Celestial Mechanics, Lecture 3: Transfer Orbits

Transfer Orbits

Racist Celestial Mechanics, Lecture 3

My purpose with this page is to provide an easy-to-understand tutorial for calculating preliminary interplanetary transfer orbits.

—Jerry Abbott

Revised: 6 August 2003

Throughout this web page, I consider the celestial mechanics of objects pursuing orbits around the sun. The sun is assumed to be the primary object, the only important source of gravitational attraction, for all the orbits treated hereafter.

The default coordinate system used on this page is classical heliocentric ecliptic coordinates. The classical HEC system has its origin at the sun, its XY plane being the plane of Earth's orbit, the X axis being positive in the direction of the Vernal Equinox, and the Z axis being positive in the direction of the angular momentum vector of Earth's orbit.

Between two predetermined points for which heliocentric ecliptic coordinates are known, there exists one or more conic sections, connecting those points, having the sun at a focus. Those conic sections correspond to transfer orbits.

I restrict my scope to elliptical and hyperbolic transfer orbits having an apside at an endpoint of the intended trajectory. That is, either perihelion or aphelion must occur at either departure or arrival.

An apside is the point on an orbit at which the distance to the sun, r, is either a maximum or a minimum. When the distance is a minimum, the apside is called perihelion. (The more general word is periapsis.) When the distance is a maximum, the apside is called aphelion. (The more general word is apapsis.)

Departure occurs at the point at which the rockets burn to enter the transfer orbit. Arrival occurs at the point where the rockets burn to match velocity with the rendezvous object.

Note that while elliptical transfer orbits have both apsides, hyperbolic ones have perihelia only.

Important note. Generally speaking, one does not immediately know where the destination planet will be when the spaceship arrives at some trial point along its orbit. Rendezvous with a moving object involves the extra problem of matching the spaceship's time of flight with that of the destination planet. I will tackle the rendezvous problem later. To start with, I only want to demonstrate how a transfer orbit can be determined when one does know the heliocentric radius vectors of departure and of arrival.

We begin with two rectangular position vectors given in heliocentric ecliptic coordinates.

Departure:	[ x₁ , y₁ , z₁ ]	@ t₁
Arrival:	[ x₂ , y₂ , z₂ ]	@ t₂

The variables r₁ and r₂ represent the heliocentric distances of departure and of arrival, respectively. The variable "d" represents the line-of-light distance between the point of departure and the point of arrival. The time difference (t₂-t₁), is the required transit time between departure and arrival. We do not know, as yet, whether any of the conic sections, having the sun at a focus and an apside at one of the endpoints, will have this transit time and thus be a valid transfer orbit. (In general, it will not.)

r₁² = x₁² + y₁² + z₁²

r₂² = x₂² + y₂² + z₂²

d² = ( x₂ - x₁ )² + ( y₂ - y₁ )² + ( z₂ - z₁ )²

The eccentricities, e, and semimajor axes, a, of the transfer orbits that are possible under the "apside restriction" can be found from the section that follows. The equations for the eccentricity were derived from a simultaneous solution of the equation of the orbit with the law of cosines. You advanced students can check my work. You niggers might as well scram.

Flying a spaceship from a point of departure in one orbit to a point of arrival in a different orbit involves a third orbit connecting the two points. This third orbit is called a transfer orbit. An interplanetary transfer orbit must be a conic section, must have the sun at a focus, and must have a transit time that will allow both the spaceship and the destination planet to reach the arrival position at the same time.

There are three general cases of transfer orbit, identifiable by inspection of the illustration above. Elliptical transfer orbits of the short path trace an ellipse around the sun through an arc of less than p radians. (I use radians for the angles throughout this paper.) Elliptical transfer orbits of the long path trace the same ellipse in the opposite angular direction, moving through an arc of more than p radians. Hyperbolic transfer orbits are open curves with only one pass by the sun; they are non-periodic.

