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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 practice, very few interplanetary transfer orbits will be Hohmann transfers. The orbits of solar system bodies are ellipses, not circles, and the planes in which the orbits exist are tilted with respect to each other. The direction in which a rocket must apply thrust, in order to enter a transfer orbit bound for an asteroid that is almost in the ecliptic plane might be, itself, in an angular sense, quite far from the ecliptic plane.

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

After Fliegel and van Flandern (1968).

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


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

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

The subscript 1 denotes "pertaining to the transfer orbit at transfer orbit insertion" or "departure."

The subscript 2 denotes "pertaining to the transfer orbit at its intersection with the destination object's orbit" or "arrival." (Whether the destination object is actually there at that same time is a question 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 equations which relates the heliocentric distances with the true anomalies,

r₁ = a (1−e²) / (1 + e cos θ₁)
r₂ = a (1−e²) / (1 + e cos θ₂)

The law of cosines,

d² = r₁² + r₂² − 2 r₁ r₂ cos(θ₂−θ₁)

In those equations, the known quantities are d, r₁, r₂, and whichever of the true anomalies, θᵦ, has the hypothetical transfer orbit's apside. The cosine of θᵦ is +1 if that apside is the perihelion, and it is −1 if it is the aphelion. The condition that one of the endpoints of the intended trajectory, either departure or arrival, occurs at one of the apsides of the hypothetical transfer orbit is why you don't need a third point on the transfer orbit to determine its elements. The three equations have three unknown quantities, namely a, e, and θᵩ.

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.


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.


Converting a velocity vector from rectangular ecliptic coordinates to spherical celestial coordinates

Let's convert the departure delta-vee in our example. Here it is in ecliptic coordinates:

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

The obliquity of the ecliptic, ε, can be estimated from the Laskar equation.

T = (t−2451545) / 3652500

Ɛ = 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

ε = Ɛ / 206264.806247

t = t₁ = JD 2457931.0
T = 0.00174839151266
Ɛ = 84373.263907"
ε = 0.409053126623 radians

The magnitude of the vector does not change by the rotation. It remains

ΔV₁' = ΔV₁ = √[ (ΔVx₁)² + (ΔVy₁)² + (ΔVz₁)² ]
ΔV₁' = 9259.4983 m/s

The velocity vector in celestial coordinates is found from

ΔVx₁' = ΔVx₁
ΔVy₁' = ΔVy₁ cos ε − ΔVz₁ sin ε
ΔVz₁' = ΔVy₁ sin ε + ΔVz₁ cos ε

ΔVx₁' = −4025.4825 m/s
ΔVy₁' = −2150.9947 m/s
ΔVz₁' = +8056.4894 m/s

The right ascension of the direction in which the velocity vector points is

α₁ = (12 hours/π) arctan( ΔVy₁' , ΔVx₁' )
α₁ = 13.8745051 hours

The declination of the direction in which the velocity vector points is

δ₁ = (180°/π) arcsin( ΔVz₁' / ΔV₁' )
δ₁ = +60.467750°


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
Jenab6

The real reason for the media attention to the Gabby Petito murder

I don't think that the Gabby Petito killing is top news because Gabby was a white girl. I think that the mainstream media have promoted it to top news status because Brian Laundrie is 𝑎 𝑤ℎ𝑖𝑡𝑒 𝑚𝑎𝑛. You know that most of the time when someone is murdered, the killer is a black guy, and the media usually ignore the killing because they want to make blacks seem to be as nice, as smart, and as worthy as white people are. You can't do that by being honest and fair, so the media proceed just as dishonestly and as unfairly as they must in service to their false racial narrative.

