# Earth Gravity Assist (EGA) maths and flight plan

#### Keithth G

##### New member
This post is going to focus on the maths behind performing an Earth Gravity Assist (EGA) manoeuvre. It will finish by setting up a basic, general purpose EGA flight plan.

The Earth Gravity Assist manoeuvre is designed to 'kick' a spacecraft from a low energy orbit around the Sun into a higher energy orbit - one that enables the spacecraft to get out to, say, Jupiter and Saturn - even though the initial Earth escape burn only adds enough energy to get out to just beyond the orbit of Mars.

For missions that are 'delta-V constrained', which is to say most missions, some gravity assist manoeuvre is necessary to get out to the outer planets. Although, the EGA manoeuvre is not the most effective way of getting the requisite energy 'kick' - gravity assist manoeuvres such as the Venus-Earth Gravity Assist (VEGA) are more effective and have smaller overall delta-V requirements than the EGA - but the EGA manoeuvre is self-contained. Whereas gravity assist techniques such as VEGA require the cooperation of other planets (they need to be in the right place at the right time for the manoeuvre to work), the EGA manoeuvre doesn't - so it can be applied at anytime. And therein lies the EGA's principal advantage.

So, how does the EGA manoeuvre work? Well, the EGA is a example of a "V-infinity Leveraged Transfer" (VILT) designed to raise the hyperbolic excess velocity of a spacecraft. It looks something like this:

In this diagram, the Sun is located in the centre of the image. The Earth traces out a (near) circular orbit marked by the dotted line. The trajectory starts at the position called "Low Encounter" in the above diagram. This is the periapsis of a blue semi-ellipse denoted by the blue arc. At the apoapsis of that ellipse, the spacecraft executes a retrograde burn (red arrow) that lowers the periapsis of its new orbit (the orange trajectory) such that that the spacecraft now encounters the Earth's orbit at two points (labelled "High Encounter (-)" and "High Encounter (+)"). If the parameters of the VILT are just right, the spacecraft will encounter the Earth again at one of these crossing points. Consequently, for the EGA, and more generally for any VILT, they always come in two flavours - one of the '+' kind, and one of the '-' kind.

The focus of this note, then, is to work out how to set up the parameters of the EGA VILT so that they are 'just right' and that there is an encounter of the spacecraft with Earth at one of the '+' and '-' crossing points. Here, we are going to just focus on just one of them - the '+' crossing - although the maths can readily be adapted to the other orbit crossing. In addition, and to keep the maths simple, we will assume that the Earth moves in a circular orbit around the Sun.

VILT mathematics
Construction of a VILT is largely about getting the timing of the intersection with Earth's orbit right. The time taken to leave Earth and then reach the (+) intersection point must exactly match the time taken for the Earth to reach the same point in space. So, let's work through the calculations to see how long it takes for the spacecraft to escape Earth ("Low Encounter" point) and to recross the Earth's orbit ("High Encounter" point).

Units
To keep the maths as simple as possible, we are going to work in length and time units such that the radius of the Earth's orbit is exactly 1; and the time taken for the Earth to complete one orbit around the Sun is [MATH]2\pi[/MATH] units. In these units, the orbital speed of the Earth around the Sun is also exactly 1. In what follows, to convert back to more conventional metres and seconds scales, when we calculate a distance (e.g., the apoapsis of an orbit), to get back to more conventional units, we have to multiply by the length of one Astronomical Unit ( ~ [MATH]1.4960\times 10^{11}[/MATH] m); and when we calculate something with the dimensions of speed (e.g., the spacecraft's speed at apoapsis), we need to remember to multiply by the Earth's mean orbital speed around the Sun (~ [MATH]29,780[/MATH] m/s).