There is also such a thing as a superperiodic elliptical transfer orbit, in which the spaceship makes more than one complete revolution around the sun before intercepting the destination planet. Superperiodic elliptical transfer orbits can be either prograde or retrograde, and they provide the spaceship pilot with more opportunities to find a transfer orbit, however at the expense of greatly increasing the transit time (always by a whole multiple of the transfer orbit's period).

In the section below, the angle q is the true anomaly, and the subscripts 1 and 2 denote departure and arrival, respectively. The true anomaly is the angle, subtended at the sun, between a ray extended through the orbit's perihelion and a ray going through the current position of the object, measured in the direction of the object's motion. The perihelion value of the true anomaly is always zero, and its aphelion value is always p radians.

I. Outward Bound. Arrival is further from the Sun than Departure: r₂ > r₁

I.A. Perihelion at Departure: q₁ = 0.
e = 2 r₁ (r₁ - r₂) / ( r₂² - r₁² - d² )

I.A.1. If e is in [0,1) then an elliptical transfer orbit exists, and a = r₁ / (1-e)
I.A.2. If e>1 then a hyperbolic transfer orbit exists, and a = r₁ / (e-1)
I.A.3. If e<0 then there is no transfer orbit having perihelion at departure.

I.B. Aphelion at Arrival: q₂ = p radians.
e = 2 r₂ (r₁ - r₂) / ( r₁² - r₂² - d² )

I.B.1. If e is in [0,1) then an elliptical transfer orbit exists, and a = r₂ / (1+e)
I.B.2. If e is not in [0,1) then there is no transfer orbit having aphelion at arrival.

II. Inward Bound. Arrival is closer to the Sun than Departure: r₂ < r₁

II.A. Perihelion at Arrival: q₂ = 0.
e = 2 r₂ (r₂ - r₁) / ( r₁² - r₂² - d² )

II.A.1. If e is in [0,1) then an elliptical transfer orbit exists, and a = r₂ / (1-e)
II.A.2. If e>1 then a hyperbolic transfer orbit exists, and a = r₂ / (e-1)
II.A.3. If e<0 then there is no transfer orbit having perihelion at arrival.

II.B. Aphelion at Departure: q₁ = p radians.
e = 2 r₁ (r₂ - r₁) / ( r₂² - r₁² - d² )

II.B.1. If e is in [0,1) then an elliptical transfer orbit exists, and a = r₁ / (1+e)
II.B.2. If e is not in [0,1) then no transfer orbit exists having aphelion at departure.

The next element of the transfer orbit that we will determine is the inclination to the ecliptic, i. We must find the unit vector normal to the orbital plane in the direction of the cross product, r₁ x r₂. We will do this first for elliptic orbits of the short path, and then (in a text box) I will show how to adjust the result for elliptic orbits of the long path.

X_N' = y₁ z₂ - z₁ y₂

Y_N' = z₁ x₂ - x₁ z₂

Z_N' = x₁ y₂ - y₁ x₂

R_N' = { (X_N')² + (Y_N')² + (Z_N')² }^1/2

X_N = X_N' / R_N'

Y_N = Y_N' / R_N'

Z_N = Z_N' / R_N'

The vector [ X_N , Y_N , Z_N ] is the unit normal vector to the transfer orbit, and

i = p/2 - arcsin(Z_N)

Attention. If the transfer orbit is an ellipse—if e is in [0,1)—then the above procedure will give the unit normal [X_N,Y_N,Z_N] for the short path. The unit normal vector for the elliptical transfer orbit of the long path is found by reversing the sign on each component of the short-path unit normal. The positive components become negative, and vice versa. The calculations for the long-path elliptical transfer orbit are otherwise carried through in exactly the same way as those of the short path.

Quantity check for long-path elliptical orbits. The sum of the inclinations of the short-path and long-path elliptical transfer orbits is p radians. I prefer to use the principal value of the arccosine [0,p] as the range of the inclination.

In order to prevent some duplication of the text, I define a subscript variable K to designate the apsidal endpoint of the trajectory. When K=1, the apside is at departure. When K=2, the apside is at arrival.

