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Astronomy and Cosmology
Astronomy and Astrophysics
Position and Velocity in Heliocentric Ecliptic Coordinates
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[QUOTE="Jenab2, post: 5479348, member: 107288"] You can also go the other way. Beginning with a position vector and a velocity vector in heliocentric ecliptic coordinates, at a time t, for an object in orbit around the sun, you can derive its Keplerian orbital elements (a, e, i, Ω, ω, T). The combination of the vectors of position and of velocity in an orbit is sometimes called the "state vector." If you are given the position in spherical ecliptic coordinates (r = distance, λ = ecliptic longitude, β = ecliptic latitude), then do this to get the position in rectangular ecliptic coordinates: x = r cos λ cos β y = r sin λ cos β z = r sin β Let's assume that x, y, and z are in meters, and that Vx, Vy, Vz are the components of the sun-relative velocity in ecliptic coordinates in meters per second. Let t be the time-of-interest in Julian date. [B]The solar gravitational parameter.[/B] GM = 1.32712440018e20 m³ sec⁻² [B]The heliocentric distance, r (in meters), of the object at time t.[/B] r = √( x² + y² + z² ) [B]The sun-relative speed, v (in m/sec), of the object at time t.[/B] v = √[ (Vx)² + (Vy)² + (Vz)² ] [B]The semimajor axis, a, of the orbit, meters.[/B] a = 1 / { 2/r − v²/(GM) } You can convert the semimajor axis to astronomical units. 1 AU = 1.49597870700e11 meters. [B]The angular momentum per unit mass (m²/sec), h, in the orbit.[/B] hx = y Vz − z Vy hy = z Vx − x Vz hz = x Vy − y Vx h = √[ (hx)² + (hy)² + (hz)² ] [B]The eccentricity, e, of the orbit.[/B] e = √[ 1 − h² / (GMa) ] [B]The inclination, i, of the orbit.[/B] i = ArcCos( hz / h ) [B]The longitude of the ascending node, Ω, of the orbit.[/B] Ω = Arctan ( hx , −hy ) [COLOR=#660000][B]Definition of the two-argument Arctan function (result in radians).[/B] atn(z) = single argument Arctan function of the argument z. Function Arctan( y , x ) if x = 0 and y > 0 then angle = +π/2 if x = 0 and y = 0 then angle = 0 if x = 0 and y < 0 then angle = −π/2 if x < 0 then angle = atn(y/x) + π if x > 0 and y > 0 then angle = atn(y/x) if x > 0 and y < 0 then angle = atn(y/x) + 2π Arctan = angle[/COLOR] Note: the math to this point works for either elliptical or hyperbolic orbits. However, some of the stuff to follow works [I]only for elliptical orbits[/I]. You'll know that the orbit is a hyperbola if (1) the semimajor axis is negative and (2) the eccentricity is greater than one. [B]The true anomaly, θ, of the object in the orbit at time t.[/B] cos θ = h²/(rGM) − 1 sin θ = h (x Vx + y Vy + z Vz) / (rGM) θ = Arctan ( sin θ , cos θ ) [B]Argument of the perihelion, ω, of the orbit.[/B] cos ω'' = (x cos Ω + y sin Ω) / r If sin i = 0 then sin ω'' = (y cos Ω − x sin Ω) / r If sin i ≠ 0 then sin ω'' = z / (r sin i) ω'' = Arctan( sin ω'' , cos ω'' ) ω' = ω'' − θ If ω' > 0 then ω = ω' If ω' < 0 then ω = ω' + 2π [B]The eccentric anomaly, u, of the object in the orbit at time t.[/B] sin u = (r/a) sin θ / √(1−e²) cos u = (r/a) cos θ + e u = Arctan( sin u , cos u ) [B]The mean anomaly, m, of the object in the orbit at time t.[/B] m = u − e sin u Note: u must be entered in radians, and m will return in radians. [B]The period of the orbit, P, days.[/B] P = (π / 43200) √[ a³/(GM) ] [B]The mean motion, μ, in the orbit, radians per day.[/B] μ = 2π / P [B]The time of perihelion passage, T, of the object in the orbit, Julian Date.[/B] T = t − m/μ [B]The Keplerian orbital elements.[/B] [ a, e, i, Ω, ω, T ] [/QUOTE]
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Position and Velocity in Heliocentric Ecliptic Coordinates
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