David Sims

Time to Fall in a Plunge Orbit

The moon doesn't fall on Earth because it has enough transverse velocity to remain in an orbit around Earth, and because angular momentum is conserved as a natural law.

Now, if you removed all of the moon's transverse velocity when the moon reached its aphelion, leaving the moon momentarily stationary with respect to Earth, then the moon and Earth would indeed fall together. How long would it take?

Two bodies having a total mass M (i.e., M=M₁+M₂) are initially at rest, separated by a distance d, in vacuum, and isolated from all forces except their mutual gravitational attraction. Find the time elapsed from the initial moment to the moment at which the separation is r, such that 0<r<d.

The problem involves the conservation of energy in a gravity dominated physical system, so we begin with the Vis Viva equation, solved for the speed in orbit:

v = √[GM(2/r − 1/a)]

where G = 6.6743e-11 m³ kg⁻¹ sec⁻²

The eccentricity of a plunge orbit is one, just as it is for a parabolic orbit. Why? Because for an elliptical orbit,

e = √(1 − b²/a²)

where b is the length of the semiminor axis, and a is the length of the semimajor axis. As b→0, the elliptical orbit becomes a plunge orbit, and e→1.

The apoapsis separation of M₁ and M₂ is therefore twice the semimajor axis.

e = 1
d = a(1+e) 
d = 2a 
a = d/2

And, since the angular momentum in a plunge orbit is zero, all of the motion therein is radial, and so

v = ∂r/∂t

Therefore, the Vis Viva equation can be rewritten as

∂r/∂t = √[2GM(1/r − 1/d)]

which is a non-linear differential equation with variables separable. We shall integrate both sides over the entire course of the fall, from the apoapsis of the plunge orbit to contact, assuming that both M₁ and M₂ are point masses.

∫(t₀,tᵢ) ∂t = − ∫(rᵢ,d) { ∂r / √[2GM(1/r − 1/d)] }

The minus sign appears because the actual motion in the plunge orbit begins at d and ends at rᵢ, and I flipped those limits. However, hereafter I'll omit the limits and treat the indefinite integral.

tᵢ−t₀ = −√[d/(2GM)] ∫ ∂r/√(d/r−1)

We make this substitution:

u = √(d/r−1)
∂u/∂r = −½ d r⁻²/√(d/r−1)
r = d/(u²+1)
∂r = (−2d) u [∂u/(u²+1)²]

Plugging them in, we get

t−t₀ = 2d √[d/(2GM)] ∫ ∂u/(u²+1)²

Integrating by parts,

t−t₀ = 2d √[d/(2GM)] { ½ ∫ [∂u/(u²+1)] + ½ u/(u²+1) }
t−t₀ = d √[d/(2GM)] { u/(u²+1) + ∫ [∂u/(u²+1)] }

We notice an integral-trigonometric identity.

∫ ∂u/(u²+1) = arctan u

So,

t−t₀ = d √[d/(2GM)] { u/(u²+1) + arctan u }

Reversing the substitution,

t−t₀ = d √[d/(2GM)] { √(d/r−1)/[(d/r−1)+1] + arctan √(d/r−1) }

After simplification, we get as the general solution for the time to fall in a plunge orbit:

tᵢ−t₀ = √[d/(2GM)] { √(rᵢd−rᵢ²) + d arccos √(rᵢ/d) }

In the case where the motion in a plunge orbit begins at the apoapsis and ends in contact (assuming point masses),

rᵢ = 0
tᵢ−t₀ = π √[d³/(8GM)]

When determining the time to fall from r₁ to r₂, such that d>r₁>r₂>0,

t₂−t₁ = (t₂−t₀) − (t₁−t₀)

Let's consider what happens when we let the moon drop straight down from its apogee to first contact between the two bodies, allowing them to keep their actual size and shape.

M = 6.0483e+24 kilograms
d = 405,503,560 meters
r₁ = 8,108,400 meters
t₁−t₀ = 450871.423 seconds = 125.242062 hours

Now let's run the fall again, from apogee to center-to-center contact, assuming that Earth and moon are both point masses.

M = 6.0483e+24 kilograms
d = 405,503,560 meters
r₂ = 0
t₂−t₀ = 451416.430 seconds = 125.393453 hours

The difference is about 9 minutes and 5 seconds.
 

As another example, let's find out how much time is required for the separation to decrease from r₁=d/2 to r₂=d/3.

t₁−t₀ = √[d/(2GM)] { √(r₁d−r₁²) + d arctan √(d/r₁−1) }
t₁−t₀ = √[d/(2GM)] { √(d²/2−d²/4) + d arctan √(2d/d−1) }
t₁−t₀ = √[d/(2GM)] { √(d²/4) + d arctan √(2−1) }
t₁−t₀ = √[d/(2GM)] { d/2 + dπ/4 }
t₁−t₀ = {1/2 + π/4} √[d³/(2GM)]

t₂−t₀ = √[d/(2GM)] { √(r₂d−r₂²) + d arctan √(d/r₂−1) }
t₂−t₀ = √[d/(2GM)] { √(d²/3−d²/9) + d arctan √(3d/d−1) }
t₂−t₀ = √[d/(2GM)] { √(2d²/9) + d arctan √(3−1) }
t₂−t₀ = √[d/(2GM)] { √(2d²/9) + d arctan √(2) }
t₂−t₀ = {√(2/9) + arctan √(2)} √[d³/(2GM)]

t₂−t₁ = {√(2/9) + arctan √(2) − 1/2 − π/4} √[d³/(2GM)]
t₂−t₁ ≈ 0.141322975518 √[d³/(2GM)]

As a teaser, I'll ask readers to calculate the fraction of d that is closed in half of the total amount of time required for two point masses to fall to contact.

Determining the Keplerian Elements of an Elliptical Orbit from Four Observations

Presented hereafter is a method for determining a preliminary heliocentric orbit from four geocentric directions of a sun-orbiting object at four distinct times of observation. I take this method after that presented in chapter six of The Determination of Orbits by A.D. Dubyago.