We must determine the sun-relative velocity of the spaceship at the apsidal end of its intended trajectory. A unit vector in the direction of the velocity is first obtained from the cross product R_N x r_K. The magnitude of the velocity V_K is found from the Vis Viva equation. The components of the velocity vector V_K are determined by multiplying the speed by the unit vector.

V_XK'' = Y_N z_K - Z_N y_K
V_YK'' = Z_N x_K - X_N z_K
V_ZK'' = X_N y_K - Y_N x_K
V_K'' = { (V_XK'')² + (V_YK'')² + (V_ZK'')² }^1/2
V_XK' = V_XK'' / V_K''
V_YK' = V_YK'' / V_K''
V_ZK' = V_ZK'' / V_K''

Elliptical orbits: V_K = [ GM ( 2/r_K - 1/a ) ]^1/2
Hyperbolic orbits: V_K = [ GM ( 2/r_K + 1/a ) ]^1/2

V_XK = V_K V_XK'
V_YK = V_K V_YK'
V_ZK = V_K V_ZK'

The GM factor in the Vis Viva equations for the orbital speed is the gravitational constant times the sun's mass. Its accepted value is

GM = 1.32712440018E+20 m³ sec^-2.

Since up to this point the distances will, most likely, be expressed in astronomical units, it is necessary to know that one astronomical unit equals 1.49597870691E+11 meters.

We now have a semimajor axis (a), an eccentricity (e), and inclination (i), and a state vector for an elliptical or hyperbolic transfer orbit:

[ x_K, y_K, z_K, V_XK, V_YK, V_ZK ]

(K=1 denotes departure, but K=2 denotes arrival.)

Quantity check for long-path elliptical orbits. The apsidal state vectors for the short path and long path elliptical orbits have the same position components, but their their velocity components will differ in sign. (Apart from the sign change, the velocity components will have the same magnitude.)

The angular momentum per unit mass, h, is equal to the cross product of the heliocentric radius vector and the velocity vector at the apsidal endpoint: r_K x V_K. Angular momentum is a conserved quantity, and the components of h are constant for the entire transfer orbit.

h_X = y_K V_ZK - z_K V_YK

h_Y = z_K V_XK - x_K V_ZK

h_Z = x_K V_YK - y_K V_XK

Quantity check for long-path elliptical orbits. The angular momentum vector for long-path elliptical orbits will have components equal in magnitude but opposite in sign to those of short-path elliptical orbits.

The longitude of the ascending node, W, of the transfer orbit can be calculated as follows:

W = arctan2(h_X , -h_Y)

Quantity check for long-path elliptical orbits. The longitude of the ascending node on the long path will be that of the short path plus or minus p radians, whichever will keep it inside the interval [0,2p).

Caveat. I solved for the following compound function from a set of equations that I found in a book. However, while visualizing different angular orientations for orbits, I became suspicious of the hz<0 part of the function. I now think that the hz>=0 part works for both prograde and retrograde orbits; i.e., regardless of the sign of hz. If I change my mind about it later, I'll restore the original function and remove this comment.

Formerly, I believed that...
If h_Z >= 0,	W = arctan2(h_X , -h_Y)
If h_Z < 0,	W = arctan2(-h_X , h_Y)

The effect of the latter portion of the compound function is to define the ascending node as the point where the orbit crosses the ecliptic (z=0) where the Vz has the same sign as hz. But that's not how the ascending node is defined. It's defined as the point where the orbit crosses the ecliptic with Vz>0.

The next element we seek is the argument of perihelion, w.

cos(w+q_K) = (x_K cos W + y_K sin W) / r_K
If i=0 or i=p radians...	sin(w+q_K) = (y_K cos W - x_K sin W) / r_K
If i is neither 0 nor p radians...	sin(w+q_K) = z_K / (r_K sin i)
w = arctan2[ sin(w+q_K) , cos(w+q_K) ] - q_K

Where q_K is the true anomaly in the transfer orbit at the apsidal endpoint. The q_K will be zero when the apside is the perihelion, but it will be or p radians for the aphelion. Correct w to the interval [0, 2p), if necessary.