Any time a white man misbehaves, it's deemed newsworthy. Any dishonor that the media can invent or magnify against white people, they do it. If Gabby had been murdered by a black, we'd probably never have heard about it. But if Brian Laundrie had killed someone else, 𝑎𝑛𝑦𝑜𝑛𝑒 else, he'd nevertheless still be headline material.
Jenab6

Homosexuality and the James Webb Space Telescope

I oppose changing the name of the James Webb Space Telescope to anything else. This is a "woke" political move, and if it is successful it will only feed a Marxist beast that is best left to starve. It's an out-growth of cancel culture. Giving this beast what it wanted is what ruined the Star Wars movie series. It can't do anything positive for the JWST, either.

How often are we led to assume that our best thinkers of yesteryear, contemplating the same social controversies that we do, somehow got it all wrong, and that the present views are more correct? Maybe they were right and we are wrong.

With regard to the consensus of thinking on homosexuality, a sea-change began in 1973. Homosexuals, together with Marxist activists, had infiltrated the American Psychiatric Association and took it over from within. They changed a long-standing conclusion that homosexuality itself was a behavioral disorder and substituted their preferred belief that homosexuality was merely an alternative lifestyle. The harm done to society through lower birthrates, higher rates of infection for certain diseases, the lowering of moral standards, the diversion of human energy from productive purposes to self-indulgent ones having (at best) zero payoff in terms of future capital... none of this was considered when homosexuality was removed from the APA's diagnostic manual.

All the other cultural dominoes that have fallen to homosexual normalization since 1973 fell from that initial woke push. We have seen the playing out of a moral and cultural slippery slope, as well as an exemplification of the principle of mass psychology described in metaphor as "Heat the frogs slowly enough, and they won't realize that they are being cooked."

Sex itself came about as a means of promoting genetic diversity in species without depending on mutations to provide variant alleles. The pleasure that people get from sex is an evolved inducement for men and women to have sex and thereby engender children. Homosexuality is a perversion of sexuality. Biologically, homosexuality is a waste of time. Additionally, homosexual acts spread diseases more efficiently than normal (heterosexual) sex does, and anal sexual penetration leads to the loss of the body’s ability to hold fecal waste, which requires the use of adult diapers.

Homosexuality is, frankly, nasty. You'd realize it if the people who have twisted our thinking did it all in a year, instead of over half a century.

NOTE: Here because censored by YouTube.
Jenab6

Gregg Popovich gets it all wrong

The "woke" US Men's Olympic basketball team has lost to France.

. . . And YouTube censored another of my comments.

I wrote:

That blasted idiot Gregg Popovich said that whites got a "head start" because they were born white, like being privileged to start a 100-meter race at the midway point instead of at the starting line.

NO! THAT IS THE WRONG COMPARISON. Whites got the same kind of lead in intellectual fields of endeavor that blacks get with a 100-meter sprint. The advantages that whites have aren't systemic; they are biological. Just as blacks sprint better, on the average, than whites do, so likewise does the higher average IQ — i.e. the better brains — of white people give them an advantage in intellectual pursuits.

These biological circumstances aren't any fault of white people. They were contrived by nature through evolution. Whites haven't done any oppressing since whites ended slavery. (Nobody else would have done that, but whites did.) Whites haven't given themselves any unfair advantage or denied to blacks anything that was properly due them. Whites are just superior in some ways than blacks are. Just as blacks have advantages versus whites in certain other ways.

The minorities of this world are RESENTFUL of the innate, heritable advantages that whites have that they do not have, and, because of this resentment, they are assigning blame to whites that whites do not deserve.

WOKE US Men's Basketball Team HUMILIATED By France!!! by Dr. Steve Turley

Jenab6

Maple-Cinnamon-Banana Cupcakes

This latest batch of cupcakes is so good that I'm posting the (updated) recipe again.

DRY INGREDIENTS
2.5 cups of whole wheat flour
4 tablespoons of Ceylon cinnamon
1/4 teaspoon salt
1/2 teaspoon baking soda
1 teaspoon cream of tartar

WET INGREDIENTS
1 mashed up banana
1 cup of dark maple syrup (formerly Grade B )
approximately 8 ounces of water

Preheat oven to 400 degrees Fahrenheit. (Or 204 degrees Celsius.)