The initial data

t₁, X⊕₁, Y⊕₁, Z⊕₁, α₁, δ₁
t₂, X⊕₂, Y⊕₂, Z⊕₂, α₂, δ₂
t₃, X⊕₃, Y⊕₃, Z⊕₃, α₃, δ₃
t₄, X⊕₄, Y⊕₄, Z⊕₄, α₄, δ₄

The times of observation, tᵢ, are given in Julian date format. The vectors [X⊕ᵢ,Y⊕ᵢ,Z⊕ᵢ] are the positions of the Earth in heliocentric ecliptic coordinates, with the components being in astronomical units. The αᵢ are the geocentric right ascensions for Ceres. The δᵢ are the geocentric declinations for Ceres.

The time intervals should be about 0.5% to 1% of the object's period (estimated as 8 to 16 days for a main belt asteroid), should be near opposition with the sun, but should NOT span an apside of the object's orbit. The precision in right ascension should be 0.01 seconds or better, and the precision in declination should be 0.1 arcseconds or better.

The Earth's orbital mean motion, κ = 0.01720209895 radians/day

We will use a single value for the obliquity of the ecliptic to transform all four of the observation angles from celestial coordinates to ecliptic coordinates. First, we find the middle of the observation time window in units of 10000 years since 1 January 2000, and then we use that number to find the obliquity using the 10-degree polynomial fit of J. Laskar.

t = (t₁ + t₄)/2
T = (t − 2451545)/3652500

The obliquity in seconds of arc is

ε" = 84381.448 − 4680.93 T − 1.55 T² + 1999.25 T³ − 51.38 T⁴ − 249.67 T⁵ − 39.05 T⁶ + 7.12 T⁷ + 27.87 T⁸ + 5.79 T⁹ + 2.45 T¹⁰

The obliquity in radians is

ε = (π / 648000) ε"

Although Earth's axial tilt (i.e. the obliquity of the ecliptic) does change over time, it changes so slowly that the difference will almost always be negligible across the t₁ to t₄ time window. However, in the rare case when this isn't true, separate evaluations of the obliquity will have to be made for each time of observation.

The geocentric positions of the sun in celestial coordinates are (for i = 1 to 4)

Xᵢ = −X⊕ᵢ
Yᵢ = −Y⊕ᵢ cos ε + Z⊕ᵢ sin ε
Zᵢ = −Y⊕ᵢ sin ε − Z⊕ᵢ cos ε

The geocentric unit vectors in the direction of the target object, in celestial coordinates, are (for i = 1 to 4)

aᵢ = cos αᵢ cos δᵢ
bᵢ = sin αᵢ cos δᵢ
cᵢ = sin δᵢ

The squares of the Sun-Earth distances at times t₁ and t₄ are

R₁² = X₁² + Y₁² + Z₁²
R₄² = X₄² + Y₄² + Z₄²

The values of 2Rᵢ cos θᵢ at times t₁ and t₄ are

2R₁ cos θ₁ = −2 ( a₁X₁ + b₁Y₁ + c₁Z₁ )
2R₄ cos θ₄ = −2 ( a₄X₄ + b₄Y₄ + c₄Z₄ )

where θ is the supplementary angle to the sun-Earth-object angle.

We find some time differences:

τ₁ = κ (t₄−t₂)
τ₂ = κ (t₂−t₁)
τ₃ = κ (t₄−t₁)
τ₄ = κ (t₄−t₃)
τ₅ = κ (t₃−t₁)

We find this pair of determinants:

Φ = a₂b₄−b₂a₄
φ = a₃b₄−b₃a₄

We calculate some intermediate quantities:

A = ( a₁b₂ − b₁a₂ ) / Φ
B = ( a₂Y₁ − b₂X₁ ) / Φ
C = ( b₂X₂ − a₂Y₂ ) / Φ
D = ( a₂Y₄ − b₂X₄ ) / Φ

A' = ( a₁b₃ − b₁a₃ ) / φ
B' = ( a₃Y₁ − b₃X₁ ) / φ
C' = ( b₃X₃ − a₃Y₃ ) / φ
D' = ( a₃Y₄ − b₃X₄ ) / φ

And then more intermediate quantities:

E = τ₁/τ₂
F = (4/3) τ₁τ₃
G = AE
H = F(A−G)
I = 4Aτ₁²
K = E (B + C) + C + D
L = F (B − C + D − K)
M = 4 (Bτ₁² + τ₁τ₂C)

E' = τ₄/τ₅
F' = (4/3) τ₄τ₃
G' = A'E'
H' = F'(A'−G')
I' = 4A'τ₄²
K' = E' (B' + C') + C' + D'
L' = F' (B' − C' + D' − K')
M' = 4 (B'τ₄² + τ₄τ₅C')

Make initial guesses for the sun-object distance r₁ at time t₁ and for the sun-object distance r₄ at time t₄. For main belt asteroids, a reasonable initial guess for both times is 2.75 AU. Then use the loop below to converge, by successive approximations, to the true sun-object distances, r, and for the Earth-object distances, ρ, distances at times t₁° and t₄° (i.e. the times for the first and fourth observations, corrected for the speed of light travel time).

