**Previous message:**Greg Roberts: "Obs 30 Oct 2005"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]

When I was compiling my old satellite databases, I included a set of "orbital elements" for each object. These were available, except for objects in the long lists of catalogued fragments from on-orbit breakups, in the semimonthly Goddard Satellite Situation Reports. The Goddard data also served as the basis for the elements used in the quarterly TRW Space Logs, and similar elements were also provided in the RAE Table of Artificial Earth Satellites. These elements were reported as period (in minutes and tenths of a minute), inclination (in degrees and decimals of a degree), and perigee and apogee (in kilometers). These are the elements I "grew up" with. Given that such elements change constantly due to atmospheric drag and gravitational perturbations, I never looked for or expected anything truly exact, just something pretty close, for purposes of comparing with the orbits of other objects and such, something that might, for example, identify a secret satellite by its orbit. It is fairly easy, using a hand calculator that can do exponentiation and extract roots, to obtain these four "basic" elements from the available 2-line elsets--at least, to a decent level of precision that provides a good and immediate idea of the shape of the object's orbit. So it's rather surprising that there seems to be no site on the Internet that tells just how to do this. Of course, I haven't the time or the means to search the whole Internet for such a website, so there's a good chance I've missed it if it does exist. But fortunately there are a number of websites that explain in detail how to interpret the 2-line elsets, and these have allowed me to use the elsets to obtain those "basic" elements. I'm sure much of what follows is old hat to list members, but in case it isn't, and just for the record (and maybe to have a convenient, handy reference), here we go. Let's use a typical 2-line elset, such as this one for the initial orbits of Molniya 1-S, 1974-060A, which I happened to be looking at on Jonathan McDowell's website: 1 07392U 74210.17851633 .00000162 00000-0 0 19 2 07392 0.0666 72.8351 0000009 27.9434 358.6246 1.00069492 00 (This is an old-format 2-line elset; the newer formats have a couple more fields, but I'll rely on list members' knowing where the fields relevant to this discussion occur in the newer formats.) [1] Converting the epoch date into month/day/year format: This is pretty easy. The epoch date is contained on the first line and in this example it reads 74210.17851633 All you need is a little table of months of the year and the number of days in each month, added up cumulatively: Jan 31 31 31 Feb 28/29 59 60 Mar 31 90 91 Apr 30 120 121 May 31 151 152 Jun 30 181 182 Jul 31 212 213 Aug 31 243 244 Sep 30 273 274 Oct 31 304 305 Nov 30 334 335 Dec 31 365 366 The first column is the three-character abbreviation of the month of the year, the second is the number of days in the month (with 29 for Feb in leap years), the third is the number of days that pass from the first day of the year through the last day of that month, and the fourth is the same as the third only for leap years. The fact that we end up at 365 and 366 in columns 3 and 4 shows that I likely did the additions correctly. So, e.g., 31 Mar is the 90th day of the year, 91st if it's a leap year. In the epoch date, the first two digits give us the year (1974), while the remaining digits (210.17851633) give us the day of the year and fraction thereof to which the elset pertains. Looking in column 3 of the table (1974 is not a leap year), we find 210 (the digits ahead of the decimal point) is greater than 30 Jun but less than 31 Jul, so the epoch date lies in July 1974. Subtracting 181 (the last day of June) from 210 gives 29, so the epoch date is 29 Jul, 1974, or 7/29/74--the very day the spacecraft was launched, incidentally. Day #210 of the year. The UT time for the epoch may be obtained by multiplying 24 by .17851933 (the digits following the decimal point) on a hand calculator to give 4.28446392, and 60 by .28446392 to give 17.0678352, so to the nearest minute the UT time is 0417, or 4:17 am. As I said, pretty easy. Ridiculously easy. Too easy. You see, I have July 29, 1974 1200 UT as the date and time of the >launch< of Molniya 1-S (this would be epoch 74210.50000000), so the 2-line elset above gives the orbit of the spacecraft at more than 7 hours before its launch(!). I have no explanation for the anomaly, only possibilities: (A) my launch date/time info are wrong; (B) the elset is wrong; or (C) the method outlined above is wrong (not very likely, as I followed several elset-explaining websites on this). I'll leave the resolution of this little problem to a future email. [2] Converting the 2-line elset orbit info into "basic" format. We begin with the easiest: inclination. This is just handed to us in the elset as the second field of line 2, to four decimal places: 0.0666. I would round this off to 0.1 or 0.07. This is a nice equatorial orbit. The second easiest is period. This is simply calculated from the second-last field of line 2: 1.00069492, which is the number of revolutions per solar day (according to