3
$\begingroup$

While reading up a bit online, I found out about Weak Stability Boundaries and the Interplanetary transport network. How is this "path" that traverses the Solar System determined?

I am aware of some relation to Lambert's problem, but am only familiar with the 2 body form of this.

ITN for reference below:

Interplanetary Transport Network

$\endgroup$
4
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – hi-bye125
    Dec 27, 2023 at 16:10
  • $\begingroup$ I agree with this answer from 2020 that ITN/WSB trajectories aren't practically useful in comparison to gravity assists like the Voyager Grand Tour trajectory pictured. Which do you want to know how to plan? $\endgroup$
    – Erin Anne
    Dec 28, 2023 at 10:04
  • $\begingroup$ Ok, in a day I’ve learnt a LOT about what I know(and dont), and it seems like I’d like to know how to plan both? Initially I thought ITN and a series of gravity assists is basically the same thing? Or at least thats what I thought. @ErinAnne $\endgroup$
    – hi-bye125
    Dec 28, 2023 at 15:44
  • $\begingroup$ I think this answer and some things linked back from there will be related $\endgroup$
    – Erin Anne
    Dec 28, 2023 at 22:28

2 Answers 2

4
+100
$\begingroup$

Mathspeak answer:

The pathways of the Interplanetary Transportation Network (ITN) are determined using numerical analysis of a dynamical system of equations of motion to find invariant manifolds which allow low-energy trajectories between Lagrange points of different celestial bodies.

What on Earth does all that mean?

Numerical analysis means you use lots of arithmetic to get an approximate solution to a problem you don’t know how to solve exactly.

For instance, say you want to know the area of a circular courtyard. A “numerical solution” is to lay down 1 meter square tiles until the courtyard is filled. This will give an approximate area, but misses a lot of small areas. You could then fill in these small areas with 10cm x 10 cm tiles and add that to the total area. Then repeat the process with 1 cm x 1 cm tiles. By repeating this algorithm, the courtyard area could be determined to any desired accuracy.

Of course, if you knew the formula for the area of a circle, you could have arrived at an exact solution much faster. If there is no exact solution (say for a winding forest path), your only choice as a numerical solution.

The differential equations which describe trajectories in a 2-body, 2-dimensional problem have exact solutions (conic sections). In 3+body problems there are few exact solutions, so numerical analysis is needed. Fortunately, computers can do the number crunching.

Dynamical Systems are mathematical functions (equations) which describe the time dependence of a point in space, like the swing of a pendulum or the position of a spacecraft.

Invariant means a quantity does not vary. For instance, when a satellite is in orbit, its total energy (kinetic + gravitational) is invariant.

A Manifold is a surface. A familiar example of a manifold is the intake manifold in an auto engine (a complex curved tube which connects the carburetor to multiple cylinders). A trombone is also a manifold. And the surface of a daffodil.

In the case of ITN, the manifold in question is a collection of contiguous state vectors with the same energy. “State vectors” means vectors of position and velocity that, together with their time, uniquely determine the state of an orbiting body. By “contiguous” I mean "side by each" in the same sense that Washington State is contiguous with Oregon. Or schooling fish are contiguous with each other.

A conventional elliptical orbit can be an example of an invariant manifold. Such an orbit can be considered the collection of all contiguous state vectors with the same total energy.

Note that for any point in the orbit, there are an infinite number of possible intersecting orbits. Even if we restrict ourselves to orbits which intersect not only at the origin, but also at a chosen orbital destination, we are still left with an infinite number of orbits. It is only if we restrict ourselves to origin, destination and contiguous state vectors do we get our “invariant” orbit.

In an equivalent 3-body, 3-dimensional example, the invariant manifold becomes 3-dimentional and assumes a more complicated, tubular “manifold” shape. However, unless we still restrict ourselves to a given origin, destination and originating state vector, there are an infinite number of “invariant manifolds” which pass through every origin point in space.

The invariant manifolds of the ITN are not, of themselves, unique structures in our solar system. They are the result of the choice we made for origin, destination and initial state vector.

By convention, ITN manifolds are chosen such that Lagrange (Libration) points are the origin and destination of each manifold section. These Lagrange (Libration) points act as interchanges between “turnpike” sections of the manifolds. Lagrange (Libration) points are chosen because

  1. They are close to massive bodies like planets, which are popular spacecraft destinations
  2. Nearby are (somewhat) stable halo orbits, where we can station spacecraft with little cost of station-keeping delta-v.
  3. These Lagrange (Liberation) points have low-energy transfers to other nearby manifolds which lead to yet more Lagrange (Libration) points near distant celestial bodies.

What are Lagrange (Libration) points? “Libration” means to wobble around, like tipsy sailors at liberty. Give them a little push and they totter off in a new direction. Lagrange points are locations near celestial bodies where a small delta-v makes a large change in direction of the state vector. This makes transfer between adjacent manifolds very low delta-v, and therefor economical on propellant.

Unfortunately, the ITN has little practical use for interplanetary spaceflight since effects between planets are small and trajectory durations are much too long for crewed flights. However, manifolds near planet/moon systems (Earth, Jupiter) have been utilized. See

ET come home. Is the inter-manifold transfer of JWST between Sun-Earth L2 and Earth-Moon L1 within the capability of existing propulsion systems? and

Can JWST come in for a pit stop? Is the fuel supply on JWST adequate for a transfer to an Earth Moon libration point to allow repairs?