Quantity check for long-path elliptical orbits. The sum of the short-path and long-path arguments of the perihelion should add up to p radians or to 3p radians. I think.

Caveat. In computer implementations of these formulae for the argument of the ascending node, roundoff error may make the inclination almost, but not exactly zero or p, which may cause an error in the program execution or output. So when the inclination, as calculated, is within an appropriate tolerance of zero or p, within perhaps a trillionth of a radian, it should be reset to exactly zero or p, respectively, for the calculation of sin(w+q_K) under the i=0 or i=p part of the function.

I define a subscript variable L to designate the non-apsidal endpoint of the intended trajectory. If the apside is at departure (K=1), then L=2. If the apside is at arrival (K=2), then L=1.

We desire a position vector for the non-apsidal endpoint of the intended trajectory, referred to the plane of the transfer orbit, such that the origin is the sun (as usual), but the x-axis runs in the positive direction through the transfer orbit’s perihelion and the z-axis runs in the positive direction with the transfer orbit's angular momentum vector. Now that we have the values of W, i, and w, it is possible to rotate r_L from heliocentric ecliptic coordinates to this coordinate system.

x_L’ = y_L sin W + x_L cos W

y_L’ = y_L cos W - x_L sin W

z_L’ = z_L

x_L’’ = x_L’

y_L’’ = z_L’ sin i + y_L’ cos i

z_L’’ = z_L’ cos i - y_L’ sin i

x_L’’’ = y_L’’ sin w + x_L’’ cos w

y_L’’’ = y_L’’ cos w - x_L’’ sin w

z_L’’’ = z_L’’

This triple-primed position vector is the one we were looking for. Important check: Within a reasonable allowance for roundoff error, the value of z_L’’’ should be zero.

q_L = arctan2 (y_L’’’ , x_L’’’)

The q_L is the true anomaly in the transfer orbit of the non-apsidal endpoint of the intended trajectory. Obtaining this angle is a milestone in solving the transfer orbit problem.

Quantity check for long-path elliptical orbits. The sum of the non-apsidal endpoint true anomalies for the short path elliptical orbit and for the long-path elliptical orbit is 2p.

So far in this analysis of transfer orbits, the elliptical and hyperbolic cases have been very similar to each other, with a few changes of sign here and there. The angular orientation of the orbital planes is identical, of course.

That’s about to change, unfortunately, as we consider the motion of a spaceship along the transfer orbit.

Elliptical orbits.

sin u_L = (r_L /a) sin q_L / (1-e²)^1/2

cos u_L = e + (r_L /a) cos q_L

u_L = arctan2(sin u_L , cos u_L)

M_L = u_L - e sin u_L

The u_L is the eccentric anomaly in the transfer orbit of the non-apsidal endpoint of the intended trajectory, and M_L is the corresponding mean anomaly. When calculating M_L it is necessary that u_L be in radians.

Quantity check for long-path elliptical orbits. The sum of the non-apsidal endpoint eccentric anomalies for the short-path elliptical orbit and for the long path elliptical orbit orbit is 2p. The same is true with respect to their mean anomalies.

Although you generally don’t intend to follow an elliptical transfer orbit around its complete circuit, its orbital period can be found from

P = (365.256898326 days) a^3/2

The mean daily motion (radians per day) of a spaceship in the transfer orbit would be

m = 2p radians / P

The short-path transit time is found from...

Apside at Departure: K=1, L=2.
Dt = M₂ / m	if q₁=0 (thus M₁=0)
Dt = (M₂-p) / m	if q₁=p (thus M₁=p)

Apside at Arrival: K=2, L=1.
Dt = (2p-M₁) / m	if q₂=0 (thus M₂=0)
Dt = (p-M₁) / m	if q₂=p (thus M₂=p)

If necessary, correct Dt to the interval [0,P) by adding or subtracting the appropriate multiple of P.

Quantity check for long-path elliptical orbits. The sum of the long-path transit time and the short-path transit time is equal to the period of the elliptical orbit, P.