VILT parameters
Before we can do calculations, let's introduce a bit of notation so that we describe the various components of the VILT. First, we note that the spacecraft's orbit upon Earth escape (the blue line in the diagram) is a semi-ellipse. We assume that we leave Earth at the periapsis of that ellipse. And because it is leaving from Earth's orbit, the periapsis of that ellipse is just 1 in the units that we have chosen. Now, let's suppose that the apoapsis of the escape orbit is [MATH]r_a[/MATH] (also labeled [MATH]r_c[/MATH] in the diagram). At the moment we don't know what value this is other than that it must be greater than 1.

At the apoapsis of the orbit, the spacecraft executes a retrograde burn of some magnitude [MATH]\Delta V[/MATH] to achieve a new orbital periapsis, [MATH]r_p[/MATH]. At the moment, we now nothing about the [MATH]r_p[/MATH] other than that it must be less that 1.

These two model parameters, [MATH]r_a[/MATH] and [MATH]r_p[/MATH] fully define the VILT. From [MATH]r_a[/MATH] and [MATH]r_p[/MATH] we can calculate [MATH]\Delta V[/MATH]. We can also determine our hyperbolic escape velocity, [MATH]v_\infty[/MATH] and we can determine the components of the escape plan that we need in order to set ourselves on the VILT trajectory. And finally we can work out when we have to leave Earth in order to rendezvous with Earth on a specified encounter date.

Time taken to reach apoapsis from Earth escape
With this choice of length/time scale and choice of parameters in mind, we can calculate the time taken to reach apoapsis. We know that the periapsis radius of our orbit is 1 and the apoapsis radius is $r_a$. So, the time taken to complete one orbit of the blue escape trajectory is:

[MATH] 2\,\pi\,\sqrt{\frac{(1+r_a)^3}{8}} [/MATH]
But the time taken to go from periapsis to apoapsis is exactly half an orbit, the time between Earth escape and orbital apoapsis is:

[MATH] \pi\,\sqrt{\frac{(1+r_a)^3}{8}} [/MATH]
Time taken to reach Earth encounter from apoapsis
This is a bit harder to calculate than the time to apoapsis from Earth escape, but let's start with the time taken to reach periapsis of the new orbit. As above, the time taken to reach periapsis from periapsis is just:

[MATH] \pi\,\sqrt{\frac{(r_p+r_a)^3}{8}} [/MATH]
But, of course, the Earth rendezvous doesn't occur at periapsis - it occurs at either the '+' position of the '-' position. Now, we calculate the mean anomaly of the '+' position. This is given by:

[MATH] M = 2\left(\tan^{-1}\left(\sqrt{\frac{1-r_{p}}{r_{a}-1}}\right)-\frac{\sqrt{\left(r_{a}-1\right)\left(1-r_{p}\right)}}{r_{a}+r_{p}}\right) [/MATH]
The mean anomaly gives us a measure of the amount of time it takes for the spacecraft to move from periapsis to the '+' orbital intersection point. So, we can write the total time to reach the '+' intersection point from orbital apoapsis as:

[MATH](\pi + M)\,\sqrt{\frac{(r_p+r_a)^3}{8}} [/MATH]
Total time from Earth escape to Earth rendezvous
Adding up the two times - Earth escape to apoapsis; and from apoapsis to the '+' intersection, we deduce that the total time to go from Earth escape to the '+' Earth rendezvous is:

[MATH] \pi\,\sqrt{\frac{(1+r_a)^3}{8}} + (\pi + M)\,\sqrt{\frac{(r_p+r_a)^3}{8}} [/MATH]
Similarly, the time taken to go from Earth escape to the '-' Earth rendezvous is:

[MATH] \pi\,\sqrt{\frac{(1+r_a)^3}{8}} + (\pi - M)\,\sqrt{\frac{(r_p+r_a)^3}{8}} [/MATH]

Position of the spacecraft at the encounter
So far, we have calculated the 'when' of a crossing of the spacecraft's trajectory with Earth's orbit. Now we need to answer the question of 'where' on Earth's orbital trajectory the intersection occurs. The 'where' is calculated from the true anomaly, [MATH]nu[/MATH]. For the '+' encounter, the true anomaly is given by the expression:

