# Tag Info

50

An elliptical one. The Wikipedia page you link gives a signfiicantly different apogee and perigee and a period of 20.8 hours. So, on average, it moves West to East a bit faster than the Earth does, but at apogee it's moving more slowly and the Earth overtakes it a bit (which are the "S-bends" in the track). It swings a little North of the equatorial plane ...

18

There are a number of software packages, many of them free, that deal with those two line elements. Use one of them. Those two line elements are not Keplerian elements. They are instead Brouwer-Lyddane mean orbital elements. Keplerian elements assume a spherical central body and no forces other than gravitation. The Brouwer-Lyddane mean orbital elements ...

14

Simply put, that's where it spends most of the time. I think the chart is a bit exaggerated, but a satellite will be at it's northern and southern extrema far more. The pattern follows roughly a sinusoidal pattern, and is further enhanced because the circles shrink in size the further you go from the equator. I can't find a plot with the orbit, but here is a ...

9

For the first part of your question, and assuming an equatorial orbit, it depends on the orbital altitude, but maths are simple enough to start with: $$T = 2\pi\sqrt{\frac{a^3}{\mu}}$$ Where $a$ is semi-major axis (in our case Moon's equatorial radius of 1,738.14 km, plus mean orbital altitude above it), and $\mu$ is standard gravitational parameter of 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.

8

tl;dr: use a parametric equation. If the Earth were not rotating, then we would have something like \begin{align} x & = \cos \omega (t-t_0)\\ y & = \sin \omega (t-t_0) \ \cos i\\ z & = \sin \omega (t-t_0) \ \sin i\\ \end{align} where the radius of the orbit is 1, $\omega$ is $2 \pi/T$ and $T$ is the period, and $i$ is the inclination of the ...

8

From the pros, from the Spaceflight 101 article Tiangong-1 Re-Entry, click for full size: For a circular LEO, for $x_p$, $y_p$ in the plane of the orbit we can just write $$x_p = \cos(\omega t)$$ $$y_p = \sin(\omega t),$$ and if it is inclined to the equator by an angle $i$, the $x$, $y$, $z$ coordinates when the $\hat{z}$ axis is parallel to the Earth'...

6

I haven't found a way to make Heavens-above draw satellite orbit images for other time periods. (It will easily draw start chart pass information for them) But APEX right now is in an orbit with a very high beta angle. So the images for its orbit should be useful. https://www.heavens-above.com/orbit.aspx?satid=23191&lat=0&lng=0&loc=...

6

TLEs are a product of fitting observations using the SGP4 propagator. So you can't really change them around the way it seems you're trying to. The TLE catalog is updated regularly, so if you want updated information for the ISS for instance, you can grab catalog data from https://www.space-track.org. There's also an API there so you can fetch the data ...

5

As @David said, those are the current and the two next orbits of the same spacecraft. On this photo you can see the ISS superimposed on one of the orbits, and the orbit numbers (4, 5, 6) to the left. As @Gerrit said, if the spacecraft isn't doing any maneuvering, the next orbit can be predicted with 100% accuracy. The only information you need is the orbit'...

4

you can use PyEphem just like this sudo apt-get install python sudo apt-get install python-dev sudo apt-get install python-pip pip install pyephem create test.py: import ephem import datetime ## [...] name = "ISS (ZARYA)"; line1 = "1 25544U 98067A 12304.22916904 .00016548 00000-0 28330-3 0 5509"; line2 = "2 25544 51.6482 170.5822 0016684 224....

4

The circumspect answer is that I think you can only find this by specific evaluation for the type of object and orbit that you have in mind. Apologies, as I enthusiastically wrote most of what follows before remembering that you said that you want to track the ISS position. The short answer is, give it a whirl, it may well be just fine! The rest of this is ...

4

This is just a supplement to the accepted answer. For a low orbits where the semi-major axis is close to the radius of the central body, the period is related to the average density of the body and unrelated to it's size. So a low orbit around a spherical asteroid (which there usually aren't) made of a mixture of rock and iron (which there usually aren't) ...

4

Interesting question! Abstracted/simplified case: Suppose a spacecraft is orbiting around a planet with no atmosphere, a perfect spherically symmetric mass distribution, such that the gravitational field can be expressed as that of a point at the origin using Newton's Shell theorem, and there are no other forces or bodies in the universe. A repeat ground ...

3

Speculation: Sanibel Island, Florida fits the bill. Like Cape Canaveral, it is on an island on the Florida coast, close enough to be reached by bridges. Its latitude is 2 degrees south of Canaveral (26°26′23″N versus 28°29′20″N), so actually slightly better. It's bigger in area than Canaveral (86 km$^2$ versus 5 km$^2$); plenty of room for the military, ...

3

From the point you are, you need to: Calculate the XYZ coordinates of your ground station Find the line in space between the satellite and the ground station Find the XYZ coordinates of the point at the intersection of that line with the ellipsoid defined by all the points with elevations of 300 km Find the latitude and longitude of that point. For (1) you ...

3

Depending on which algorithm/set of equations you are using to convert, you may need to convert the TLE parameters into ECEF coordinates, then convert that into latitude, longitude, and altitude. Here is a page that explains the ECEF-to-LLA conversion: http://www.gmat.unsw.edu.au/snap/gps/clynch_pdfs/coordcvt.pdf A common math difficulty is that the true ...

3

This is necessarily a short answer, I don't doubt there will be other more extensive answers later. The location of a spacecraft in space is changing due to gravity forces and perturbations from drag and solar pressure. The location of DSN stations in inertial reference frame is not constant either. Earth's orbital motion is the first component to consider....

2

I had a similair question, and using pyephem as suggested by zdRan worked great, except for one thing: Those are instructions for Debian/Ubuntu/etc. distributions and I was on a bare-bones CentOS install. In case anyone else runs into this, here's the install instructions, to be used in place of the first block in zdRan's post: sudo yum install python ...

2

I don't fully agree with the explanation given so far. Here's my additional guesses: 2) The outer edges: This is not a binning artifact as stated by @PearsonArtPhoto. Debris doesn't only fall along the path of satellite, there is a region with a certain width around the path where we have to expect debris coming down. It's reasonable to assume the ...

2

ICESAT-2 The POD team developed the ICESat-2 orbit with the initial requirements of: (1) a 92°inclination orbit for coverage of polar ice and sea ice while still producing orbit-crossings for altimeter cross-over observations, (2) a frozen orbit to limit altitude variation at any given latitude in order to maintain beam pattern geometry on the surface, (3) ...

1

So why do rockets from Baikonur bound for GTO launch to the northeast, rather than due east? A quick look at a map shows that if you aim any further south than Altai you risk dumping your space junk in China and Mongolia. China in particular probably wouldn't like that, so I'm guessing they steer well clear.

1

My guess is the way that they did this is to take the orbital paths that might happen, and figured out where it would reenter if that was the case. The artifact at the edges is most likely a binning size. Only the very middle part (Top and bottom) is a possibility , but the bin is still there, so the actual amount is small. A simpler way to think of this is ...

1

Here's how I would solve it: First, the math is easier if proper spherical coordinates $(r, \theta, \phi)$ are used instead of longitude and latitude are used. The difference is latitude is the angle from the equator, whereas $\phi$ is the angle from the north pole. To find $\theta$ and $\phi$ from the longitude and latitude, $$\theta = \text{Long}$$ \phi ...

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