The speed of the spaceship in the transfer orbit at the non-apsidal endpoint of the transfer orbit is found from the Vis Viva equation:

V_L = [ GM (2/r_L - 1/a) ]^1/2

We now find the velocity vector, V_L, in the transfer orbit at the non-apsidal endpoint of the intended trajectory.

V_XL’’’ = -sin q_L { GM / [ a (1-e²) ] }^1/2
V_YL’’’ = (e + cos q_L) { GM / [ a (1-e²) ] }^1/2
V_ZL’’’ = 0
V_XL’’ = V_XL’’’ cos w - V_YL’’’ sin w
V_YL’’ = V_XL’’’ sin w + V_YL’’’ cos w
V_ZL’’ = V_ZL’’’
V_XL’ = V_XL’’
V_YL’ = V_YL’’ cos i
V_ZL’ = V_YL’’ sin i
V_XL = V_XL’ cos W - V_YL’ sin W
V_YL = V_XL’ sin W + V_YL’ cos W
V_ZL = V_ZL’

The velocity [ V_XL , V_YL , V_ZL ] is that of the spaceship in the transfer orbit at the non-apsidal endpoint, in heliocentric ecliptic coordinates. You can check this vector by comparing its magnitude with the speed calculated from the Vis Viva equation. Be careful about units. The velocity components will have whichever unit used for the length divided by days (if you’re using the radians/day value for m that I provided). Usually, we like to have speeds in meters per second, and there are 86400 seconds in each day.

Quantity check for long-path elliptical orbits. The components of the velocity in the transfer orbit at the non-apsidal endpoint for long-path elliptical orbits will be equal in magnitude, but opposite in sign, to those of short-path elliptical orbits.

The magnitude of the velocity in the transfer orbit calculated by using the Pythagorean theorem on the components [V_XL,V_YL,V_ZL] should be equal to the speed V_L found from the Vis Viva equation. Be careful to adjust the units for the distances to MKS (meter-kilogram-second) if you are using the MKS value of the gravitational parameter GM.

Hyperbolic orbits.

Q = cosh u_L = (1/e) (1 + r_L /a)
u_L’ = ln [ Q + (Q² - 1)^1/2 ]
If sin q_L < 0 then u_L = -u_L’ else u_L = u_L’
M_L = e sinh u_L - u_L
m = (86400 sec/day) ( GM / a³ )^1/2

The hyperbolic mean motion, m, is given in radians per day.

Perihelion at Departure: K=1, L=2.
Dt = M₂ / m	if q₁=0 (thus M₁=0)

Perihelion at Arrival: K=2, L=1.
Dt = -M₁ / m	if q₂=0 (thus M₂=0)

The speed of the spaceship in the transfer orbit at the non-apsidal endpoint of the transfer orbit is found from

V_L = [ GM (2/r_L + 1/a) ]^1/2

We now find the velocity vector, V_L, in the transfer orbit at the non-apsidal endpoint of the intended trajectory.

V_XL’’’ = -(a/r_L) { GM / a }^1/2 sinh u_L
V_YL’’’ = +(a/r_L) { GM / a }^1/2 (e² - 1)^1/2 cosh u_L
V_ZL’’’ = 0
V_XL’’ = V_XL’’’ cos w - V_YL’’’ sin w
V_YL’’ = V_XL’’’ sin w + V_YL’’’ cos w
V_ZL’’ = V_ZL’’’
V_XL’ = V_XL’’
V_YL’ = V_YL’’ cos i
V_ZL’ = V_YL’’ sin i
V_XL = V_XL’ cos W - V_YL’ sin W
V_YL = V_XL’ sin W + V_YL’ cos W
V_ZL = V_ZL’

You can check the velocity vector [ V_XL , V_YL , V_ZL ] by comparing its magnitude with the speed calculated from the Vis Viva equation. As with the elliptical case, be careful about your units.

Delta-Vees.

The delta-vee for departure is found by subtracting the pre-rocketburn velocity vector at the departure position from the HEC velocity vector in the transfer orbit at departure.