O = 9.999e+99
N = r₁ + r₄

while |N−O|/N > 1ᴇ-11 do
ξ = (r₁ + r₄)⁻³
η = (r₄ − r₁) / (r₁ + r₄)
P = G + ξH + ηξI
Q = K + ξL + ηξM
P' = G' + ξH' + ηξI'
Q' = K' + ξL' + ηξM'
ρ₁ = (Q'−Q)/(P−P')
ρ₄ = Pρ₁ + Q
r₁ = √[R₁² + (2R₁cos θ₁)ρ₁ + ρ₁²]
r₄ = √[R₄² + (2R₄cos θ₄)ρ₄ + ρ₄²]
O = N
N = r₁ + r₄
endwhile

The positions of the object at times t₁ and t₄ in geocentric celestial coordinates are

x₁ = a₁ρ₁ − X₁
y₁ = b₁ρ₁ − Y₁
z₁ = c₁ρ₁ − Z₁

x₄ = a₄ρ₄ − X₄
y₄ = b₄ρ₄ − Y₄
z₄ = c₄ρ₄ − Z₄

The reciprocal of the speed of light
ç = 0.00577551833 days/AU

The sun's gravitational parameter
μ = 1.32712440018ᴇ20 m³ sec⁻²

The conversion factor from AU to meters is
U = 1.495978707ᴇ11

The conversion factor from AU/day to m/sec is
β = 1731456.8368

Correcting times of observation for planetary abberation.

t₁° = t₁ − çρ₁
t₄° = t₄ − çρ₄

The nominal time associated with the forthcoming state vector is

t₀ = ½ (t₁° + t₄°)

The nominal heliocentric distance of the object at time t₀ is

r₀ = ½ (r₁ + r₄)

Find the heliocentric position vector [x',y',z'] for the object at time t₀ in celestial coordinates.

x" = ½ (x₁ + x₄)
y" = ½ (y₁ + y₄)
z" = ½ (z₁ + z₄)
r" = √[(x")²+(y")²+(z")²]
x' = (r₀/r") U x"
y' = (r₀/r") U y"
z' = (r₀/r") U z"

Find the sun-relative velocity vector for the object at time t₀ in celestial coordinates.

S = √[ (x₄ − x'/U)² + (y₄ − y'/U)² + (z₄ − z'/U)² ]
s = √[ (x'/U − x₁)² + (y'/U − y₁)² + (z'/U − z₁)² ]
Ψ = S + s
ψ = √[(x₄−x₁)² + (y₄−y₁)² + (z₄−z₁)²]
Vx' = β (Ψ/ψ) (x₄−x₁) / (t₄°−t₁°)
Vy' = β (Ψ/ψ) (y₄−y₁) / (t₄°−t₁°)
Vz' = β (Ψ/ψ) (z₄−z₁) / (t₄°−t₁°)

The object's position in heliocentric ecliptic coordinates

x₀ = x'
y₀ = y' cos ε + z' sin ε
z₀ = −y' sin ε + z' cos ε

The object's sun-relative velocity in ecliptic coordinates

Vx₀ = Vx'
Vy₀ = Vy' cos ε + Vz' sin ε
Vz₀ = −Vy' sin ε + Vz' cos ε

The object's speed relative to the sun

V₀ = √[(Vx₀)² + (Vy₀)² + (Vz₀)²]

The semimajor axis of the object's orbit, in AU

a = (2/r₀ − V₀²/μ)⁻¹ / U

The angular momentum per unit mass in the object's orbit

hx = y₀ Vz₀ − z₀ Vy₀
hy = z₀ Vx₀ − x₀ Vz₀
hz = x₀ Vy₀ − y₀ Vx₀

h = √[(hx)² + (hy)² + (hz)²]

The eccentricity of the object's orbit

e = √[1 − h²/(aμU)]

The inclination of the object's orbit

i = arccos(hz/h)

The longitude of the ascending node of the object's orbit

Ω' = arctan(−hx/hy)
if hy>0 then Ω = Ω' + π
If hy<0 and hx<0 then Ω = Ω' + 2π

The true anomaly at time t₀

sin θ₀ = h ( x₀ Vx₀ + y₀ Vy₀ + z₀ Vz₀ ) / (r₀μ)
cos θ₀ = h²/(r₀μ) − 1
θ₀' = arctan( sin θ₀ / cos θ₀ )
If cos θ₀ < 0 then θ₀ = θ₀' + π
If cos θ₀ > 0 and sin θ₀ < 0 then θ₀ = θ₀' + 2π

The sum of the true anomaly at time t₀ and the argument of the perihelion of the object's orbit

sin(θ₀+ω) = z₀ / (r₀ sin i)
cos(θ₀+ω) = ( x₀ cos Ω + y₀ sin Ω ) / r₀
(θ₀+ω)' = arctan[ sin(θ₀+ω) / cos(θ₀+ω) ]
If cos(θ₀+ω) < 0 then θ₀ = θ₀' + π
If cos(θ₀+ω) > 0 and sin(θ₀+ω) < 0 then (θ₀+ω) = (θ₀+ω)' + 2π

The argument of the perihelion of the object's orbit

ω' = (θ₀+ω) − θ₀
if ω'<0 then ω=ω'+2π else ω=ω'

The eccentric anomaly of the object at time t₀

cos u₀ = 1 − r₀/(aU)
sin u₀ = (x₀ Vx₀ + y₀ Vy₀ + z₀ Vz₀) / √(aμU)
u₀' = arctan( sin u₀ / cos u₀ )
If cos u₀ < 0 then u₀ = u₀' + π
If cos u₀ > 0 and sin u₀ < 0 then u₀ = u₀' + 2π

The mean anomaly of the object at time t₀

M₀ = u₀ − e sin u₀

The period of the object's orbit in days

P = 365.256898326 a¹·⁵

The object's time of perihelion passage

T = t₀ − PM₀/(2π)

 

Example problem. Find the orbit of Ceres.

Observation #1
t₁ = JD 2457204.625
X⊕₁ = +0.155228396 AU
Y⊕₁ = −1.004732775 AU
Z⊕₁ = +0.00003295786 AU
α₁ = 20h 46m 57.02s
δ₁ = −27°41′33.9″

Observation #2
t₂ = JD 2457214.625
X⊕₂ = +0.319493277 AU
Y⊕₂ = −0.965116604 AU
Z⊕₂ = +0.0000311269 AU
α₂ = 20h 39m 57.10s
δ₂ = −28°47′21.5″

Observation #3
t₃ = JD 2457224.625
X⊕₃ = +0.4747795623 AU
Y⊕₃ = −0.8983801739 AU
Z⊕₃ = +0.00002841127 AU
α₃ = 20h 31m 22.81s
δ₃ = −29°49′22.7″