[MATH] \nu = 2\tan^{-1}\left(\sqrt{\frac{r_{a}}{r_{p}}\,\frac{1-r_{p}}{r_{a}-1}}\right) [/MATH]
Ensuring that there is an Earth encounter
For there to be an encounter with Earth, we need to make sure that the Earth is at the '+' crossing point at exactly the same time as the spacecraft passes through the same point. To achieve this, Earth has to go around its orbit approximately an integer number of times in the time it takes the spacecraft to return to re-cross Earth's orbit. In this note, we are going to assume that this integer is 2 - i.e., the Earth completes approximately two orbits before the spacecraft encounters the Earth. If we target the '+' intersection, Earth will have to complete slightly more than two orbits; and if we target the '-' intersection, Earth will have to complete slightly less than two orbits.

And what is the offset? Well, that's just the true anomaly. So, the time taken to for the Earth to reach the '+' intersection is:

[MATH] 2\times 2\,\pi + \nu [/MATH]
So, the condition for there to be an intersection is of the spacecraft with Earth at the '+' intersection point is:

[MATH] 2\times 2\,\pi + \nu = \pi\,\sqrt{\frac{(1+r_a)^3}{8}} + (\pi + M)\,\sqrt{\frac{(r_p+r_a)^3}{8}} [/MATH]
And, also for an intersection at the '-' point, we require that:

[MATH] 2\times 2\,\pi - \nu = \pi\,\sqrt{\frac{(1+r_a)^3}{8}} + (\pi - M)\,\sqrt{\frac{(r_p+r_a)^3}{8}} [/MATH]
V-infinity constraint
The above isn't quite enough to specify the VILT. We need to supply one more piece of information. The whole purpose of the VILT is to use a small retrograde burn at apoapsis to achieve a much larger increase in hyperbolic excess velocity at Earth encounter. It is this increase in [MATH]v_\infty[/MATH] that gives the spacecraft the 'kick' needed to reach Jupiter, say. So, when we construct a VILT, we have a target [MATH]v_\infty[/MATH] in mind. But we already know the equation for [MATH]v_\infty[/MATH] (from "A Very Boring Post" in the Maths/Physics section):

[MATH] v_{\infty} = \sqrt{3-\frac{2}{r_{a}+r_{p}}-2\,\sqrt{\frac{2\,r_{a}\,r_{p}}{r_{a}+r_{p}}}} [/MATH]
Putting it all together
OK, so we can now put all of the bits and pieces together so that we can see the system of equations that we need to solve in order to set up a VILT. For the '+' encounter, we need to solve the following system of equations:

[MATH] M = 2\left(\tan^{-1}\left(\sqrt{\frac{1-r_{p}}{r_{a}-1}}\right)-\frac{\sqrt{\left(r_{a}-1\right)\left(1-r_{p}\right)}}{r_{a}+r_{p}}\right) [/MATH]
[MATH] \nu = 2\tan^{-1}\left(\sqrt{\frac{r_{a}}{r_{p}}\,\frac{1-r_{p}}{r_{a}-1}}\right) [/MATH]
[MATH] 2\times 2\,\pi + \nu = \pi\,\sqrt{\frac{(1+r_a)^3}{8}} + (\pi + M)\,\sqrt{\frac{(r_p+r_a)^3}{8}} [/MATH]
[MATH] v_{\infty} = \sqrt{3-\frac{2}{r_{a}+r_{p}}-2\,\sqrt{\frac{2\,r_{a}\,r_{p}}{r_{a}+r_{p}}}} [/MATH]
If we treat [MATH]M[/MATH], [MATH]\nu[/MATH, [MATH]r_a[/MATH] and [MATH]r_p[/MATH] as unknowns, we have a system of four equations in four unknowns. True, as systems of equations go, this is an ugly set of equations to solve. But, the equations can be put into a spreadsheet without much difficulty and one can use Excel's 'Solver' utility to numerically solve these equations if one wants. Or one can simply use trial and error. Either way, if one solves this set of equations for $v_\infty$ = 9.2 km/s (or 9.2/29.78 in our dimensionless units) - a typical value to reach Jupiter fromo Earth, we find that:

