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Jenab2
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There have been many questions on this forum about celestial mechanics in general, and concerning position and velocity in an orbit in particular. So I offer this post as a summary and reference.
Here's a method for finding heliocentric position and sun-relative velocity in ecliptic coordinates from the Keplerian orbital elements (a, e, i, Ω, ω, T) at a time of interest, t, for both elliptical and hyperbolic orbits relative to the sun.
a : semimajor axis
* here usually in astronomical units
e : eccentricity
i : inclination to the ecliptic
* entered in degrees, usually converted to radians
Ω : longitude of the ascending node
* entered in degrees, usually converted to radians
* Ω is an angle, subtended at the sun, measured in the ecliptic plane in the counterclockwise direction from the perspective of someone positioned far along the North Ecliptic Pole, from a ray extending toward the Vernal Equinox to another ray extending toward the place where the orbit crosses the ecliptic going from south to north.
ω : argument of the perihelion
* entered in degrees, usually converted to radians
* ω is an angle, subtended at the sun, measured in the plane of the orbit, in the angular direction of motion, from a ray extending toward the ascending node to another ray extending toward the perihelion of the orbit.
T : time of perihelion passage
* here usually in Julian date, with differences in time being usually in days
.
For ELLIPTICAL ORBITS.
An elliptical orbit has a positive semimajor axis and an eccentricity strictly between zero and one. It is a bound orbit with a period. The planets, the asteroids, and some comets are in elliptical orbits.
The period of the orbit, P.
P = 365.256898326 a^(1.5)
The mean anomaly, m.
m''' = (t−T)/P
m'' = m''' − integer(m''')
If m'' is less than 0, then m'=m''+1 else m'=m''
m = 2πm'
The eccentric anomaly, u.
The mean anomaly must be in radians when finding the eccentric anomaly.
u₀ = m + (e − e³/8 + e⁵/192) sin(m) + (e²/2 − e⁴/6) sin(2m) + (3e³/8 − 27e⁵/128) sin(3m) + (e⁴/3) sin(4m)
i = 0
Repeat
i = i+1
E = uᵢ − e sin uᵢ − m
F = 1 − e cos uᵢ
G = e sin uᵢ
H = e cos uᵢ
A = −E/F
B = −E/(F + ½ AG)
C = −E/(F + ½ AG + ⅙ B²H)
uᵢ₊₁ = uᵢ + C
Until |uᵢ₊₁−uᵢ| < 1e-12
u = uᵢ₊₁
The eccentric anomaly, u, will be in radians.
* u is an angle, subtended at the geometric center of the elliptical orbit (not at the sun), measured in the plane of the orbit in the direction of motion, from a ray extending through the orbit's perihelion, to another ray extending through a point on a circle that circumscribes the orbital ellipse which is in the +y direction of the position of the body in its orbit at time t.
The canonical heliocentric position vector [x''',y''',z'''].
x''' = a (cos u − e)
y''' = a sin u √(1− e²)
z''' = 0
The canonical coordinate system has the sun at the origin, the XY plane in the plane of the orbit, with the orbit's perihelion being on the +x axis, and the +z axis such that an observer somewhere along it would observe a counterclockwise direction of motion.
The true anomaly, θ.
θ = arctan( y''' , x''' )
* θ is an angle, subtended at the sun, measured in the plane of the orbit, in the angular direction of motion, from a ray extending toward the perihelion of the orbit to another ray extending toward the orbiting body's position at time t.
Rotate the coordinates from canonical to ecliptic.
x'' = x''' cos ω − y''' sin ω
y'' = x''' sin ω + y''' cos ω
z'' = z'''
x' = x''
y' = y'' cos i
z' = y'' sin i
Position in heliocentric ecliptic coordinates [x,y,z].
x = x' cos Ω − y' sin Ω
y = x' sin Ω + y' cos Ω
z = z'
The canonical heliocentric velocity vector [Vx''',Vy''',Vz'''].
Vx''' = (−29784.6918 m/s) √{1/[a(1−e²)]} sin θ
Vy''' = (29784.6918 m/s) √{1/[a(1−e²)]} (e + cos θ)
Vz''' = 0
Enter the semimajor axis, a, in astronomical units. The dimension of length has already been extracted from the square root.
Rotate the coordinates from canonical to ecliptic.
Vx'' = Vx''' cos ω − Vy''' sin ω
Vy'' = Vx''' sin ω + Vy''' cos ω
Vz'' = Vz'''
Vx' = Vx''
Vy' = Vy'' cos i
Vz' = Vy'' sin i
Sun-relative velocity in ecliptic coordinates [Vx,Vy,Vz].
Vx = Vx' cos Ω − Vy' sin Ω
Vy = Vx' sin Ω + Vy' cos Ω
Vz = Vz'
Sun-relative speed, v.
v = √[(Vx)²+(Vy)²+(Vz)²]
v = (29784.6918 m/s) √(2/r−1/a)
Enter a,r in astronomical units. The dimension of length has already been extracted from the square root.
