Step-by-Step Guide to Generating an Ephemeris
SSP, July 2009
1. Get the classical orbital elements (a,e,Mp,i,) and determine M at ephemeris
time:
Using k=0.01720209894 and =1.000000 for an asteroid orbiting the sun, with time
units in days:
M  n(t  TE ) +Mp where, using Kepler’s 3rd:
nk

a3
Note that Mp is the published mean anomaly and TE is the ephemeris date. t is the
time of observation. For most of our asteroids TE=2455000.5 Julian date, which is
June 18, 2009 at 00:00 UT.
2. Get E from M inverting the Kepler equation, M  E  e sin E using the Newton
method (covered by Prof Mason later on) as follows: Your first guess for E is M.
You then use f  M  ( E  e sin E ) you find f’, and so your second guess of E
 M  ( E0  e sin( E0 )) 
f ( E0 )
 E0  
 . Repeat this step
f '( E0 )
e cos( E0 )  1


several times for E2 etc. until the M you get in Kepler’s equation differs little from
the M you had to begin with.
3. Get the Cartesian coordinates of the position vector:
 x   a cos E  ae 
  

2
 y    a 1  e sin E 
z 

0
  

That’s the vector describing the asteroid position in orbit coordinates with the x axis
coinciding with the perihelion.
4. Now start spinning that vector. First rotation is by (-), then by (i) and then by
() to get the r vector in ecliptic coordinates:
0
0  cos( )  sin(  ) 0  x 
 x1   cos()  sin( ) 0  1
  


 
 y1    sin( ) cos() 0  0 cos(i)  sin( i)  sin(  ) cos( ) 0  y 
z   0
0
1  0 sin( i) cos(i)  0
0
1  z 
 1 
5. Convert r to equatorial by rotating the coordinates by the tilt of earth relative to
the ecliptic :
0
0  x1 
 x2   1
  
 
 y 2    0 cos( )  sin(  )  y1 
 z   0 sin(  ) cos( )  z 
 2 
 1 
would be: E1  E0 
NEXT STEPS TO BE CONTINUED AT A LATER DATE:
6. Determine the earth-sun vector from the JPL site
  
 
7. Determine the range vector  from   r  R , and also ˆ    .
8. Solve for RA and Dec from : sin(  )  ˆ z , and cos( )  ˆ x cos( )