Full Thread: Voyager 1 View Single Post August 18th, 2019 #1 Jerry Abbott Senior Member   Join Date: Nov 2007 Location: In the hills north of Hillsboro WV Posts: 996  Voyager 1 NASA's veteran space probe, Voyager 1, is the bit of white culture that has gone the greatest distance away from Earth. Using its hyperbolic orbital elements from JPL's HORIZONS website, I calculated that Voyager 1 will be exactly one light-day (1440 light-minutes, or 25.9 trillion meters) from the sun sometime on 5 December 2026, the day that would have been my sister's 60th birthday. The Keplerian orbital elements for a hyperbolic orbit. [ a, e, i, Ω, ω, T ] @ time = t Notes: a<0, e>1 Sun's gravitational parameter, GM๏. GM๏ = 1.32712440018e20 m³ sec⁻² Conversion between astronomical units and meters. 1 AU = 1.49597870691e11 meters Hyperbolic mean motion, m, of the comet in radians per day. m = (86400 sec/day) √[GM๏/(−a)³] Choose time of observation, t, and convert it to Julian Date. Examples: 28 November 2013 at 12h GMT is JD 2456625.0 1 January 2014 at 18h GMT is JD 2456659.25 Hyperbolic mean anomaly, M, of the comet at time t. M = m (t−T) Eccentric anomaly, u, of the comet. u₀ = 0 i = −1 Repeat i = i + 1 f₀ = e sinh uᵢ − uᵢ − M f₁ = e cosh uᵢ − 1 f₂ = e sinh uᵢ f₃ = e cosh uᵢ d₁ = −f₀ / f₁ d₂ = −f₀ / [ f₁ + ½ d₁ f₂ ] d₃ = −f₀ / [ f₁ + ½ d₁ f₂ + ⅙ d₂² f₃ ] uᵢ₊₁ = uᵢ + d₃ Until |uᵢ₊₁−uᵢ| < 1e-12 u = uᵢ₊₁ True anomaly, θ, of the comet. θ' = arccos { (e − cosh u) / (e cosh u − 1) } if u>0 then θ = θ' if u=0 then θ = 0 if u<0 then θ = 2π − θ' Heliocentric distance, r, of the comet. r = a (1 − e cosh u) Canonical position vector. x''' = r cos θ y''' = r sin θ z''' = 0 Rotate the canonical position by the argument of the perihelion. x'' = x''' cos ω − y''' sin ω y'' = x''' sin ω + y''' cos ω z'' = z''' Rotate the double-primed vector by the inclination. x' = x'' y' = y'' cos i z' = y'' sin i Rotate the single-primed vector by the longitude of ascending node to get the heliocentric position of the comet in ecliptic coordinates. x = x' cos Ω − y' sin Ω y = x' sin Ω + y' cos Ω z = z' The canonical velocity vector is Vx''' = 29784.6918 m/s (a/r) √(−1/a) sinh u Vy''' = −29784.6918 m/s (a/r) √[(1−e²)/a] cosh u Vz''' = 0 And to convert that velocity vector to heliocentric ecliptic coordinates, you'd put it through the same rotations that applied to the canonical position vector. Last edited by Jerry Abbott; August 18th, 2019 at 01:59 PM.