[MATH] r_a = 2.25503 [/MATH]
[MATH] r_p = 0.903067 [/MATH]
[MATH] M = 18.4067 \,\text{deg.} [/MATH]
[MATH] \nu = 47.4185 \,\text{deg.} [/MATH]
So, when we escape Earth, we need initially to target a orbit around 2.255 AU, and when we reach apoapsis, we need to lower our periapsis from 1.000 AU to 0.903 AU. Doing this means that we should have an encounter with Earth with at the '+' intersection point which well be 47.42 degrees after our initial departure point from Earth. Notice that this VILT requires an initial aphelion that is greater than the mean orbital radius of Mars (around 1.52 AU).

And what about the '-' solution? For this, we need to solve the same set of equations - except that a few of the signs have changed from '+' to '-':

[MATH] M = 2\left(\tan^{-1}\left(\sqrt{\frac{1-r_{p}}{r_{a}-1}}\right)-\frac{\sqrt{\left(r_{a}-1\right)\left(1-r_{p}\right)}}{r_{a}+r_{p}}\right) [/MATH]
[MATH] \nu = 2\tan^{-1}\left(\sqrt{\frac{r_{a}}{r_{p}}\,\frac{1-r_{p}}{r_{a}-1}}\right) [/MATH]
[MATH] 2\times 2\,\pi - \nu = \pi\,\sqrt{\frac{(1+r_a)^3}{8}} + (\pi - M)\,\sqrt{\frac{(r_p+r_a)^3}{8}} [/MATH]
[MATH] v_{\infty} = \sqrt{3-\frac{2}{r_{a}+r_{p}}-2\,\sqrt{\frac{2\,r_{a}\,r_{p}}{r_{a}+r_{p}}}} [/MATH]
And for the same hyperbolic excess velocity of 9.2 km/s, this set of equations has solution:

[MATH] r_a = 2.19225 [/MATH]
[MATH] r_p = 0.897773 [/MATH]
[MATH] M = 19.6954 \,\text{deg.} [/MATH]
[MATH] \nu = 49.1751 \,\text{deg.} [/MATH]
So, what can information can we learn from from these numeric solutions? Well, just about anything we might want to know. For example, for the '+' EGA solution the hyperbolic excess velocity at Earth departure is given by:

[MATH] 29.78\,\sqrt{3-\frac{2}{1 + r_a}-2\,\sqrt{\frac{2\,r_a}{1 + r_a}}} \to 5.274\,\text{km/s} [/MATH]
and the [MATH]\Delta V[/MATH] required at apoapsis to lower orbital periapsis from 1.0 to 0.903067 is:

[MATH] 29.78 \left( \sqrt{\frac{2}{r_a \left(r_a+1\right)}}- \sqrt{\frac{2\,r_p}{r_a \left(r_a+r_p\right)}}\right) \to 0.5476\,\text{km/s} [/MATH]
So, by executing a retrograde burn of 548 m/s at apoapsis, this is sufficient to increased the hyperbolic excess velocity of the spacecraft at Earth encounter from the departure value of 5,274 m/s to the targeted 9,200 m/s - i.e., an increase of 3,924 m/s. This is the 'leveraging' effect at work: a relatively small apoapsis burn has led to a much larger increase in the hyperbolic excess velocity.

We can, of course, also calculate (from Earth escape) the time to apoapsis and the time to Earth encounter. The time to apoapsis is just:

[MATH] \frac{365.25}{2\,\pi}\,\pi\,\sqrt{\frac{(1+r_a)^3}{8}} \,\text{days} \to 379.182\,\text{days}[/MATH]
and the time from apoapsis to Earth encounter is:

[MATH] \frac{365.25}{2\,\pi}\,(\pi + M)\,\sqrt{\frac{(r_p+r_a)^3}{8}} \to 399.427 \, \text{days}[/MATH]
The total time to execute the EGA (+) manoeuvre is 778.609 days - or 2 years 48 days and 2.6 hours.