DV_X1 = V_X1 - V_X,preburn

DV_Y1 = V_Y1 - V_Y,preburn

DV_Z1 = V_Z1 - V_Z,preburn

The components of the preburn velocity vector can be calculated from the orbital elements of the spaceship before departure and a mean anomaly for the spaceship in its pre-departure orbit.

The delta-vee for arrival is found by subtracting the HEC velocity vector in the transfer orbit at arrival from the velocity of the rendezvous object at the time of arrival.

DV_X2 = V_X,rendezvous - V_X2

DV_Y2 = V_Y,rendezvous - V_Y2

DV_Z2 = V_Z,rendezvous - V_Z2

The components of the rendezvous velocity vector can be calculated from the orbital elements of the destination planet and the mean anomaly for the destination planet (in its own orbit) at the time of arrival.

The sum of the magnitudes of the two delta-vees will be roughly proportional to the cost of the transit, if your rocket's engines are any good.

Transfer Orbit, Worked Example.

We begin with two rectangular position vectors given in heliocentric ecliptic coordinates.

Departure (Vesta):

x₁ = +0.603288669
y₁ = -2.093171651
z₁ = -0.010132931
t₁ = JD 2453040.3

Arrival (Earth):

x₂ = +1.000217362
y₂ = -0.098879727
z₂ = 0
t₂ = JD 2453265.4

Notice that the time of "arrival" ( t₂ - t₁ ) is 225.1 days later than the time of departure. During that time, both the destination planet (Earth) and the spaceship are moving along their respective orbits. Since they must reach the same place, r₂, at the same time, t₂, two conditions must be satisfied: (1) there must be a transfer orbit between r₁ and r₂ having transit time, Dt=225.1 days and (2) Earth must have been at the proper position at departure time to have reached r₂ at the time of arrival.

Those requirements constitute the problem of rendezvous, which I mentioned in the text box at the top of this page. In general, you don't know right away which configurations for the spaceship and the destination planet at the time of departure will be associated with transfer orbits that permit rendezvous. When I prepared this example, I hunted around until I found a good case.

r₁ = 2.178400205

r₂ = 1.005093016

d = 2.033434371

Since r₂ < r₁, the transfer orbit is inward bound.

There are two conic sections, having the sun at a focus and having an apside at either r₁ or r₂, linking r₁ with r₂.

(1) A hyperbola, perihelion at arrival, a=0.205048715, e=5.901727932.

(2) An ellipse, aphelion at departure, a=1.320616879, e=0.649532304.

The unit normal vector for the hyperbolic orbit and the short-path elliptic orbit is

X_N = -0.000492597

Y_N = -0.004982860

Z_N = +0.999987464

The unit normal vector for the long-path elliptic orbit is

X_N = +0.000492597

Y_N = +0.004982860

Z_N = -0.999987464

For the hyperbolic orbit and the short-path elliptic orbit, the inclination to the ecliptic is

i = 0.005007179 radians

For the long-path elliptic orbit, the inclination is

i = 3.136585474 radians

The hyperbolic orbit.

Perihelion at arrival, K=2, L=1, q₂=0.

The eccentricity is

e = 5.901727932

The semimajor axis is

a = 0.205048715 AU

The inclination is

i = 0.005007179 radians

The arrival velocity is

V_X2	= +7678.289 m/sec
V_Y2	= +77669.693 m/sec
V_Z2	= +390.804 m/sec

The angular momentum vector for the orbit is

h_X = -5.780855978E+12 m² / sec

h_Y = -5.847621845E+13 m² / sec

h_Z = +1.173532509E+16 m² / sec

The longitude of the ascending node is

W = 6.184647238 radians

The longitude of the ascending node, W, should be the same for both the hyperbolic and the elliptic orbit of the short path.

The argument of the perihelion is

w = 0.000000000 radians

The zero value for w is not to be wondered at. The destination planet being Earth means that the arrival is in the ecliptic plane, so z₂=0. At arrival, the spaceship is crossing the ecliptic with V_Z2 positive; i.e., arrival occurs at the ascending node of the transfer orbit, which is also the perihelion of the transfer orbit, and therefore the argument of the perihelion is zero.

