Orbite

Effemeridi Asteroidi e Comete

 
Neocp and MPC
 
   1   | 2  |  3 |    4   |  5 |   6  |  7   |  8    | 9 | 10 | 11 |
	      
Object   H     G    Epoch    M   Peri.  Node   Incl.   e    n    a      NObs NOpp   Arc    r.m.s.       Orbit ID

Thesky6
Nome asteroide       | epoca        | e eccen. | a sem a|i incl. | Node    | Peri    |     |M mean ano|H    | G   |
164222 2004 RN9      |2007 10 27.000|0.247494  |1.305053| 16.1112|  1.0362 | 67.1019 | 2000|343.5438  |20.8 | 0.15|   0.00

Xephem
      1       | 2 |  3   |  4   |  5   |    6   |  7    |     8  |   9    |   10       | 11 | 12  |  13 |
Nome asteroide, e, I inc , Node , Peri  , a semi , n     , e ecc. , M      , Epoca      , J  , H   , G 
164222 2004 RN9,e,16.1127,1.0156,67.1149,1.304844,0.6612504,0.247282,152.5206,01/04.0/2010,2000,H20.8,0.15

Comandi Utili

cat MPCORB.DAT | sed '1,41d' | sed '/^$/d'| sed 's/          K/00.0   0.00 K/g' | sed 's/          J/00.0   0.00 J/g' > mpcorb.dat
# elimino le prime righe di testo e le riche vuote e creo file per elaborazione 
# inserico nel file mpcorb 00.0 0.00 ad H e G dove non sono presenti

cat mpcorb.dat |awk '{ print (5.204/$11)+2*cos($8*3.14/180)*sqrt((1-$9*$9)*($11/5.204)), $0}'| awk '{if ($1 > 3 && $1 > 2.8) print $0}
# trova i T3

cat mpcorb.dat | awk '{ if ($11 > 10) print $0}'
# filtra parametri orbitali

The primary The primary orbital elements are here denoted as:
    N = longitude of the ascending node
    i = inclination to the ecliptic (plane of the Earth's orbit)
    w = argument of perihelion
    a = semi-major axis, or mean distance from Sun
    e = eccentricity (0=circle, 0-1=ellipse, 1=parabola)
    M = mean anomaly (0 at perihelion; increases uniformly with time)
 Related orbital elements are: 
    w1 = N + w   = longitude of perihelion
    L  = M + w1  = mean longitude
    q  = a*(1-e) = perihelion distance
    Q  = a*(1+e) = aphelion distance
    P  = a ^ 1.5 = orbital period (years if a is in AU, astronomical units)
    T  = Epoch_of_M - (M(deg)/360_deg) / P  = time of perihelion
    v  = true anomaly (angle between position and perihelion)
    E  = eccentric anomalyorbital elements are here denoted as:
    N = longitude of the ascending node
    i = inclination to the ecliptic (plane of the Earth's orbit)
    w = argument of perihelion
    a = semi-major axis, or mean distance from Sun
    e = eccentricity (0=circle, 0-1=ellipse, 1=parabola)
    M = mean anomaly (0 at perihelion; increases uniformly with time)

 q  Perihelion distance  = a * (1 - e)

    Q  Aphelion distance    = a * (1 + e)

    P  Orbital period       = 365.256898326 * a**1.5/sqrt(1+m) days,
       where m = the mass of the planet in solar masses (0 for
       comets and asteroids). sqrt() is the square root function.

    n  Daily motion         = 360_deg / P    degrees/day

    t  Some epoch as a day count, e.g. Julian Day Number. The Time
       at Perihelion, T, should then be expressed as the same day count.

    t - T   Time since Perihelion, usually in days

    M  Mean Anomaly         = n * (t - T)  =  (t - T) * 360_deg / P
       Mean Anomaly is 0 at perihelion and 180 degrees at aphelion

    L  Mean Longitude       = M + w + N

    E  Eccentric anomaly, defined by Kepler's equation:   M = E - e * sin(E)
       An auxiliary angle to compute the position in an elliptic orbit

    v  True anomaly: the angle from perihelion to the planet, as seen
       from the Sun

    r  Heliocentric distance: the planet's distance from the Sun.
 This relation is valid for an elliptic orbit:

        r * cos(v) = a * (cos(E) - e)
        r * sin(v) = a * sqrt(1 - e*e) * sin(E)

    x,y,z  Rectangular coordinates. Used e.g. when a heliocentric
           position (seen from the Sun) should be converted to a
           corresponding geocentric position (seen from the Earth).

Comete

Un interessante post di Charles Bell sul calcolo della velocità di un copro celeste

Someone asked if I could calculate to velocity of comet ISON for today. So I did and thought that since this is an interesting topic and a frequently asked question I share with everyone here.

A simple derivation using basic kinetic and potential energy distance in AU, and time in days
k = k_Gauss = 0.0172 AU/day
and dividing each energy term by mass
potential energy = - k^2 / r
kinetic energy = v^2 / 2
1 AU = 149597870.691 km
1 day = 24 hours
1 hour = 3600 secs

for an object with eccentricity ~=1,
when solar distance r --> oo, 1/r --> 0, potential energy --> 0, velocity v --> 0, kinetic energy --> 0, thus total energy at r =oo is 0.

at any arbitrary solar distance on the orbit for this object, total energy is conserved, total energy = 0
v^2/2 - k^2/r = 0
v = sqrt(2k^2 / r)
v = k sqrt(2 / r)
today for r = 0.737237538914 AU,
v = 0.0172 AU/day x sqrt( 2/ 0.737237538914)
v = 0.02832955315529571984981631585637 AU/day
v = 49 km/sec

at perihelion, solar distance = q = 0.0124464 AU, total energy is conserved, total energy = 0, and v^2/2 - k^2/q = 0
v = sqrt(2k^2/q) = k sqrt(q / 2)
v = 0.0172 AU/day x sqrt( 2/ 0.0124464)
v = 0.21803266725477195495803895356869 AU/Day
v = 377 km/sec

To check these result, I use Minor Planet Ephemeris for comet C/2012 S1 (ISON) with output to display vector output, x,y,z, vx, vy, vz
for Nov 11, 2013 00:00:00 UT
vx = 0.010739453 AU/day
vy = -0.022680545 AU/day
vz = -0.013154187 AU/day

seconds per day = 3600 * 24 = 86400
v =v = sqrt(vx * vx + vy * vy + vz * vz)
* (149597870.691 km/AU ) / 86400 sec/day
v = 49.05787740461514 km/sec

At perihelion on 2013 Nov. 28.7750
vx = 0.114893032d;
vy = -0.088597017d;
vz = 0.162412063d;
v =v = sqrt(vx * vx + vy * vy + vz * vz)
* (149597870.691 km/AU ) / 86400 sec/day
v = 377.07424511550096 km/sec

Both values are very close to basic energy result.

So using conservation of energy and convenient units and Gaussian constant, it is easy to derive an expression to calculate the velocity of a comet with eccentricity ~= 1.