And what about the Earth escape burn? From a 300 x 300 km circular orbit where the orbital speed is 7.730 km/s, the escape burn needed to achieve an initial hyperbolic excess velocity of 5.274 km/s is:
[MATH] \sqrt{5.274^2 + 2\times 7.730^2} - 7.730\,\text{km/s} \to 4.408 \, \text{km/s} [/MATH]
In total, then, the dV requirements for the EGA (+) manoeuvre - Earth escape and apoapsis retrograde burn - is 4.955 km/s.

We can go through a similar analysis for the EGA (-) solution. If we can do that, we can complete the following table for the EGA (+) and EGA (-) solitions:

| EGA (+) | EGA (-)
Earth escape burn | 4.408 km/s | 4.342 km/s
Initial hyp. escape velocity | 5.274 km/s | 5.121 km/s
Time to apoapsis | 379.182 days | 368.265 days
dV of retrograde burn | 0.548 km/s | 0.588 km/s
Time to Earth encounter | 399.427 days | 312.343 days
Total manoeuvre time | 778.609 days | 680.608 days
Total dV requirements | 4.955 km/s | 4.930 km/s
Encounter hyp. escape velocity | 9.200 km/s | 9.200 km/s

The Earth's orbit isn't circular
Strictly speaking, the above solutions to the EGA VILT problem are valid if the Earth rotates the Sun in a circular orbit. The Earth's orbit is slightly elliptical so this VILT solution needs a slight modification for it to work for Earth's real orbit motion around the Sun. Nonetheless, so long as one is prepared to be a little adaptable in implementing this basic flight plan, this basic VILT scheme is perfectly 'fly-able' on Orbiter without further finesse.

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#### Col_Klonk

##### Member
Don't forget the moon and other big objects floating around the sun

#### Keithth G

##### New member
Don't forget the moon and other big objects floating around the sun

Sorry, 'Col_Klonk' - that's a bit too cryptic for me.

---------- Post added at 05:53 AM ---------- Previous post was at 12:46 AM ----------

While I think about it, here's a nice animation of an EGA manoeuvre to Jupiter - as used by the Juno mission (launched mid-2011 - arrival Jupiter mid-2016).

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#### Col_Klonk

##### Member
The earth moon system has it's COG offset from the earth centre.. added to this, other planets and their gravitational effects... which probably would change the actual COG for the EGA...
Just thinking, but haven't looked into it. It might help in getting that little extra 'boost'...

Maybe it's possible to extend for workings to a multiple body system - this would be a nice touch.
:thumbup:

#### Keithth G

##### New member
The earth moon system has it's COG offset from the earth centre.. added to this, other planets and their gravitational effects... which probably would change the actual COG for the EGA...
Just thinking, but haven't looked into it. It might help in getting that little extra 'boost'...

Maybe it's possible to extend for workings to a multiple body system - this would be a nice touch.

Oh, yes - there's a bunch of things that effect these somewhat idealised results. In no particular order, these are:

1. The eccentricity of planetary orbits

2. The Earth orbits the Earth-Moon Barycentre

3. The Sun orbits the Solar System Barycentre

4. There are small, but significant gravitational influences from other planets

5. And so on.

But for all that, the simple circular theory provides a reasonable starting point. In the normal scheme of things, the simple circular theory is used to prune the tree of possible trajectories into something manageable. It also serves as a starting point for a more comprehensive linked-conics (i.e., TransX-like) two-body model. In turn, the linked conics model is used as a starting point for a full n-body analysis.

As for the extending to multiple bodies, yes, it is possible - and that's the goal. But one has to start somewhere.

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#### Keithth G

##### New member
This post is really a continuation of the opening post in this thread. The point of this addition is to expand a little on the theory of VILTs so that they cover trajectories that exploit different resonances - e.g., 2:1, 3:1, 3:2 and 4:3 and so on - and to make a comparison of the effectiveness of these EGA schemes in reducing delta-V requirements in getting out to, say. Jupiter.