The true anomaly of departure is

q₁ = 5.091535143 radians

The eccentric anomaly of departure is

u₁ = -1.299203324 radians

The mean anomaly of departure is

M₁ = -8.714924194 radians

The mean motion is

m = 0.185266060 radians / day

The transit time is

Dt = 47.040 days

Whoops! This transit time does not equal the required time difference of 225.1 days! So, although the hyperbolic orbit does connect r₁ with r₂, a spaceship travelling along it would go from r₁ to r₂ too quickly, and the destination planet would not have arrived yet. So this orbit won't work.

But I'll provide the departure velocity just so you can verify that you implemented the procedure properly. The departure velocity is

V_X1	= +17432.112 m/sec
V_Y1	= +69547.802 m/sec
V_Z1	= +355.138 m/sec

The elliptical orbit [of the short path].

Aphelion at departure, K=1, L=2, q₁=p radians.

The eccentricity is

e = 0.649532305

The semimajor axis is

a = 1.320616880 AU

The inclination is

i = 0.005007171 radians

The departure velocity is

V_X1	= +11479.434 m/sec
V_Y1	= +3308.466 m/sec
V_Z1	= +22.141 m/sec

The angular momentum vector for the orbit is

h_X = -1.917798322E+12 m² / sec

h_Y = -1.939947891E+13 m² / sec

h_Z = +3.893192782E+15 m² / sec

The longitude of the ascending node is

W = 6.184647238 radians

The longitude of the ascending node, W, should be the same for both the hyperbolic and the elliptic orbit of the short path.

The argument of the perihelion is

w = 1.949942489 radians

The true anomaly of arrival is

q₂ = 4.333242818 radians

The eccentric anomaly of arrival is

u₂ = 5.089070117 radians

The mean anomaly of arrival is

M₂ = 5.693064072 radians

The mean motion is

m = 0.011334860 radians / day

The transit time is

Dt = 225.1 days

Ahhh! This transit time equals the required time difference of 225.1 days! What amazing good fortune that is. Not only does the elliptical orbit connect r₁ with r₂, a spaceship travelling along it would go from r₁ to r₂ in exactly the right amount of time to rendezvous with the destination planet. Since the elliptic orbit satisfies both the geometrical conditions and the requirement of transit time, it is therefore a valid transfer orbit from Vesta to Earth having a departure on Julian Date 2453040.3 (4 February 2004, 19h 12m UT).

The arrival velocity is

V_X2	= -17921.699 m/sec
V_Y2	= +27790.438 m/sec
V_Z2	= +129.649 m/sec

Vesta's HEC velocity at the time of departure is

V_X,PREBURN	= +20209.120 m/sec
V_Y,PREBURN	= +4861.494 m/sec
V_Z,PREBURN	= -2601.720 m/sec

The departure delta-vee is

DV_X1 = V_X1 - V_X,PREBURN	= -8729.686 m/sec
DV_Y1 = V_Y1 - V_Y,PREBURN	= -1553.028 m/sec
DV_Z1 = V_Z1 - V_Z,PREBURN	= +2623.861 m/sec
DV₁	= 9246.835 m/sec

At departure, the pilot will have aimed his spaceship at the star Vindemiatrix, in Virgo.

Earth's HEC velocity at the time of arrival is

V_X,RENDEZVOUS	= +2445.653 m/sec
V_Y,RENDEZVOUS	= +29532.286 m/sec
V_Z,RENDEZVOUS	= 0

The arrival delta-vee is

DV_X2 = V_X,RENDEZVOUS - V_X2	= +20367.352 m/sec
DV_Y2 = V_Y,RENDEZVOUS - V_Y2	= +1741.848 m/sec
DV_Z2 = V_Z,RENDEZVOUS - V_Z2	= -129.649 m/sec
DV₂	= 20442.110 m/sec

The arrival delta-vee is applied with the spaceship's nose pointing in the direction of Pisces. If no rendezvous delta-vee is applied, the spaceship may strike Earth at slightly more than DV₂, having been somewhat accelerated by Earth's gravity.

