# Tag Info

23

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 ...

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 ...

18

Given the mass costs in terms of consumables and the risk and support costs of keeping humans in space for longer, it seems unlikely that the multiple Earth-Venus flybys used by a lot of robot probes to get out to Jupiter or in to Mercury will ever be a sensible choice for humans. A Jupiter flyby on the way to Saturn is probably a no-brainer apart from maybe ...

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 ...

11

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: $a=\frac{\mu^{1/3}}{\frac{2n\pi}{86400}^{2/3}}$. For $n=15.5918272 \,\, \... 11 There is an international network of observers of classified satellites, organized around the Seesat-L mailing list: http://www.satobs.org/seesat/seesatindex.html They typically look for satellite passes using binoculars and a stopwatch, or using a camera. Then they fit TLEs to those observation, to be able to find the satellite on a later pass. You can get ... 11 Not surprisingly, one needs to use hyperbolic functions as opposed to trigonometric functions with regard to hyperbolic trajectories. The motivation is simple. Let's start with Kepler's equation,$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 ... 11 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 ... 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 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 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 ... 9 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 ... 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 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 ... 8 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 ... 8 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 ... 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. 8 If you define perfection as absolutely zero eccentricity then perfection is impossible. There will always be eccentricity in an orbit, even if it is very small. Orbits vary due to: Spacecraft system inaccuracy: no spacecraft is perfect, no matter how accurate Changes in the density of the orbited planet Gravitational influence of other celestial bodies: my ... 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 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=\...

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

As a new user I cannot comment or tweak the original answer, so I'll try my own. Happy update per any recommendations that come up. @uhoh's answer is close, but a few things to note. The TLE tells us the number of orbits per day is 15.50995519 (line 2 columns 53–63)  \frac{24 \frac{hours}{day} * 60 \frac{mins}{hour} }{15.50995519 \frac{orbits}{day}} = ...

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 first key to figuring this out is getting your coordinate system correct. There are two commonly used coordinate systems for such things. They are the Earth Centered Earth Fixed (ECEF), and Earth Centered Interial (ECI) frames. At midnight, these two line up exactly, but they diverge in other times, based on the rotation of the Earth. ECEF works best for ...

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