(Just as a quick reminder, a 2:1 resonance is one for which the spacecraft does 1 revolutions around the Sun while the Earth completes 2 revolutions around the Sun; and a 3:2 resonance is one for which the spacecraft does two orbits of the Sun while Earth completes three. And so on...)

First, let's start with a picture of the 2:1 EGA scheme of the original post.

This is the gravity assist manoeuvre used by the the Galileo and Juno missions to Jupiter. The basic idea is that at Earth departure, the spacecraft is injected into a near 2:1 resonant orbit with the Earth and traces out the blue arc to aphelion There, the spacecraft executes a retrograde burn which lowers orbital perihelion from 1.0 AU to about 0.90 AU sending the spacecraft on the green arc to an Earth encounter. At the encounter, the spacecraft gets an energy 'kick' from Earth giving it sufficient energy to reach Jupiter on an near-Hohmann transfer trajectory.

The question is though: what's so special about a 2:1 resonant orbit? Couldn't one have used, say, a 3:1 resonant orbit or perhaps a 3:2 energy orbit? And the answer to that question is yes, once could - but it turns out the the 2:1 resonant orbit represents the best balance between fuel savings and the overall mission time.

Just for reference, here is the plot of the 3:1 EGA scheme:

And here's the plot for the 3:2 EAG scheme:

In all three cases, I've targeted the same final hyperbolic escape velocity of 9,200 m/s after encountering Earth and starting on the Jupiter transfer trajectory.

A couple of things to note about these graphs:

1. Both the 3:1 EGA scheme and the 3:2 EGA scheme take a little over three years to complete whereas the 2:1 EGA scheme takes only two. So, unless the 3:1 or 3:2 EGA schemes offer significant fuel savings, it may be hard to justify the additional one year in mission duration.

2. The 3:1 EGA scheme requires that the spacecraft be sent out to over 3 AU from the Sun - twice Mars mean orbital radius before executing a Deep Space Manoeuvre to lower the periapsis. In addition to a longer duration mission, this requires more delta-V just to escape Earth. On the other hand, the spacecraft only needs to lower its periapsis to around 0.95 AU at aphelion, cutting the DSM costs by about a factor of two.

3. The 3:2 EGA scheme only requires that the spacecraft's aphelion only be lifted to about 1.7 AU (just beyond the orbit of Mars) - but on the other hand the periapsis now needs to be lowered to 0.82 AU in order to get the same energy kick from the Earth.

4. All three schemes - the 2:1, the 3:1 and the 3:2 EGA schemes - are in competition with the Oberth effect when it comes to reducing overall mission costs. Although, the hyperbolic escape velocity in the 3:2 scheme works out at around 3.0 km/s less than for the 3:1 EGA scheme, when leaving from Low Earth Orbit (LEO), because of the efficiency of the Oberth effect, the incremental delta-V cos of that 3.0 km/s is perhaps only half of that.

So, when you take into account all of the various factors affecting the fuel budget, how do these three mission scenarios stack up against each other (and a direct Hohmann transfer to Jupiter)?

| 2:1 resonance | 3:1 resonance | 3:2 resonance | Hohmann Transfer |
Earth escape burn | 4.408 km/s | 5.242 km/s | 3.864 km/s | 6.558 km/s |
Initial hyp. escape velocity| 5.274 km/s| 6.983 km/s| 3.862 km/s | 9.200 km/s |
dV of DSM(s)| 0.548 km/s| 0.214 km/s| 1.027 km/s | 0.0 km/s |
Total manoeuvre duration| 2 yr 48 days| 3 yr 30 days| 3 yr 67 days | 0 years |
Total dV requirrments| 4.956 km/s| 5.456 km/s| 4.891 km/s | 6.558 km/s |
Final hyp. escape velocity| 9.200 km/s| 9.200 km/s| 9.200 km/s | 9.200 km/s

So, what observations can we make:

1. The direct Hohmann transfer is the clear winner on speed. It is between two and three years faster than a trajectory based on an EGA manoeuvre. However, the fuel cost of the Hohmann transfer is high at 6.558 km/s. This probably exceeds the capabilities of most modern transfer systems - which explains why EGA manoeuvres are used in practice.