Grease a twelve-hole metal cupcake pan with coconut oil. (Each hole in the cupcake pan should hold about 3.25 fluid ounces.)

Mix the dry ingredients together in a bowl. Then add the maple syrup and the mashed up banana, and mix again. Then slowly add the water while mixing to form the cupcake dough.

Dollop the dough into the greased holes of the cupcake pan. There should be just enough dough to fill 11 of the holes, with a little left over for an undersized cupcake in hole number 12.

Put the pan into the oven and bake at 400F for 16 minutes, then turn off the heat and let the pan sit in the oven for another half minute. Remove the pan from the oven and let the cupcakes cool for about 30 minutes.

You can serve these Maple-Cinnamon-Banana cupcakes with chilled whipped cream, or just plain. They go well with espresso.

Ceylon cinnamon is the best kind of cinnamon, and it comes from Sri Lanka. A much more common kind of cinnamon is called Cassia cinnamon, which is made from the inner bark of an evergreen tree native to China. Ceylon cinnamon is sweeter than Cassia cinnamon is, and I prefer Ceylon cinnamon in my baking.

Grade B or "dark" maple syrup is better for cooking than is Grade A or "table" maple syrup. Some people incorrectly believe that these grades are indicators of quality, but that is not true. What they indicate is how much the sap was cooked down and thickened thereby, with Grade B having an increase in density, viscosity, and flavor.

A mixture of two parts cream of tartar with one part baking soda makes three parts BAKING POWDER. By itself, baking soda emits some carbon dioxide when it is heated, but in combination with cream of tartar it chemically reacts (when wet) to produce even more CO2, making for fluffier baked goods.

The mashed-up banana is the additional sweetener used in this recipe. The maple syrup is a constant. But you can replace the banana with whatever other sweetening ingredient that you might want to experiment with: chocolate syrup, peach yogurt, cheesecake filling, or whatever you like.
Jenab6

Why is race and IQ such a controversial topic?

The races of mankind differ in average intelligence, and IQ is a numerical index of intelligence. Western Civilization, built by white people of European descent, is the world’s first civilization based upon high technology. It is the civilization that first developed the use of electric power. It is the civilization that first developed radio and laser for communication and for data processing. Any other group that uses electrically powered devices is culturally appropriating from white people.

Because of the West's reliance upon advanced technology, many of its best jobs a high level of intelligence for satisfactory job performance. The distributed racial differences in intelligence will cause the free market racial distribution in hiring for those jobs to deviate with the racial demographic proportions in the country. Demagogues among the less intelligent races will insidiously cry foul, even though no foul play ever really happened.

It is very easy to provide a mathematical explanation of how the races differ in IQ.

While it is true that someone’s race doesn’t determine his individual IQ, it does determine the probability for a randomly chosen member of a race having an average IQ being at, or above, a specified value.

The normal distribution that most closely matches the IQ distribution of white male US citizens is 103.08±14.54 (Jensen & Reynolds, "Sex Differences on the WISC-R," Personality & Individual Differences, volume 4, number 2, pp. 223-226, 1983).

The normal distribution that most closely matches the IQ distribution of US-resident mulattoes (usually called "blacks" or "African-Americans") is 85.0±13.0 (a typical finding of studies since 1950).

A good approximation of the fraction, f, of a race having an average IQ of x̄ and a standard deviation in IQ of σ, which is above the minimum IQ of μ, can be found as follows:

f(μ) = [σ√(2π)]⁻¹ ∫(μ,∞) exp{ −[(x−x̄)/σ]²/2 } dx

Taking advantage of the normal distribution's symmetry, we make it more easily integrable.

f(μ) = ½ − [σ√(2π)]⁻¹ ∫(x̄,μ) exp{ −[(x−x̄)/σ]²/2 } dx

You can avoid integrating the probability density function if you have a handy error function to call.

f(μ) = 1 − ½ { 1 + erf [(μ−x̄)/(σ√2)] }

Let us suppose that an employer wants to hire workers for a job that, in his opinion, requires a minimum IQ of 130 for satisfactory performance. He lives in an area that is demographically typical for the United States, where whites outnumber blacks by a ratio of five.