Since the components of the delta-vees are in the HEC system, they form a vector by which a spaceship pilot can direct his thrust, presumably guided by a star atlas marked with ecliptic coordinates, leaving him only the thrust magnitude and duration to fuss over with his instruments.

Position of the Destination Planet at the Time of Departure.

This is an important detail, which must be determined in order to pursue the transfer orbit problem in regard to determining when there exists a transfer orbit that will reach the arrival position at the same time as the destination planet does. It serves no purpose to get there when the destination planet is at another point along its own orbit.

We saw in the section just above that there exists a valid transfer orbit, an ellipse with aphelion at departure, from Vesta on JD 2453040.3 (4 February 2004, 19h 12m UT) to Earth on JD 2453265.4 (16 September 2004, 21h 36m UT).

Although I didn't show the work (on this page), the arrival position [ x₂ , y₂ , z₂ ] was calculated from Earth's orbital elements at the arrival time. On JD 2453265.4, the Earth's mean anomaly will be

M_{Earth,arrival} = 4.418560393 radians

The mean motion of Earth is

m_Earth =	2p / { 365.256898326 days (a_Earth)^1.5 }
m_Earth =	0.017202096 radians per day

The valid transfer orbit that we found had a transit time of Dt=225.1 days. We wind the mean anomaly backwards by the corresponding angle:

M_{Earth,departure} =	M_{Earth,arrival} - m_Earth Dt_transfer
M_{Earth,departure} =	0.546174097 radians

If it had been necessary, we would have corrected M_{Earth,departure} to the interval [0,2p).

What we've done on this page is sort of reverse-engineer the problem. In practice, what you can see are the positions of your spaceship and your destination planet at departure, and you want to find out whether a valid transfer orbit exists at a particular time. That's a tough problem because there's no general solution in closed form. So what you do, basically, is work out all possible solutions (or representative samples thereof) and construct a table of departure and arrival positions from which you can "read off" whatever planetary configuration happens to prevail when you want to depart.

You'll probably use the mean anomaly of the preburn orbit and the mean anomaly of the destination planet's orbit as indices going forward. You'll determine for each pair of mean anomalies all possible conic sections having the sun at a focus and an apside at one end of the planned trajectory. For each of these conic sections, you'll output certain important data, such as the orbital elements (a,e,i,W,w), the time of flight (Dt), the delta-vee's on each end ( DV₁ , DV₂ ), the heliocentric longitude of the departure position, and the heliocentric longitude of the destination planet at the time of departure.

Getting the heliocentric longitude of the departure point (on Vesta, in the example) is easy:

Longitude₁ =	arctan2( y₁ , x₁ )
Longitude₁ =	4.993001385 radians

Once you have the mean anomaly of the destination planet at the time of departure, you use it with the destination planet's orbital elements to get its heliocentric position at the time of departure.

[ x_{Earth,departure}, y_{Earth,departure}, z_{Earth,departure} ]
[ -0.701051407, +0.693059582, 0 ]
Longitude_{Earth,departure} =	arctan2( y_{Earth,departure} , x_{Earth,departure} )
Longitude_{Earth,departure} =	2.361926988 radians

The table you construct will be huge because it will contain up to four orbits (with all the relevant data for each) for each pair of sampled mean anomalies. If I were to continue developing the example problem that I began above, I might sample Vesta's orbit at intervals of 0.01 radians of mean anomaly, and likewise for Earth. The output file will be about 100 megabytes long if there are 85 output characters per orbit.

But once you have constructed the table, you simply look up whatever arrangement of heliocentric longitude occurs around the time you want to move your spaceship, or whatever, from Vesta to Earth. Once you get an idea about what travel paths are available, and at what cost in energy, you can use another program to fine-tune your transfer orbit by "reverse engineering."

Elliptical orbits:	V_K = [ GM ( 2/r_K - 1/a ) ]^1/2
Hyperbolic orbits:	V_K = [ GM ( 2/r_K + 1/a ) ]^1/2