2. Of these three scenarios, the delta-V cheapest is the 3:2 EGA manoeuvre. However, this beats the shorter duration 2:1 EGA manoeuvre by a mere 65 m/s. Since the 65 m/s doesn't really justify the extra one year in mission duration, the 3:2 EGA scheme is not favoured.

3. The 3:1 EGA scheme requires more fuel than the 2:1 EGA scheme - and it takes a year longer. It's pretty clear that the 3:1 scheme is not going to be anyone's favourite.

So, overall, the 2:1 EGA scheme is looking pretty good.

Composite EGA schemes
But what about more complicated EGA schemes? For example, one could use a 3:2 EGA scheme to kick a spacecraft up to a hyperbolic excess velocity needed to execute a 2:1 EGA scheme; and then the 2:1 EGA could then kick the energy of the spacecraft up to a level needed to get to Jupiter. How does this stack up against the vanilla 2:1 EGA scheme?

Well, this is a trajectory plot of this composite scheme:

If one grinds one's way through the somewhat tedious calculations, this is what one finds: the total mission duration is (of course) a little over five years (a long time!); the initial Earth escape burn is a modest 3.746 km/s and the two subsequent periapsis lowering DSMs (the first for the 3:2 scheme; and the second for the 2:1 scheme) require 0.265 km/s and 0.548 km/s. The total fuel requirements of this scheme are 4.559 km/s - around 2.0 km/s less than for the basic Hohmann transfer. Of course, here the mission duration is the problem: at over five years, unless the mission were highly fuel constrained, the 400 m/s delta-V saving doesn't quite warrant the three additional years added to the mission duration. Nonetheless, it useful to know that composite schemes can further lower mission delta-V requirements.

So, overall the 2:1 EGA scheme looks the best.

(In a sequel to this post, I'll post the maths behind these calculations; and run the numbers through PyKEP so that for those interested in 'flying' these missions, they have a mission script to follow.)

---------- Post added 05-15-16 at 12:47 AM ---------- Previous post was 05-14-16 at 11:24 AM ----------

Just as a bit of an addendum:

With three possible 'base' resonant EGA sequences, it is possible to construct more composite trajectories. In fact, the full set of possible composite trajectories is:

1. 'Do nothing' EGA sequence (i.e., a straight Hohmann transfer to Jupiter)
2. A singlet 3:2 EGA - Jupiter manoeuvre (examined above)
3. A singlet 2:1 EGA - Jupiter manoeuvre (examined above)
4. A singlet 3:1 EGA - Jupiter manoeuvre (examined above)
5. A doublet EGA sequence: 3:2 EGA -> 2:1 EGA -> Jupiter (examined above)
6. A doublet EGA sequence: 3:2 EGA -> 3:1 EGA -> Jupiter
7. A doublet EGA sequence: 3:2 EGA -> 3:1 EGA -> Jupiter
8. A triplet EGA sequence: 3:2 EGA -> 2:1 EGA -> 3:1 EGA -> Jupiter

I total there are 2^3 = 8 possibilities. Now, one could have thrown other resonant orbits into the mix - e.g. 5:2 and 4:3 to name but a few - and if we allow ourselves, say, 10 possible resonant orbits to, then we will have 2^10 = 1024 possibilities to examine and quantify.

It is this rapid escalation in the number of permutations that makes working with VIT sequences problematic. Although the problem isn't particularly acute for working with planetary VILT sequences (such as composite EGA manoeuvres) where the protracted flight durations for composite sequences makes many possibilities somewhat moot because no-one would be prepared to wait long enough for them to unfold, the same mathematics can be applied to endgame sequences around the moos of Jupiter and Saturn. Here, endgame sequences can be protracted involving many flybys and the number of possibilities can quickly run into the tens of millions.