The fraction of whites who are qualified for the job on the basis of IQ is

f( μ=130.0, x̄=103.08, σ=14.54 ) = 0.0320528311

The fraction of blacks who are qualified for the job on the basis of IQ is

f( μ=130.0, x̄=85, σ=13 ) = 0.0002685491

If the population of whites and of blacks were equal in size, then the ratio of mentally qualified whites to mentally qualified blacks would be 119.355755.

Since whites outnumber blacks in the area where the employer's business is, by a ratio of five, the actual ratio of mentally qualified whites to mentally qualified blacks is 596.778775.

If the employer needs fewer than 100 new employees, it could very easily turn out that he will hire no blacks at all, even if he uses no racism whatsoever in selecting his hirelings. In fact, of the occasions in which this scenario plays out, and exactly 100 new workers are hired, the employer will have hired...

100 whites and 0 blacks on 84.5% of occasions
99 whites and 1 black on 14.2% of occasions
98 whites and 2 blacks on 1.2% of occasions
97− whites and 3+ blacks on 0.1% of occasions

Because the United States is a First World country in which most of the best jobs are mentally challenging jobs, purely free-market hiring practices will exclude a demographically disproportionately high fraction of low-average-IQ races from those jobs. This is normal and natural. It is the only way by which a country can remain competitive internationally, especially with countries that don't engage in politically correct tampering with the free market for labor. A focus on merit is a good thing. It is in conflict with diversity and inclusion. Therefore, diversity and inclusion are bad ideas, and to focus on them, instead of on merit, is harmful.
Jenab6

What the "Swamp" Costs Us All

The "swamp" is a euphemism that refers to the bureaucratic, unelected part of our federal government. It refers most particularly to the arrogant, legalistic part of the Executive Branch, but it has its minions in the Legislative and Judicial Branches of government also.

A bureaucracy may be necessary for governing a country as large as the United States, but bureaucratic organization is vulnerable to corrupt influences, to infiltration by power-mongers pretending to be civil servants, to barratry by lawyers pretending to have the public good uppermost in mind, and so on.

A swamp develops within a bureaucracy over time, as venal, arrogant evil-doers and their lackeys gradually become the dominant presence among the officials thereof.

I recently had a lesson in why you don't want your country's government offices to become a swamp.

I had it in mind to trademark the phrase "Solar System Empire," so that I could put it on items of merchandise, such as tee-shirt and coffee mugs, and sell them for a bit more than they cost me to make, and in so doing earn a little money.

I didn't know at the outset how much the process would cost. By the time I was made to understand that the $1175 in "filing fees" that I had already paid was just the beginning, that further fees would be forthcoming, I was very sorry that I'd lowered my dignity to the point of doing business with the US Patent and Trademark Office (USPTO).

After I'd paid all that money (about a month's worth of my income) for what was surely a very, very little work on the part of anyone in the Office, I was told, via an "Office Action" — mmm, how dynamic that sounds — that my trademark application was put on hold because of alleged imperfections in my paperwork.

Now, I am not a stupid man. But nobody gets a complicated course of government paperwork right on the first go, if he has never had any prior experience with that specific kind of government paperwork. For example, there are terms and numerical codes with which the novice is unfamiliar. Further, the USPTO didn't specify the nature of those alleged imperfections. They merely told me that what I'd submitted wasn't good enough.

It was clear that I was being pressured into hiring an attorney to represent me to the government. Apparently, a common citizen isn't good enough to speak for himself, and there are penalties awaiting any citizen who tries to do so.

