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.
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:
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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