.
For HYPERBOLIC ORBITS.
A hyperbolic orbit has a negative semimajor axis and an eccentricity greater than one. It is unbound and not periodic, but one-pass only. Some comets are in hyperbolic orbits.
The hyperbolic mean anomaly, m.
m = 0.01720209895 (t−T) √[1/(−a)³]
The hyperbolic eccentric anomaly, u.
The hyperbolic mean anomaly must be in radians when finding the hyperbolic eccentric anomaly.
u₀ = 0
i = 0
Repeat
i = i+1
E = e sinh uᵢ − uᵢ − m
F = e cosh uᵢ − 1
G = e sinh uᵢ
H = e cosh uᵢ
A = −E/F
B = −E/(F + ½ AG)
C = −E/(F + ½ AG + ⅙ B²H)
uᵢ₊₁ = uᵢ + C
Until |uᵢ₊₁−uᵢ| < 1e-12
u = uᵢ₊₁
Distance from the sun, r.
r = a (1 − e cosh u)
True anomaly, θ.
θ = arccos{ (e − cosh u) / (e cosh u − 1) }
The canonical heliocentric position vector [x''',y''',z'''].
x''' = r cos θ
y''' = r sin θ
z''' = 0
Rotate the coordinates from canonical to ecliptic.
x'' = x''' cos ω − y''' sin ω
y'' = x''' sin ω + y''' cos ω
z'' = z'''
x' = x''
y' = y'' cos i
z' = y'' sin i
Position in heliocentric ecliptic coordinates [x,y,z].
x = x' cos Ω − y' sin Ω
y = x' sin Ω + y' cos Ω
z = z'
The canonical heliocentric velocity vector [Vx''',Vy''',Vz''').
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
Enter a,r in astronomical units. The dimension of length has already been extracted from the square root.
Rotate the coordinates from canonical to ecliptic.
Vx'' = Vx''' cos ω − Vy''' sin ω
Vy'' = Vx''' sin ω + Vy''' cos ω
Vz'' = Vz'''
Vx' = Vx''
Vy' = Vy'' cos i
Vz' = Vy'' sin i
Sun-relative velocity in ecliptic coordinates [Vx,Vy,Vz].
Vx = Vx' cos Ω − Vy' sin Ω
Vy = Vx' sin Ω + Vy' cos Ω
Vz = Vz'
Sun-relative speed, v.
v = √[(Vx)²+(Vy)²+(Vz)²]
v = (29784.6918 m/s) √(2/r−1/a)
Enter a,r in astronomical units. The dimension of length has already been extracted from the square root.
Here's a method for finding heliocentric position and sun-relative velocity in ecliptic coordinates from the Keplerian orbital elements (a, e, i, Ω, ω, T) at a time of interest, t, for both elliptical and hyperbolic orbits relative to the sun.
a : semimajor axis
* here usually in astronomical units
e : eccentricity
i : inclination to the ecliptic
* entered in degrees, usually converted to radians
Ω : longitude of the ascending node
* entered in degrees, usually converted to radians
* Ω is an angle, subtended at the sun, measured in the ecliptic plane in the counterclockwise direction from the perspective of someone positioned far along the North Ecliptic Pole, from a ray extending toward the Vernal Equinox to another ray extending toward the place where the orbit crosses the ecliptic going from south to north.
ω : argument of the perihelion
* entered in degrees, usually converted to radians
* ω is an angle, subtended at the sun, measured in the plane of the orbit, in the angular direction of motion, from a ray extending toward the ascending node to another ray extending toward the perihelion of the orbit.
T : time of perihelion passage
* here usually in Julian date, with differences in time being usually in days
.
For ELLIPTICAL ORBITS.
An elliptical orbit has a positive semimajor axis and an eccentricity strictly between zero and one. It is a bound orbit with a period. The planets, the asteroids, and some comets are in elliptical orbits.
The period of the orbit, P.
P = 365.256898326 a^(1.5)
The mean anomaly, m.
m''' = (t−T)/P
m'' = m''' − integer(m''')
If m'' is less than 0, then m'=m''+1 else m'=m''
m = 2πm'
The eccentric anomaly, u.
The mean anomaly must be in radians when finding the eccentric anomaly.
u₀ = m + (e − e³/8 + e⁵/192) sin(m) + (e²/2 − e⁴/6) sin(2m) + (3e³/8 − 27e⁵/128) sin(3m) + (e⁴/3) sin(4m)
i = 0
Repeat
i = i+1
E = uᵢ − e sin uᵢ − m
F = 1 − e cos uᵢ
G = e sin uᵢ
H = e cos uᵢ
A = −E/F
B = −E/(F + ½ AG)
C = −E/(F + ½ AG + ⅙ B²H)
uᵢ₊₁ = uᵢ + C
Until |uᵢ₊₁−uᵢ| < 1e-12
u = uᵢ₊₁
The eccentric anomaly, u, will be in radians.