$\endgroup$
10
  • $\begingroup$ "the manifold in question is a collection of contiguous state vectors with the same energy" Isn't it better to describe it as a collection of contiguous trajectories with the same energy? e.g. i.stack.imgur.com/FrJtf.jpg $\endgroup$
    – uhoh
    Dec 31, 2023 at 22:58
  • $\begingroup$ @uhoh ... true. I thought "state vectors" dovetailed better with the rest of the paragraph. Isn't time included in a "state vector" even though it is scalar? Two trajectories could be the same, but if separate in time, they could not be substituted for contiguous state vectors. Contiguous state vectors would need to be simultaneous. $\endgroup$
    – Woody
    Dec 31, 2023 at 23:28
  • $\begingroup$ These manifolds exist in the CR3BP which is a rotating frame, this means that the manifolds are static. They're rock-solid fixed and timeless "objects" mathematically at least. The manifolds represent a collection of trajectories, in other words they represent the paths an object would take if you put it somewhere on the trajectory moving in the direction of the trajectory with the energy for which the manifold was calculated. You can place the object moving on the trajectory any time you like, but the trajectory itself (a collection of state vectors) is timeless. At least that's how I see it. $\endgroup$
    – uhoh
    Dec 31, 2023 at 23:35
  • 1
    $\begingroup$ @uhoh ... Invariant manifolds are fixed in the CR3BP because of the restrictions the problem imposes on itself. Invariant manifolds also exist in unrestricted 3 body problems, and in 4 body problems. Like the real earth/moon system where orbits are ellipses rather than circles. Or the Sun/Earth/Mars system. These manifolds are constantly changing as the relative locations of the bodies change. $\endgroup$
    – Woody
    Dec 31, 2023 at 23:47
  • 1
    $\begingroup$ @uhoh ... "invariant" refers to a defined property (in this case energy) which is invariant when operated on by specified operations (in this case, laws of motion and gravitation). Your bank balance is invariant with respect to the phase of the moon, but not with respect to withdrawals. $\endgroup$
    – Woody
    Jan 1 at 2:04
0
$\begingroup$

If I understand it correctly, you are trying to calculate the time it will take to get to the destination and then figuring out when a rocket should fire its engines, aka launch window, to reach that destination.

Kepler's third law can be used to calculate the time it will take to get to the destination.

the squares of the orbital periods of the planets are directly proportional to the cubes of the semi-major axes of their orbits

This is the formula for the velocity of an object in a circular orbit:

$$v^2= \frac{G \cdot M}{r}$$

Where

  • r = radius of the orbit
  • G = Gravitational constant
  • v = velocity of the spaceship
  • M = Mass of the body it is orbiting (In your case the Sun)

To find out the velocity of a spacecraft in an elliptical orbit, one has to replace the value for r with r1 or r2

enter image description here

$$v^2_{periapsis}= 2 \cdot G \cdot M \cdot \left( \frac{1}{r1}- \frac{1}{r1+r2} \right)$$

$$v^2_{apoapsis}= 2 \cdot G \cdot M \cdot \left( \frac{1}{r2}- \frac{1}{r1+r2}\right)$$

The velocity a spacecraft travels in a circle is measured in how many times around the circumference it can go in a given time

$$v= \frac{circumference}{time}=\frac{2 \cdot pi \cdot r}{t}$$

If one places v in the formula for a circular orbit, one gets this: $$v^2= \frac{G \cdot M}{r}$$ $$ \left( \frac{2 \cdot pi \cdot r}{t}\right)^2= \frac{G \cdot M}{r}$$ $$\frac{4 \cdot pi^2 \cdot r^2}{t^2}= \frac{G \cdot M}{r}$$ $$\frac{4 \cdot pi^2 \cdot r^3}{t^2}= G \cdot M$$ $$t^2= \left( \frac{G \cdot M}{4 \cdot pi^2} \right) \cdot r^3 $$

Now to calculate the time it takes in an elliptical orbit one has to replace r with the semi-major axis. $$ semi\ major\ axis = \frac{r_1 + r_2}{2} $$

So one gets this for the time required to complete a whole orbit:

$$t^2= \left(\frac{G \cdot M}{4 \cdot pi^2}\right) \cdot \left(\frac{r_1 + r_2}{2}\right)^3 $$

If you want to calculate the time it takes to get to a lower orbit to a higher orbit where the spacecraft will pass the planet to do a fly by, one cannot use the value t. Instead one has to use the half of t, since t is the time needed to complete a whole orbit.

enter image description here

Now when you wanted to calculate when the spacecraft should fire it engines to go to the other planet, you have to calculate how long it needs to get to the other planet. In other words, how long until point T1-satellite to T2 satellite. Then you need to calculate how far the planet will travel in the time frame.

If the time from T1-Satellite to T2-Satellite and the time from T1-planet to T2-planet are the same, then the rocket should fire its engines since that is when there is a launch window.

$\endgroup$
3
  • 1
    $\begingroup$ Thanks for the attempt to answer, unfortunately I do not think this answer answers my question as there is nothing about the ITN. Nonetheless, thank you for your time. $\endgroup$
    – hi-bye125
    Dec 28, 2023 at 1:23
  • 1
    $\begingroup$ @uhoh I didn't know about those functions, great tip though! I edited the answer now. $\endgroup$ Dec 28, 2023 at 15:42
  • $\begingroup$ and v= \frac{\text{circumference}}{\text{time}}=\frac{2\pi r}{t} gives $$v= \frac{\text{circumference}}{\text{time}}=\frac{2 \pi r}{t}$$ $\endgroup$
    – uhoh
    Dec 29, 2023 at 4:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.