Observation #4
t₄ = JD 2457234.625
X⊕₄ = +0.616702829 AU
Y⊕₄ = −0.8063620175 AU
Z⊕₄ = +0.00002486325 AU
α₄ = 20h 22m 06.57s
δ₄ = −30°41′57.3″

Converting the observation angles to radians:

α₁ = 5.44084723
δ₁ = −0.483329666

α₂ = 5.41030979
δ₂ = −0.502468171

α₃ = 5.37290956
δ₃ = −0.520509058

α₄ = 5.33245865
δ₄ = −0.53580299

Continuing through the procedure,

t = 2457219.63
T = 0.001553628
ε = 0.409057547 radians

X₁ = −0.155228396
Y₁ = +0.921851498
Z₁ = +0.399597005

X₂ = −0.319493277
Y₂ = +0.885503087
Z₂ = +0.383841559

X₃ = −0.474779562
Y₃ = +0.824271594
Z₃ = +0.357299982

X₄ = −0.616702829
Y₄ = +0.739843884
Z₄ = +0.320703493

a₁ = +0.589463371
b₁ = −0.660726081
c₁ = −0.464730008

a₂ = +0.563195268
b₂ = −0.671477524
c₂ = −0.4815901

a₃ = +0.532276132
b₃ = −0.685093499
c₃ = −0.497321844

a₄ = +0.49965704
b₄ = −0.69978587
c₄ = −0.510531662

R₁² = 1.03358381
R₄² = 1.03054208

2R₁ cos θ₁ = 1.772595
2R₄ cos θ₄ = 1.97920299

τ₁ = 0.344041979
τ₂ = 0.17202099
τ₃ = 0.516062969
τ₄ = 0.17202099
τ₅ = 0.344041979

Φ = −0.058607619
φ = −0.030167527

A = +0.404275125
B = −7.08013785
C = +4.84883362
D = −0.043927505

A' = +1.72864027
B' = −12.7399767
C' = +3.76138574
D' = +0.951283093

E = +2
F = +0.236729767
G = +0.80855025
H = −0.095703956
I = +0.191407912
K = +0.342297675
L = −2.91537363
M = −2.20429551

E' = +0.5
F' = +0.118364883
G' = +0.864320133
H' = +0.102305152
I' = +0.204610303
K' = +0.223373365
L' = −1.86702289
M' = −0.617533885

r₁ = 2.75 (initial guess)
r₄ = 2.75 (initial guess)

1st approximation
ρ₁ = 1.97723208
ρ₄ = 1.9223289
r₁ = 2.90652064
r₄ = 2.92071388

2nd approximation
ρ₁ = 2.001149
ρ₄ = 1.94460337
r₁ = 2.93008666
r₄ = 2.94292188

3rd approximation
ρ₁ = 2.00417695
ρ₄ = 1.94741724
r₁ = 2.93307059
r₄ = 2.94572742

4th approximation
ρ₁ = 2.00455348
ρ₄ = 1.94776713
r₁ = 2.93344165
r₄ = 2.94607627

5th approximation
ρ₁ = 2.0046002
ρ₄ = 1.94781054
r₁ = 2.93348769
r₄ = 2.94611956

6th approximation
ρ₁ = 2.00460599
ρ₄ = 1.94781593
r₁ = 2.9334934
r₄ = 2.94612492

7th approximation
ρ₁ = 2.00460671
ρ₄ = 1.9478166
r₁ = 2.93349411
r₄ = 2.94612559

8th approximation
ρ₁ = 2.0046068
ρ₄ = 1.94781668
r₁ = 2.9334942
r₄ = 2.94612567

9th approximation
ρ₁ = 2.00460681
ρ₄ = 1.94781669
r₁ = 2.93349421
r₄ = 2.94612568

10th approximation (final, converged)
ρ₁ = 2.00460681
ρ₄ = 1.94781669
r₁ = 2.93349421
r₄ = 2.94612568

HEC positions in AU at t₁ & t₄

x₁ = +1.33687069
y₁ = −2.2463475
z₁ = −1.33119794

x₄ = +1.58994315
y₄ = −2.10289848
z₄ = −1.31512559

The reciprocal of the speed of light
ç = 0.00577551833 days/AU

Aberration corrections to time

çρ₁ = 0.011577643 days
t₁° = 2457204.61

çρ₄ = 0.011249651 days
t₄° = 2457234.61

Epoch of state vector
t₀ = 2457219.61 JD

The object's state vector in heliocentric ecliptic coordinates

x₀ = +1.46520344 AU
y₀ = −2.52458426 AU
z₀ = −0.349479243 AU
Vx₀ = +14610.4367 m/s
Vy₀ = +7967.42879 m/s
Vz₀ = −2442.63758 m/s

The object's distance from the sun at t₀
r₀ = 2.93980995 AU

Sun-relative speed
V₀ = 16819.9661 m/s

The sun's gravitational parameter
μ = 1.32712440018ᴇ20 m³ sec⁻²

The semimajor axis of the object's orbit
a = 2.76694735 AU

The angular momentum per unit mass in the object's orbit

hx = +1.3390648ᴇ15 m²/sec
hy = −2.2844842ᴇ14 m²/sec
hz = +7.26435028ᴇ15 m²/sec

h = 7.39026848ᴇ15 m²/sec

The eccentricity of the object's orbit
e = 0.076026341

The inclination of the object's orbit
i = 10.5918141°

The longitude of the ascending node of the object's orbit
Ω = 80.3183813°

The true anomaly at time t₀
θ₀ = 147.669798°

The sum of the true anomaly at time t₀ and the argument of the perihelion of the object's orbit
(θ₀+ω) = 220.296384°

The argument of the perihelion of the object's orbit
ω = 72.6265867°

The eccentric anomaly of the object at time t₀
u₀ = 145.259666°

The mean anomaly of the object at time t₀
M₀ = 142.777370°

The period of the object's orbit
P = 1681.12408 days

Times of perihelion passage

T₀ = 2456552.87
T₁ = 2458234.01 = T₀+P

 

A summary of the calculation results and a comparison with JPL's numbers

Orbital elements (as calculated)

a = 2.76694735 AU
e = 0.076026341
i = 10.5918141°
Ω = 80.3183813°
ω = 72.6265868°
T₀ = JD 2456552.87
T₁ = JD 2458234.01

Orbital elements (From the JPL Small-Body Database)

a = 2.76916515 AU
e = 0.076009027
i = 10.5940672°
Ω = 80.3055309°
ω = 73.5976947°
T = JD 2458238.75

Elliptical Transfer-Intercept Orbits

Non-Hohmann Transfer Orbits that Take You Somewhere

Most of the time, when someone speaks of transfer orbits, he's referring to a special case known as a Hohmann transfer orbit. Hohmann transfer orbits have a departure occurring at one of its apsides (perihelion or aphelion) and an arrival occurring at the other apside. Thus, I could describe Hohmann transfer orbits as transfer orbits having two anchored apsides.

