Parameters for Solar system simulation

AI Thread Summary
The discussion focuses on constructing an N-body simulation of the solar system in Java, specifically starting with the Sun, Mercury, and Venus. Participants discuss how to determine the initial positions and velocities of these bodies, using the formula for velocity derived from gravitational parameters. They highlight the importance of using actual ephemeris data for accurate initial conditions, as well as the distinction between modeling the Sun at the solar system barycenter versus its center. The conversation also touches on the implications of using a unit system where G=1, and the need to express planetary masses in terms of length and time for the simulation. Accurate initial conditions are crucial for realistic orbital modeling, and using the JPL Horizons Web Interface is recommended for obtaining necessary data.
  • #51
Thanks,
gneill said:
The eccentricity vector can be determined starting with any r,v pair. As a bonus you also get the eccentricity of the orbit and the direction to perihelion.
But what is special about this vector (i.e what makes it different from the position vector of the perihelion). Do I extract the ecliptic angle in the same way I did previously, but this time using the eccentricity vector?
 
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  • #52
CAF123 said:
Thanks,

But what is special about this vector (i.e what makes it different from the position vector of the perihelion). Do I extract the ecliptic angle in the same way I did previously, but this time using the eccentricity vector?

Nothing special other than it hands you the extra information, which can be useful if you're trying to determine the canonical orbital elements for an orbit from position and velocity.

Sure, that works.
 
  • #53
gneill said:
Nothing special other than it hands you the extra information, which can be useful if you're trying to determine the canonical orbital elements for an orbit from position and velocity.

Sure, that works.
I am using data from this page: http://ssd.jpl.nasa.gov/horizons.cgi#results
This gives position and velocity in km and km/s respectively so to compute the eccentricity, I converted everything to SI. My problem is when I do the computation of the eccentricity, via the two equations in Wikipedia, I end up with both times e = <-0.545, -0.746, 0.22> (which I confirmed with Wolfram alpha) and this means the arcsin does not exist.
 
  • #54
CAF123 said:
I am using data from this page: http://ssd.jpl.nasa.gov/horizons.cgi#results
This gives position and velocity in km and km/s respectively so to compute the eccentricity, I converted everything to SI. My problem is when I do the computation of the eccentricity, via the two equations in Wikipedia, I end up with both times e = <-0.545, -0.746, 0.22> (which I confirmed with Wolfram alpha) and this means the arcsin does not exist.

Can you post the position and velocity values here? I won't see the "results" that you see on the link you gave; it'll show a results page from my last visit there.
 
  • #55
gneill said:
Can you post the position and velocity values here? I won't see the "results" that you see on the link you gave; it'll show a results page from my last visit there.
Sorry about that, here they are:
-3.056663785023214E+09 3.715913718019426E+09 -1.453046492475981E+09 (Position x,y,z km)
-2.107989578404891E-01 1.729355547234978E+00 -3.417437459119960E-01 (Velocity vx,vy,vz km/s)
 
  • #56
Okay, using those values I find ## \vec{e} = < 0.54757, -0.74966, 0.27407 >##. Its magnitude is 0.96796. arcsin(0.27407/0.96796) certainly has a value...
 
  • #57
gneill said:
Okay, using those values I find ## \vec{e} = < 0.54757, -0.74966, 0.27407 >##. Its magnitude is 0.96796. arcsin(0.27407/0.96796) certainly has a value...
Sorry gneill for wasting your time, I was doing arcsin(0.96796/0.27401) despite the fact that I had written it down correctly. 16o is quite a good result, I believe the actual result is about 18o. Given that the method used to find this angle is not an approximation, the differences can be attributed to the integration algorithm I used?
 
  • #58
CAF123 said:
Sorry gneill for wasting your time, I was doing arcsin(0.96796/0.27401) despite the fact that I had written it down correctly. 16o is quite a good result, I believe the actual result is about 18o. Given that the method used to find this angle is not an approximation, the differences can be attributed to the integration algorithm I used?

Well, I assume that these initial position and velocity values are the real deal from Horizons, and they never get near your integrator for this determination of the angle. So any discrepancy with a 'real' value would have to have another source. A couple of things come to mind. First, the inclination of a comet can change over time due to perturbations (particularly by Jupiter). Second, the coordinate system used may not correspond to that in which the original 18° figure was determined; in particular the Fundamental Plane (ecliptic plane) changes over time as the Earth's orbit evolves, and the Horizons values were probably for a barycentric coordinate system rather than a Sun-centered one. Any and all of these things could contribute to the difference. They'd have to be tracked down individually to stamp them out.
 

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