Fun with 2-line elsets

Date: Mon Oct 31 2005 - 16:29:13 EST

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    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
    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  
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