In this essay, I will treat a more general case of transfer orbits that have only one anchored apside, which turns out to be enough to close the equation set and permit the Keplerian elements of the transfer orbit to be found, as well as the changes of velocity required for transfer orbit insertion and, later, for matching velocity with the destination object.

In what follows, the following example problem will be used for illustration:

A spaceship is initially in Earth's orbit, but is on the opposite side of the sun from Earth. Its captain wants to enter a transfer orbit, bound for Vesta, at 12h UT on 26 June 2017. The navigator does some trial runs on a computer and discovers an elliptical transfer orbit having its aphelion at Vesta upon arrival at 4h 45m 36.036s UT on 12 June 2018. Check the navigator's work to ensure that an elliptical transfer orbit does exist for these times for departure and arrival. Show the elements of the transfer orbit and the delta-vees required for transfer orbit insertion (departure) and for matching velocity with Vesta at arrival.
 

Spaceship initial orbit.
a = 1.000002 AU
e = 0.016711
i = 0.0°
Ω = 0.0°
ω = 103.095°
T = JD 2454285.96

Vesta's orbital elements.
a = 2.36126914 AU
e = 0.089054753
i = 7.13518389°
Ω = 103.91484282°
ω = 149.85540185°
T = JD 2454267.1969204

Departure time,
t₁ = 12h UTC, 26 June 2017

Arrival time,
t₂ = 4h 45m 36.036s UTC, 12 June 2018

It is convenient to convert t₁ and t₂ from calendar date format to Julian date format.

Converting from Calendar Date to Julian Date

The time zone must be Greenwich, Zulu, UT, UTC (all the same zone)

Y = the four-digit year
M = the month of the year (1=January... 12=December)
D = the day of the month
Q = the time of the day in decimal hours

A = integer [ (M−14) / 12 ]
B = integer { [ 1461 (Y + 4800 + A) ] / 4 }
C = integer { [ 367 (M − 2 − 12A) ] / 12 }
E = integer [ (Y + 4900 + A) / 100 ]
F = integer [ (3E) / 4 ]
t = B + C − F + D − 32075.5 + Q/24

Converting the time of departure, t₁, from calendar date to Julian date

t₁ = 12h UTC, 26 June 2017
Y = 2017
M = 6
D = 26
Q = 12
A = 0
B = 2489909
C = 122
E = 69
F = 51
t₁ = JD 2457931.0

Converting the time of arrival, t₂, from calendar date to Julian date

t₂ = 4h 45m 36.036s UTC, 12 June 2018
Y = 2018
M = 6
D = 12
Q = 4.76001
A = 0
B = 2490274
C = 122
E = 69
F = 51
t₂ = JD 2458281.69833375

Instead of having the initial position vectors given to us, we must calculate them by reducing the elements of the spaceship's initial orbit (around the sun) and the time of departure therefrom, t₁, in order to obtain the position vector r₁, and by reducing the elements of Vesta's orbit and the time of arrival thereto, t₂, in order to obtain the position vector r₂.

For what passes below, the Sun's gravitational parameter,

GM = 1.32712440018ᴇ20 m³ sec⁻²

The ratio of the astronomical unit to the meter,

AU = 1.495978707ᴇ11 m au⁻¹

And the
 

Definition of the two-dimensional arctangent function

atn(z) = single argument arctangent function of the argument z.

Function arctan( y , x )
. if x = 0 and y greater than 0 then angle = +π/2
. if x = 0 and y = 0 then angle = 0
. if x = 0 and y less than 0 then angle = −π/2
. if x greater than 0 and y greater than 0 then angle = atn(y/x)
. if x less than 0 then angle = atn(y/x) + π
. if x greater than 0 and y less than 0 then angle = atn(y/x) + 2π
arctan = angle

Unless otherwise indicated, the coordinate system to which all unprimed vectors in this essay refer is ecliptic coordinates — heliocentric for position, and sun-relative for velocity.

Reducing Keplerian orbital elements and a time to position and velocity in heliocentric ecliptic coordinates

Find the period, P, in days.

P = (365.256898326 days) a¹·⁵

Find the mean anomaly, m, in radians.

m₀ = (t − T) / P
m = 2π [ m₀ − integer(m₀) ]

Find the eccentric anomaly, u, in radians.

The Danby first approximation for the eccentric anomaly, u, in radians.

u' = m
+ (e − e³/8 + e⁵/192) sin(m)
+ (e²/2 − e⁴/6) sin(2m)
+ (3e³/8 − 27e⁵/128) sin(3m)
+ (e⁴/3) sin(4m)

The Danby's method refinement for the eccentric anomaly.

u = u'

REPEAT
U = u
F₀ = U − e sin U − m
F₁ = 1 − e cos U
F₂ = e sin U
F₃ = e cos U
D₁ = −F₀ / F₁
D₂ = −F₀ / [ F₁ + D₁F₂/2 ]
D₃ = −F₀ / [ F₁ + D₁F₂/2 + D₂²F₃/6 ]
u = U + D₃
UNTIL |u−U| is less than 1ᴇ-14

The loop, just above, converges u to the correct value of the eccentric anomaly. Usually. However, when e is near one and the orbiting object is near the periapsis of its orbit, there is a chance that this loop will fail to converge. In such cases, a different root-finding method will be needed.

