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.