector is the satellite's current position relative to the center of the Earth. The second formula is the altitude of the periapsis (lowest point) of an elliptical orbit. 7 To revisit this question, it does appear that the during the 1950s & 1960s, a number of extreme inclination and polar launches did indeed take place from Cape Canaveral in Florida. The range safety limitations placed on launches from Florida was only enacted after a number of polar launches had already taken a place (possibly for good reason at the time ... 7 CelesTrack has Mir ephemerides (as TLE, two-line element sets) in its NORAD Two-Line Element Sets Historical Archives. It's a 755 KB in size ZIP archive (direct link) packing a text file with 22,333 TLE spanning time period from February 19, 1986 to March 23, 2001 when it was deorbited. If these don't go far enough back in time for your needs, you can ... 7 The Brouwer-Lyddane Transformation is based on two articles: "Solution of the Problem of Artificial Satellite Theory Without Drag," D. Brouwer, The Astronomical Journal, Nov. 1959, pp.378-396 "Small Eccentricities or Inclinations in the Brouwer Theory of the Artificial Satellite," R. H. Lyddane, The Astronomical Journal, Oct. 1963, pp.555-558 There are ... 7 According to Wikipedia, field 8 of TLE line 2 is the "mean motion in revolutions per day"; you can determine the orbital period from this. For geosynchronous orbit, you should expect 1.0 revolutions per day (in fact, due to complexity in the definition of "day", they're very close to 1.0027 as a rule). LEO defined as < 2000 km altitude should get ... 7 An exercise that was left unsolved from last year's class gives me this equation : $$t-t_{p} = \sqrt{\frac{a^3}{\mu}}*(\arcsin(X) - e*X)$$ where : $$X = \frac{\sqrt{1-e^2}*\sin(v)}{1+e*\cos(v)}.$$ This is just Kepler's equation$M = E-e\sin E$, but written in terms of$X = \sin E$, where$E$is the eccentric anomaly. We don't have a derivation of ... 6 Given: 16031.25992506 The 16 corresponds to 2016. As 1957 was the first year with satellites launched, 57 would be 1957, and in 2057 this might change, as there will be an issue. The 31 means the 31st day of the year (January 31) The .25992506 is the fractional day from midnight. This means 6.2382 hours, 14.292 minutes, 17.52 seconds, basically ... 6 The example initial conditions do not result in an orbit. The energy is positive and the eccentricity is greater than one. It is a hyperbola. Also you over-specified the problem by claiming it is in perifocal coordinates, but then providing initial conditions that result in a periapsis that is not on the p-axis. The computed eccentricity is close, but ... 6 You seem to be looking for the time equation: $$t=\sqrt{a^3\over\mu}\left(\tau-e\sin\tau\right)$$ where$t$is time and$\tau$is the eccentric anomaly. It looks like you have the eccentric anomaly at the "current" time. You can just plug that in to see the time relative to periapsis, which is at$t=0$and$\tau=0$. For the time at apoapsis, plug in$\tau=\...

6

There are a few ways. Visual tracking Tracking of it's RF emissions Radar tracking. I suspect it's the first two. If you know about where something is, you can see it visually, and track it's RF emissions to boot. People make a hobby of such things. There are a number of web sites that track this information. https://www.prismnet.com/~mmccants/tles/ is ...

6

If we assume a perfect two-body problem, absent perturbations from external bodies or non-spherical gravity sources (i.e., perfect conic orbits with no precession or variation), your constraints regarding inclination are actually unnecessary, as we may, without loss of generality, examine this problem in a perifocal reference frame. Perifocal coordinates ...

6

The Wikipedia article on Hubble provides the apogee and perigee height with respect to the Earth's surface not its center. As a consequnce the semi-major axis should be equal to the sum of the Earth's radius and the perigee or apogee respectively (of course only for cirular orbits).

6

@Jack's answer is excellent, I'll just address @Niket's question in this comment a little further. Just to be sure, is the data realistic? Like does the day to day variation seem realistic? tl;dr: The drift makes sense and is within 1% of what we can easily calculate. There are some of the orbital parameters plotted at the bottom of the page. The RA of ...

5

The longitude of the ascending node $\Omega$, the inclination $i$, and the argument of periapsis $\omega$ are simply the classic Euler angles $\phi$, $\theta$, and $\psi$, in that order. Then the classic Euler transform applied to an orbit in the X-Y plane with the periapsis on the +X-axis gives you the actual orbit in X-Y-Z. There is probably no reason ...

Only top voted, non community-wiki answers of a minimum length are eligible