Find the canonical position vector of the object in its orbit at time t.

x''' = a (cos u − e)
y''' = a sin u √(1−e²)
z''' = 0

Find the true anomaly, θ. We'll use it below when we find the velocity.

θ = arctan( y''' , x''' )

Rotate the triple-prime position vector by the argument of the perihelion, ω.

x'' = x''' cos ω − y''' sin ω
y'' = x''' sin ω + y''' cos ω
z'' = z''' = 0

Rotate the double-prime position vector by the inclination, i.

x' = x''
y' = y'' cos i
z' = y'' sin i

Rotate the single-prime position vector by the longitude of the ascending node, Ω.

x = x' cos Ω − y' sin Ω
y = x' sin Ω + y' cos Ω
z = z'

The unprimed position vector [x,y,z] is the position in heliocentric ecliptic coordinates.

Find the canonical (triple-prime) heliocentric velocity vector.

k = √{ GM / [ a AU (1 − e²) ] }

k is a speed in meters per second.

Vx''' = −k sin θ
Vy''' = k (e + cos θ)
Vz''' = 0

Rotate the triple-prime velocity vector by the argument of the perihelion, ω.

Vx'' = Vx''' cos ω − Vy''' sin ω
Vy'' = Vx''' sin ω + Vy''' cos ω
Vz'' = Vz''' = 0

Rotate the double-prime velocity vector by the inclination, i.

Vx' = Vx''
Vy' = Vy'' cos i
Vz' = Vy'' sin i

Rotate the single-prime velocity vector by the longitude of the ascending node, Ω.

Vx = Vx' cos Ω − Vy' sin Ω
Vy = Vx' sin Ω + Vy' cos Ω
Vz = Vz'

The unprimed velocity vector [Vx,Vy,Vz] is the sun-relative velocity in ecliptic coordinates.
 

Calculate the position and velocity of the spaceship in its initial orbit at the time of departure

P = 365.257994
m₀ = 9.97935722
m = 6.15348288
u' = 6.15128508
u = 6.15128508
x''' = +0.974604719
y''' = −0.131499998
θ = 6.14906877
x'' = −0.092732158
y'' = +0.979054316
x' = −0.092732158
y' = +0.979054316
z' = 0
xi = −0.092732158
yi = +0.979054316
zi = 0
k = 29788.8217
Vx''' = +3983.20734
Vy''' = +30019.1146
Vx'' = −30140.9504
Vy'' = −2921.69307
Vx' = −30140.9504
Vy' = −2921.69307
Vz' = 0
Vxi = −30140.9504
Vyi = −2921.69307
Vzi = 0

Calculate the position and velocity of Vesta at the time of arrival

P = 1325.30752
m₀ = 3.02910935
m = 0.182899417
u' = 0.200646945
u = 0.200648459
x''' = +2.10361404
y''' = +0.468742457
θ = 0.219245394
x'' = −2.0545179
y'' = +0.651051227
x' = −2.0545179
y' = +0.646009389
z' = +0.080867606
xf = −0.13298229
yf = −2.14957848
zf = +0.080867606
k = 19460.2928
Vx''' = −4232.48025    
Vy''' = +20727.481
Vx'' = −6748.92645
Vy'' = −20049.7967
Vx' = −6748.92645
Vy' = −19894.5281
Vz' = −2490.40168
Vxf = +20933.6861
Vyf = −1766.64767
Vzf = −2490.40168

We will refer to a "hypothetical" transfer orbit until we have assured ourselves that it satisfies the condition that the calculated transit time be equal, or very nearly equal, to the required transit time.

The required transit time is the amount of time that the destination object (Vesta, in our example) takes to go from where it is at t₁ to where it is at t₂. This time difference is, of course, t₂−t₁.

The calculated transit time is the amount of time, Δt, that the spaceship takes to travel, along the hypothetical transfer orbit, from where it is at t₁ to the intersection of the hypothetical transfer orbit with the orbit of the destination object.

In general, Δt will differ substantially from t₂−t₁. It is necessary that t₁ and t₂ be chosen such that Δt is nearly equal to t₂−t₁. Once we know that to be the case, we can drop the word "hypothetical," for we will have determined that the transfer orbit does, indeed, exist.

The determination of an elliptical transfer-intercept orbit from a position and time of departure and from a position and time of arrival

At time t₁ a spaceship in free orbit around the sun (i.e. there is no planet nearby) has this state vector:

xi , yi , zi , Vxi , Vyi , Vzi

At time t₂ (such that t₂>t₁) an asteroid in free orbit around the sun (i.e. there is no significant perturbing third mass) has this state vector:

xf , yf , zf , Vxf , Vyf , Vzf

We want to find out whether or not there exists a transfer orbit between the position elements of those two state vectors, such that

x₁ = xi
y₁ = yi
z₁ = zi

x₂ = xf
y₂ = yf
z₂ = zf

Where the subscript 1 denotes "pertaining to the transfer orbit at transfer orbit insertion," and the subscript 2 denotes "pertaining to the transfer orbit when the spaceship crosses the destination object's orbit." (Whether the destination object is actually there at t₂ is a question that we will answer presently.)

r₁ = √[ x₁² + y₁² + z₁² ]
r₂ = √[ x₂² + y₂² + z₂² ]
d = √[ (x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)² ]

We define the integer variable β and permit it to have only the values 1 and 2.

If β=1, an apside (perihelion or aphelion) of the transfer orbit occurs at departure.
If β=2, an apside (perihelion or aphelion) of the transfer orbit occurs at arrival.