* u is an angle, subtended at the geometric center of the elliptical orbit (not at the sun), measured in the plane of the orbit in the direction of motion, from a ray extending through the orbit's perihelion, to another ray extending through a point on a circle that circumscribes the orbital ellipse which is in the +y direction of the position of the body in its orbit at time t.
The canonical heliocentric position vector [x''',y''',z'''].
x''' = a (cos u − e)
y''' = a sin u √(1− e²)
z''' = 0
The canonical coordinate system has the sun at the origin, the XY plane in the plane of the orbit, with the orbit's perihelion being on the +x axis, and the +z axis such that an observer somewhere along it would observe a counterclockwise direction of motion.
The true anomaly, θ.
θ = arctan( y''' , x''' )
* θ is an angle, subtended at the sun, measured in the plane of the orbit, in the angular direction of motion, from a ray extending toward the perihelion of the orbit to another ray extending toward the orbiting body's position at time t.
Rotate the coordinates from canonical to ecliptic.
x'' = x''' cos ω − y''' sin ω
y'' = x''' sin ω + y''' cos ω
z'' = z'''
x' = x''
y' = y'' cos i
z' = y'' sin i
Position in heliocentric ecliptic coordinates [x,y,z].
x = x' cos Ω − y' sin Ω
y = x' sin Ω + y' cos Ω
z = z'
The canonical heliocentric velocity vector [Vx''',Vy''',Vz'''].
Vx''' = (−29784.6918 m/s) √{1/[a(1−e²)]} sin θ
Vy''' = (29784.6918 m/s) √{1/[a(1−e²)]} (e + cos θ)
Vz''' = 0
Enter the semimajor axis, a, in astronomical units. The dimension of length has already been extracted from the square root.
Rotate the coordinates from canonical to ecliptic.
Vx'' = Vx''' cos ω − Vy''' sin ω
Vy'' = Vx''' sin ω + Vy''' cos ω
Vz'' = Vz'''
Vx' = Vx''
Vy' = Vy'' cos i
Vz' = Vy'' sin i
Sun-relative velocity in ecliptic coordinates [Vx,Vy,Vz].
Vx = Vx' cos Ω − Vy' sin Ω
Vy = Vx' sin Ω + Vy' cos Ω
Vz = Vz'
Sun-relative speed, v.
v = √[(Vx)²+(Vy)²+(Vz)²]
v = (29784.6918 m/s) √(2/r−1/a)
Enter a,r in astronomical units. The dimension of length has already been extracted from the square root.
.
For HYPERBOLIC ORBITS.
A hyperbolic orbit has a negative semimajor axis and an eccentricity greater than one. It is unbound and not periodic, but one-pass only. Some comets are in hyperbolic orbits.
The hyperbolic mean anomaly, m.
m = 0.01720209895 (t−T) √[1/(−a)³]
The hyperbolic eccentric anomaly, u.
The hyperbolic mean anomaly must be in radians when finding the hyperbolic eccentric anomaly.
u₀ = 0
i = 0
Repeat
i = i+1
E = e sinh uᵢ − uᵢ − m
F = e cosh uᵢ − 1
G = e sinh uᵢ
H = e cosh uᵢ
A = −E/F
B = −E/(F + ½ AG)
C = −E/(F + ½ AG + ⅙ B²H)
uᵢ₊₁ = uᵢ + C
Until |uᵢ₊₁−uᵢ| < 1e-12
u = uᵢ₊₁
Distance from the sun, r.
r = a (1 − e cosh u)
True anomaly, θ.
θ = arccos{ (e − cosh u) / (e cosh u − 1) }
The canonical heliocentric position vector [x''',y''',z'''].
x''' = r cos θ
y''' = r sin θ
z''' = 0
Rotate the coordinates from canonical to ecliptic.
x'' = x''' cos ω − y''' sin ω
y'' = x''' sin ω + y''' cos ω
z'' = z'''
x' = x''
y' = y'' cos i
z' = y'' sin i
Position in heliocentric ecliptic coordinates [x,y,z].
x = x' cos Ω − y' sin Ω
y = x' sin Ω + y' cos Ω
z = z'
The canonical heliocentric velocity vector [Vx''',Vy''',Vz''').
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
Enter a,r in astronomical units. The dimension of length has already been extracted from the square root.
Rotate the coordinates from canonical to ecliptic.
Vx'' = Vx''' cos ω − Vy''' sin ω
Vy'' = Vx''' sin ω + Vy''' cos ω
Vz'' = Vz'''
Vx' = Vx''
Vy' = Vy'' cos i
Vz' = Vy'' sin i
Sun-relative velocity in ecliptic coordinates [Vx,Vy,Vz].
Vx = Vx' cos Ω − Vy' sin Ω
Vy = Vx' sin Ω + Vy' cos Ω
Vz = Vz'
Sun-relative speed, v.
v = √[(Vx)²+(Vy)²+(Vz)²]
v = (29784.6918 m/s) √(2/r−1/a)
Enter a,r in astronomical units. The dimension of length has already been extracted from the square root.
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