The swamp is mostly a collection of lawyers, who have perverted the offices of government in ways that benefit themselves and that hinder or burden others, especially when those others cannot easily afford to hire attorneys.

That's why the swamp is bad. It is unjust by its very nature. Donald Trump was right to fight the swamp while he was President. The only flaw in his plan was that he wasn't aggressive enough and he didn't move quickly enough.

By the way, the USPTO rejected my application to trademark the phrase

Solar System Empire

I paid them money. They denied me service. A private company would not be permitted to do business in such a sleazy manner.
Jenab6

Race and Public Assistance (Welfare) in the United States

The mainstream media in the United States lies a lot, and one of the subjects upon which they lie frequently is the racial breakdown of people who subsist on public assistance programs, which are run by the government and which are subsidized by taxing the earnings of working citizens.

Here are the population demographic percentages.

Of US citizens,
60.4% are white
13.4% are black
18.3% are mestizos
5.9% are Asians
1.3% are Injuns

To follow are the percentages of each race who receive benefits from the largest public assistance programs. By dividing the percentages of these recipients by the demographic percentages shown above, you can calculate the per capita rates with which each race benefits from each program.

Of US food stamp recipients (SNAP),
35.8% are white (1.000)
25.4% are black (3.198)
16.5% are mestizos (1.521)
3.2% are Asians (0.9151)
1.4% are Injuns (1.817)
Total SNAP beneficiaries in 2017: 42.1 million
Total cost of benefits in 2017: $68 billion
Race ranking, least dependent to most dependent: Asians, Whites, Mestizos, Injuns, Blacks

For example, 35.8% of SNAP beneficiaries are white, but 60.4% of the US population is white, so the relative rate by which whites benefit from SNAP is 35.8/60.4 = 0.5927. Meanwhile, 25.4% of SNAP beneficiaries are black, but only 13.4% of the US population is black, so the relative rate by which blacks benefit from SNAP is 25.4/13.4 = 1.896. This means that the black-to-white per capita rate ratio for receiving SNAP benefits is 1.896/0.5927 = 3.198. In other words, blacks are more than three times more likely than whites to be getting SNAP benefits.

The same math applies when comparing any two races in any of the welfare programs. The numbers in parentheses are these "Race X" to White per capita rate ratios.

Of US recipients of Temporary Assistance for Needy Families (TANF),
27.2% are white (1.000)
28.9% are black (4.789)
37.8% are mestizos (4.587)
1.9% are Asians (0.7151)
1.5% are Injuns (2.562)
Total recipients in 2018: 2.27 million
Total benefits in 2018: $17 billion
Race ranking, least dependent to most dependent: Asians, Whites, Injuns, Mestizos, Blacks

Of US Medcaid beneficiaries,
41% are white (1.000)
30% are black (3.298)
20% are mestizos (1.610)
5% are Asians (1.248)
1% are Injuns (1.133)
Total beneficiaries: about 74 million in 2017
Total cost per annum: about $600 billion in 2018
Race ranking, least dependent to most dependent: Whites, Injuns, Asians, Mestizos, Blacks

Of US residents living in Section 8 or other public housing,
30% are white (1.000)
42% are black (6.310)
23% are mestizos (2.530)
2% are Asians (0.6825)
1% are Injuns (1.549)
Total households getting assistance of some kind: about 5 million
Total persons living in public housing: about 1 million
Race ranking, least dependent to most dependent: Asians, Whites, Injuns, Mestizos, Blacks

Of US residents receiving housing assistance of any kind,
29% are white (1.000)
46% are black (7.150)
20% are mestizos (2.276)
2% are Asians (0.7060)
1% are Injuns (1.602)
Total persons receiving such assistance: 2.8 million
Race ranking, least dependent to most dependent: Asians, Whites, Injuns, Mestizos, Blacks

The average welfare recipient receives $1000/month in benefits.

