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Old August 18th, 2019 #1
Jerry Abbott
Senior Member
Join Date: Nov 2007
Location: In the hills north of Hillsboro WV
Posts: 996
Jerry Abbott
Default 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 m sec⁻

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)]

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
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 + d₁ f₂ ]
d₃ = −f₀ / [ f₁ + d₁ f₂ + ⅙ d₂ 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−e)/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 12:59 PM.


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