the elset-explaining websites), so, since a solar day is exactly 1440 minutes long (by definition), the period of the satellite in minutes is 1440/1.00069492 = 1439.00001 on my hand calculator, which I'd round off to two decimal places: 1439.00. This is a few minutes longer than geostationary, so one could argue that Molniya's first orbit was its "station-hunting" orbit until it reached wherever it would be stationed for its mission. Perigee and apogee are the most tedious items to calculate. Here "perigee" means nearest point of the orbit to earth's surface, and apogee means farthest point of the orbit from earth's surface. For my purposes, I'm happy to consider the earth as a perfect sphere rather than the messy, mountain-and-valley-covered, pear-shaped oblate spheroid that it is. We are given only the orbit's eccentricity e, as the fourth field of line 2: 0000009. This is expressed as a decimal when the decimal point is placed at the head of the string of digits: e = 0.0000009. To retrieve the orbit's size, we need to return to Kepler's laws, which relate the object's period to the major axis of the orbit. Way back in the early 1600s, Kepler discovered that for any satellite orbiting a planet (or planet orbiting a star), the square of the period is directly proportional to the cube of the orbit's major axis. The constant of proportionality is a geophysical property of the planet (or planetary system, or primary star) and doesn't change, regardless of the size or shape of the satellite's orbit. We have to calculate this constant K for the earth, and once we do so, we can use it for all satellite orbits around the earth. As long as the satellites are fairly close to earth, we can neglect the influence of the moon and other perturbative agents, and K may be treated as a true constant. To obtain K in the equation a^3 = K*p^2 (where a is the major axis in kilometers and p is the period in minutes), I used the nice orbit for Molniya 1-S provided in Jonathan McDowell's table of orbited objects. On August 20, 1977, the orbit had period 1435.88 min, perigee 35751 km, apogee 35814 km, and inclination 2.4 degrees. Adding 35751, 35814, and the earth's average diameter 12756 km (another constant that I found on the Internet with a bit of snooping) gives the major axis of the orbit as 84321 km, and plugging this and the period into the above equation gives K = 290784298.7 km^3/min^2 Keep this number handy if you ever need to calculate the major axis from the period for any earth-orbiting satellite(!). It is a fundamental earth constant, along with earth's diameter, mass, and so forth. Just square the period in minutes, multiply by K, take the cube root, and bang, you have the major axis in kilometers. In particular, for a period of 1439 minutes, the major axis works out to be qbrt (1439*1439*290784298.7) = 84443.10229 which I round off to 84443 km. At this point, we still don't have the perigee and apogee. For these we need the eccentricity, which, I will remind the reader, for the orbit we are working with was e=0.0000009. This is a very small eccentricity; the orbit is a practically circular ellipse. One definition of eccentricity is the ratio of the distance between the foci of an ellipse to the major axis. (This gives, in particular, 0 if the foci are coincident and the ellipse is a perfect circle, and 1 if the ellipse is completely stretched out into a line segment--a parabola viewed from infinitely far away, so to speak.) A little geometry provides the following nicely symmetric expressions for perigee and apogee in terms of the eccentricity e, major axis a, and earth's diameter D: perigee = ((1-e)*a)-D)/2 apogee = ((1+e)*a)-D/'2 (To check, note that perigee plus apogee plus earth's diameter must equal the major axis, since perigee and apogee occur on opposite sides of the earth. Adding the two expressions above, plus D, gives a.) For our orbit of eccentricity e=0.0000009, the hand calculator yields: perigee = ((.9999991*84443)-12756)/2 = 35843.462 apogee = ((1.0000001*84443)-12756)/2 = 35843.50422 or rounded to the nearest kilometer, perigee = 35843 apogee = 35844 As expected, virtually circular. Just to check, let's see what this procedure does with another of Jonathan's orbits, this one for the Molniya 1-S third stage 1974-060B, whose parking orbit on 7/30/74 is given as period = 88.07 min perigee = 173 km apogee = 185 km inclination = 51.4 degrees I don't happen to have the 2-line elsets for this object, so we're merely going to check Kepler: a period of 88.07 minutes gives a major axis of qbrt (88.07*88.07*290784298.7) = 13114.21547 Subtracting the earth's diameter 12756 gives 358.2154728 (the extra two decimal places at the end were hidden by my calculator in the previous line). Subtracting the perigee 173 gives an apogee of 185.2154728, which compares nicely with the 185 km as given. Kinda inspires confidence in the method! Or at least, the Kepler part. Additions, corrections, refinements to the foregoing accepted with gratitude. ------------------------------------------------------------------------- Subscribe/Unsubscribe info, Frequently Asked Questions, SeeSat-L archive: http://www.satobs.org/seesat/seesatindex.html

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