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#### boogabooga

##### Bug Crusher
3. The 3:1 EGA scheme requires more fuel than the 2:1 EGA scheme - and it takes a year longer. It's pretty clear that the 3:1 scheme is not going to be anyone's favourite.

Well, hold on. The 3:1 requires more fuel overall.

I think one has to consider the hardware involved. What 3:1 has going for it vs. 2:1 is that it requires less delta-V from the probe itself.

There are a lot of high-energy upper stages and "cheap" solid kick motors floating around. Meanwhile, the probe may be small and have a relatively inefficient hypergolic or monopropellant engine. Don't discount the possibility that it would be cheaper to come up with 830 m/s on launch day than it would be to come up with 330 m/s a year after launch.

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#### Keithth G

##### New member
The above theory is all based on circular orbits. In reality, the planets move in pronounced ellipses, and this can through of the circular orbit theory dV calculations by a few hundred m/s.

To get a more precise picture of the EGA, one can use a tool better suited to optimising trajectories for elliptical orbits - e,g, PyKEP / PyGMO. If now runs the basic 2:1 EGA scheme through PyKEP, this is the trajectory plan that results:

Code:
Date of Earth departure:  2009-Jun-04 00:16:31.575584
Date of DSM:              2010-Jun-23 14:03:35.636933
Date of Earth encounter:  2011-Jul-23 04:58:27.856113
Date of Jupiter arrival:  2014-Feb-02 01:47:33.162571
Transfer time from Earth to DSM:      384.57  days
Transfer time from DSM to Earth:      394.62  days
Transfer time from Earth to Jupiter:  924.87  days
Total mission duration:               1704.06  days

TransX escape plan - Earth escape
--------------------------------------
MJD:                 54986.0115
Outward:                -39.841 m/s
Plane:                   -0.200 m/s
Hyp. excess velocity:  5338.508 m/s
Earth escape burn:     4434.623 m/s

DSM
--------------------------------------
MJD:                 55370.5858
Outward:                 -4.090 m/s
Plane:                   -0.003 m/s
dV needed:              540.285 m/s

Earth encounter
--------------------------------------
MJD:                 55765.2073
Approach velocity:     9215.852 m/s
Departure velocity:    9215.852 m/s
Outward angle:           16.408 deg
Inclination:             -3.078 deg
Turning angle:           47.259 deg
Periapsis altitude:     637.800 km
dV needed:                0.000 m/s

Jupiter arrival
--------------------------------------
MJD:                 56690.0747
Hyp. excess velocity:  5610.532 m/s
Orbit insertion burn    264.345 m/s - (C3 = 0)
Total fuel cost:       5239.254 m/s

Of course, this still doesn't take into account the motion of the Earth around the Earth-Moon Barycentre (~ 10 m/s error in dV), nor does it take into account various three-body effects and assorted perturbations. But the trajectory plan is a fair representation.

In setting up the above trajectory plan, I have set up the Earth-Jupiter leg of the flight to coincide with that of the challenge http://www.orbiter-forum.com/showthread.php?t=36402 "Fully loaded Shuttle-A from Earth to Ganymede".

---------- Post added at 01:32 AM ---------- Previous post was at 01:30 AM ----------

Well, hold on. The 3:1 requires more fuel overall.

I think one has to consider the hardware involved. What 3:1 has going for it vs. 2:1 is that it requires less delta-V from the probe itself.

There are a lot of high-energy upper stages and "cheap" solid kick motors floating around. Meanwhile, the probe may be small and have a relatively inefficient hypergolic or monopropellant engine. Don't discount the possibility that it would be cheaper to come up with 830 m/s on launch day than it would be to come up with 330 m/s a year after launch.

Yes, a valid point.

The only way of matching hardware with trajectory requirements is to enumerate all EGA possibilities and then seeing which match the available hardware.

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