β = either 1 or 2
φ = 3 − β
N = (−1)ᵠ

The variables β and φ will usually be subscripts. The variable N is a sign toggle factor.

m : mean anomaly
u : eccentric anomaly
θ : true anomaly

If the apside at the apsidal endpoint of the intended trajectory is the perihelion, then

mᵦ = uᵦ = θᵦ = 0

If the apside at the apsidal endpoint of the intended trajectory is the aphelion, then

mᵦ = uᵦ = θᵦ = π radians

The eccentricity of a conic section, having the sun at a focus, which includes the point of departure and the point of arrival, is found by solving, simultaneously,

The polar equation which relates the heliocentric distance with the true anomaly,

cos θ₂ − cos θ₁ = { [a (1−e²) / r₂ − 1] − [a (1−e²) / r₁ − 1] } / e

The law of cosines,

d² = r₁² + r₂² − 2 r₁ r₂ cos(θ₂−θ₁)

We can eliminate the semimajor axis by recalling that rᵦ is equal to either a(1+e), the transfer orbit aphelion, or to a(1−e), the transfer orbit perihelion. In general, the semimajor axis can always be eliminated because one or the other endpoints of the intended trajectory occurs at one or the other apside of the transfer orbit. This is why you don't need a third point on the transfer orbit to determine its elements.

After some algebra, we get the eccentricity of the hypothetical transfer orbit.

e = 2 (cos θᵦ) rᵦ (rᵦ−rᵩ) / (rᵩ² − rᵦ² − d²)

The semimajor axis of the hypothetical transfer orbit is found from

a = rᵦ / (1 − e cos θᵦ)

The true anomaly in the hypothetical transfer orbit at the non-apsidal endpoint of the intended trajectory is found as follows:

θᵩ = θᵦ + N arccos{(rᵦ² + rᵩ² − d²) / (2rᵦrᵩ)}

The eccentric anomaly in the hypothetical transfer orbit at the non-apsidal endpoint of the intended trajectory is found as follows:

sin uᵩ = (rᵩ/a) sin θᵩ / √(1−e²)
cos uᵩ = (rᵩ/a) cos θᵩ + e
uᵩ = arctan(sin uᵩ , cos uᵩ)

The mean anomaly in the hypothetical transfer orbit at the non-apsidal endpoint of the intended trajectory is found as follows:

mᵩ = uᵩ − e sin uᵩ

The period of the hypothetical transfer orbit is

P = (365.256898326 days) a¹·⁵

The mean motion in the hypothetical transfer orbit is

μ = 2π/P

For short path trajectories (for which the arc of true anomaly going from departure to arrival is less than π radians), the calculated transit time in the hypothetical transfer orbit is

Δt = (N/μ) [mᵩ − π sin(θᵦ/2)]

Here's the test that determines whether a transfer orbit exists between heliocentric position r₁ at time t₁ and heliocentric position r₂ at time t₂. It is necessary that

Δt ≈ t₂ − t₁

And the match should be a close one, ideally a small fraction of a second. In general, this will not be the case. If the difference in the required and the calculated transit times is unacceptably large, then the spaceship pilot will have to choose either a different departure time, or a different arrival time, or both, and try again.

The procedure being demonstrated here finds elliptical transfer orbits of the short path, by which it is meant that the arc of true anomaly along the intended trajectory, from departure to arrival, is strictly less than π radians. One of the transfer orbit's apsides will occur at either the position of departure or at the position of arrival, but the other apside will not occur at all within the intended trajectory. To be complete about things, we will calculate the time of perihelion passage in the transfer orbit, whether or not the spaceship is ever there.

T = tᵦ − mᵦ/μ

The inclination of the transfer orbit is found from the cross product of the heliocentric position vectors of departure and arrival, r₁ x r₂.

Xn' = y₁ z₂ − z₁ y₂
Yn' = z₁ x₂ − x₁ z₂
Zn' = x₁ y₂ − y₁ x₂

Rn' = √[ (Xn')² + (Yn')² + (Zn')² ]

Xn = Xn' / Rn'
Yn = Yn' / Rn'
Zn = Zn' / Rn'

The vector Rn is a unit normal to the transfer orbit in the direction of the orbit's angular momentum.

i = arccos(Zn)

Having found the components of the vector normal to the transfer orbit (in the direction of the angular momentum), we now use it to find the velocity in the transfer orbit at the apsidal endpoint of the intended trajectory.

Vxᵦ'' = Yn zᵦ − Zn yᵦ
Vyᵦ'' = Zn xᵦ − Xn zᵦ
Vzᵦ'' = Xn yᵦ − Yn xᵦ

Vᵦ'' = √[ (Vxᵦ'')² + (Vyᵦ'')² + (Vzᵦ'')² ]

Vxᵦ' = Vxᵦ'' / Vᵦ''
Vyᵦ' = Vyᵦ'' / Vᵦ''
Vzᵦ' = Vzᵦ'' / Vᵦ''

Vᵦ = √[ (GM/AU) (2/rᵦ − 1/a) ]

Vxᵦ = Vᵦ Vxᵦ'
Vyᵦ = Vᵦ Vyᵦ'
Vzᵦ = Vᵦ Vzᵦ'

Now we find the angular momentum per unit mass in the transfer orbit.

hx = AU (yᵦ Vzᵦ − zᵦ Vyᵦ)
hy = AU (zᵦ Vxᵦ − xᵦ Vzᵦ)
hz = AU (xᵦ Vyᵦ − yᵦ Vxᵦ)

From here, we find the longitude of the ascending node of the transfer orbit.

Ω = arctan( hx , −hy )

Notice that hx is proportional to sin Ω, while −hy is proportional to cos Ω.