The race ranking for SNAP is

Asians, the least dependent race
Whites
Mestizos
Injuns
Blacks, the most dependent race

The race ranking for TANF, for Public Housing, and for Housing Assistance is

Asians, the least dependent race
Whites
Injuns
Mestizos
Blacks, the most dependent race

The race ranking for Medicaid is

Whites, the least dependent race
Injuns
Asians
Mestizos
Blacks, the most dependent race

Public assistance, or welfare programs, are a systematic looting from the better races, in order to subsidize the lives of members of inferior races.
Jenab6

The lies of BLM and the MSM are exposed by the body cam video of the officer who shot Ma'Khia Bryant

Police in Columbus, Ohio, responded to a call about a knife fight. When they arrived, the officer saw a teenage female black, holding a knife. Her name was Makiyah Bryant. Makiyah had just finished knocking down a white girl or woman and, brandishing the knife, attacked the black woman who is wearing a pink sports suit and white sneakers.

Meanwhile, an adult male black ran up and kicked the white woman in the back of the head while she was trying to stand up, knocking her back to the ground.

The officer who shot Makiyah probably saved the life of the woman in the pink sports suit. But he probably also should have shot the adult male black, too. What really bothers me is how the white police officer defended the black woman in pink from attack, but he did not defend the white woman who had been knocked down from her second attacker.

Here's a link to the officer's body cam video. You probably won't find it on YouTube because the MSM narrative is that Makiyah is "an innocent victim who got shot for no reason." Watch the video and observe how blacks excuse and/or justify their misbehavior and how egregiously the leftist media lie.

The embed link doesn't work here. So click the link to go to air.tv where the video is hosted.

https://www.air.tv/watch?v=e4u0DpYpRp6xw0tLwv_v1Q

The video is on YouTube, after all, but it is a less complete version in which the action takes place much more quickly.

https://www.youtube.com/watch?v=tpIOWBgQy1Q

PART 2.

Blacks don't always win when they attack their victims. Observe this robbery scene in South Africa.

https://media.gab.com/system/media_attachments/files/071/357/185/original/afb0e95e373a22e1.mp4

I'll bet that armed black robber was surprised when the white guy with the gun showed up. Now there's one less of them to worry about. Robbers, I mean.
Jenab6

Reply to RamzPaul regarding the real difference between religion and science

Concerning a RamzPaul video:

This is a false portrayal of the dark matter vs modified gravity theory controversy. No scientist is saying that the universe is wrong, as that would be equivalent to saying the the truth is false. Scientists don't know for certain why the stars in the spiral arms of galaxies orbit the center of their galaxy faster than the standard theory of gravity would predict. They are trying to find out what the reason is, and they are reserving final judgment on that reason until enough evidence has become available.

That's how science works. Having to accept uncertainty, learning to live without "absolute" certainty, is a strength that is often misconstrued as a weakness.

The better measure of a purported method for seeking truth, whether it is scientific or religious, is how much it has increased the powers of mankind over time. Valid methods for seeking truth do that because useful truths are a subset of all truths, and it is a subset in which people have a particular interest and to which they devote a considerable amount of their time. Any method for seeking truth that really does work will discover useful truths often, and the powers of men will grow thereby.

Conversely, if all that a purported method for finding truth has ever done is spin unprovable fantasies about a deity and an afterlife, intended to comfort people as they are dying, but which fantasies are in conflict with each other about fundamental issues (e.g. Christian doctrine vs. Muslim doctrine), and they do not discover any useful truths that increase the powers of men, then most likely that method for seeking truth doesn't really work.

The basic difference between science and religion is that science treats truth as something to be discovered, whereas religion treats truth as something to be decided (or revealed by other people who did the deciding). The practical consequence of that difference is that different religious groups will come up with religious metaphysics that disagree with each other on the most fundamental level, whereas science will lead different groups, working independently of each other, to the same conclusions, after enough data has been gathered to identify the hypothesis that is most likely correct.