We find the argument of the perihelion of the transfer orbit as follows:

cos ω'' = (xᵦ cos Ω + yᵦ sin Ω) / rᵦ

If sin i = 0 then sin ω'' = (yᵦ cos Ω − xᵦ sin Ω) / rᵦ
If sin i ≠ 0 then sin ω'' = zᵦ / (rᵦ sin i)

ω'' = arctan( sin ω'' , cos ω'' )

ω' = ω'' − θᵦ

If ω' ≥ 0 then ω = ω'
If ω' < 0 then ω = ω' + 2π

The two known points on the hypothetical transfer orbit are

x₁ = −0.092732158 AU
y₁ = +0.979054316 AU
z₁ = 0

x₂ = −0.13298229 AU
y₂ = −2.14957848 AU
z₂ = +0.080867606 AU

The sides of the sun-departure-arrival triangle are

r₁ = 0.98343612 AU
r₂ = 2.15520567 AU
d = 3.129936551 AU

We are given that the apoapsis of the hypothetical transfer orbit occurs at the arrival position, so

β = 2
φ = 1
N = −1
m₂ = u₂ = θ₂ = π radians

The eccentricity and the semimajor axis of the hypothetical transfer orbit are

e = 0.37484849
a = 1.56759505 AU

The true, eccentric, and mean anomalies in the hypothetical transfer orbit at the non-apsidal endpoint of the intended trajectory (i.e. the departure position) are

θ₁ = 0.16062918 radians
u₁ = 0.10844236 radians
m₁ = 0.067872532 radians

The period and mean motion of the hypothetical transfer orbit are

P = 716.884602 days
μ = 0.00876457005 radians/day

The calculated and required transit times in the hypothetical transfer orbit, from departure to arrival, are

Δt = 350.698335 days
t₂−t₁ = 350.69833375 days

Subject to roundoff error, the difference is about one-tenth of a second. That's close enough. The transfer orbit exists.

The time of perihelion passage in the transfer orbit (although the spaceship is never there) is

T = JD 2457923.256033 = 18h 8m 41s UTC on 18 June 2017

Find the unit normal vector to the plane of the transfer orbit

Xn' = +0.079173779
Yn' = +0.0074990276
Zn' = +0.329531936

Rn' = 0.338992654

Xn = +0.233556030
Yn = +0.0221215048
Zn = +0.972091673

The transfer orbit's inclination to the ecliptic

i = 13.56812324°

Find the velocity in the transfer orbit at the apsidal endpoint of the intended trajectory

Vx₂'' = +2.091376254
Vy₂'' = −0.148158093
Vz₂'' = −0.499105248

V₂'' = 2.155205676

Vx₂' = +0.970383605
Vy₂' = −0.068744295
Vz₂' = −0.231581261

V₂ = 16041.367805 m/s

Vx₂ = +15566.280326 m/s
Vy₂ = −1102.752513 m/s
Vz₂ = −3714.880181 m/s

The angular momentum per unit mass

hx = +1.207943483ᴇ15 m²/s
hy = +1.144116363ᴇ14 m²/s
hz = +5.027623568ᴇ15 m²/s

The transfer orbit's longitude of the ascending node

Ω = 95.41068849°

Find the transfer orbit's argument of the perihelion

cos ω'' = −0.987126840
sin ω'' = +0.159939380
ω'' = 2.980963411 radians
ω' = −0.160629243 radians
ω = 350.79662233°

Keplerian elements of the transfer orbit

a = 1.56759505 AU
e = 0.37484849
i = 13.56812324°
Ω = 95.41068849°
ω = 350.79662233°
T = JD 2457923.256033

Now that you have the elements of the transfer orbit, you can calculate the changes-of-velocity needed for transfer orbit insertion (departure) and for matching velocity with the target asteroid at arrival.

Calculation of the delta-vees for departure and for arrival

When we reduce the elements of the transfer orbit with the time of departure, t₁, we find that the velocity of the spaceship in the transfer orbit is

Vx₁ = −34166.4329 m/s
Vy₁ = −1690.83202 m/s
Vz₁ = +8247.34992 m/s

The velocity of the spaceship in its initial orbit at t₁ was

Vxi = −30140.9504 m/s
Vyi = −2921.69307 m/s
Vzi = 0.0 m/s

So the change of velocity required at departure for transfer orbit insertion is

ΔVx₁ = −4025.4825 m/s
ΔVy₁ = +1230.8611 m/s
ΔVz₁ = +8247.3499 m/s

ΔV₁ = 9259.4983 m/s

The velocity of Vesta, when it is intercepted by the spaceship, is

Vxf = +20933.6861 m/s
Vyf = −1766.64767 m/s
Vzf = −2490.40168 m/s

When we reduce the elements of the transfer orbit with the time of arrival, t₂, we find that the velocity of the spaceship in the transfer orbit is

Vx₂ = +15566.2801 m/s
Vy₂ = −1102.75259 m/s
Vz₂ = −3714.88014 m/s

So the change of velocity required of the spaceship at arrival to match velocity with Vesta is

ΔVx₂ = +5367.4060 m/s
ΔVy₂ = −663.8951 m/s
ΔVz₂ = +1224.4785 m/s

ΔV₂ = 5545.1917 m/s

Remember that all of the unprimed vectors in this tutorial are referred to ecliptic coordinates. If you want them in celestial coordinates (so that you can use a star chart to show you the right ascension and declination in which to point the nose of your spaceship when you apply thrust), you'll still have that to do.

A check on the accuracy of the method by a numerical evolution of the state vector

As calculated from the Keplerian elements of the transfer orbit, at time t₁=JD 2457931.0, the spaceship's heliocentric state vector is

x₁ = −0.092732158 AU
y₁ = +0.979054316 AU
z₁ = 0
Vx₁ = −34166.4329 m/s
Vy₁ = −1690.83202 m/s
Vz₁ = +8247.34992 m/s

As calculated from the Keplerian elements of the transfer orbit, at time t₂ = JD 2458281.69833375, the spaceship's heliocentric state vector is

x₂ = −0.13298229 AU
y₂ = −2.14957848 AU
z₂ = +0.080867606 AU
Vx₂ = +15566.2801 m/s
Vy₂ = −1102.75259 m/s
Vz₂ = −3714.88014 m/s

The transit time of the spaceship in the transfer orbit is
t₂−t₁ = 30300336.036 sec

If we take the state vector at t₁ and numerically walk it forward in time by 1 second intervals for 30300336 seconds, we get this result.

x = −0.132983124 AU
y = −2.149579643 AU
z = +0.080867833 AU
Vx = +15566.27354 m/s
Vy = −1102.76889 m/s
Vz = −3714.87818 m/s