The Reduced n-body Problem

To find the most fuel efficient route from Earth to Moon, we need to predict the future. Given some initial configuration of the planets and our spaceship, we want to know where we will end up. Let’s start with a simple example.

A Single Planet

Here, we have a single planet and life is easy. Let’s give ourselves some initial velocity.

And we get a nice circular orbit around our planet. If we now increase our initial velocity we get the usual Kepler orbits. First, we have the ellipse, then the parabola, and finally, the hyperbola.

Multiple Planets

Consider a slightly more complicated setup. Here, we have our starship in a three planet system. Let’s see what happens this time.

And right away, out trajectory looks a lot more complicated. Now, here comes the fun part. Let’s vary our initial velocity again.

The system is chaotic! Chaotic here means that a tiny variation in the initial velocity \(v\) dramatically affects where we end up over a big enough distance. This is a problem for us. In chaotic systems, it becomes basically impossible to predict where we will end up over a long period of time.

However, it is also possible for only certain regions of the system to be chaotic. For instance, given a high enough initial velocity in this example, we enter into a stable orbit around all three bodies. This then becomes a case where we can calculate accurate predictions far into the future.

Out goal then, will be to find where these regions of stable motion exist and exploit them to reach the moon.

From LEO to the Moon

Now back to our original problem.

Here, we have a few thousand spaceships all currently sitting in Low Earth Orbit (LEO). To get to the Moon, we must leave Low Earth Orbit. So let’s give ourselves a little push.

We get this ring of spaceships as they all fly away from Earth. And when this ring intersects with Moon, we get a possible trajectory for getting to Moon. This first intersection, which is the fastest route, is our traditional Hohmann transfer orbit. But we are moving too fast relative to Moon and must slow down which uses a lot of fuel. This is no good. Let’s run our simulation a little longer

Pay attention to the green trajectory. It flies all the way away from Earth to near the sphere of influence of Earth. Then it falls back in and is ballistically captured by Moon! There is no second burn here. This is interesting for us since this will save us a lot of fuel.

Let’s take a closer look at the trajectory. We start off in an elliptical orbit around Earth. But we don’t stick to it. Why not? Because of the Sun! The force from the Sun changes our orbit and lets us catch up to Moon. This is our low-energy ballistic capture trajectory. Now let’s try to explain where it comes from.

Lagrange Points

Let’s consider just the Earth and Moon for now which we place on the x-y plane. Each body exerts a gravitational influence so they both have a gravitational potential energy.

The gravitational potential energy is given by:

$$ U_g = -G \frac{m_E}{r_E} - G \frac{m_M}{r_M}. $$

where \(m_E\) and \(m_M\) are the masses and \(r_E\) and \(r_M\) are the distances to the Earth and Moon respectively (we are setting the mass of our small test object \(m = 1\) for simplicity). This is plotted as the blue surface.

The Moon is also orbiting around the Earth. But for us, it will be simpler to consider the Moon as stationary. To compensate for this, we must introduce fictitious forces.

Fictitious Forces

Newton’s first law of motion, also known as the law of inertia, states that every object will remain at rest or in uniform motion. However, this law only applies in what is called an inertial frame (or perhaps, inertial frames are to be defined as those frames in which the first law holds, but that is a topic for another time).

Galileo discovered that they are in fact an infinite number of inertial frames and that they are all mutually related by a simple set of transformations, called the Galilean transformations. These transformations include:

• Translations, i.e., moving the origin of the coordinate system by some distance.
• Velocity “boosts”, i.e., moving the origin of the coordinate system with some uniform velocity.

This means that once you have found one inertial frame, you have essentially found all of them!

Clearly, this means that some (most) frames are not inertial. One example is that of an accelerating car. When pressing down on the accelerator pedal, you can feel yourself being pushed back into your seat. Furthermore, if we imagine a small box on a frictionless surface in our car, it would also begin sliding backwards as the car moved forwards. Yet, there is no “real” force pushing on you, or the block. So Newton’s first law does not apply.

However, we can still continue to make predictions in a non-inertial frame if we start with an inertial frame and transform our equations of motion to that non-inertial frame. In the inertial frame, motion is described by Newton’s second law:

$$ \mathbf{F} = m\mathbf{a} = m \frac{\mathrm{d}^2 \mathbf{x}}{\mathrm{d} t^2}. $$

Performing the transformation \(\mathbf{x'}(t) = \mathbf{x}(t) - \frac{\mathbf{c}t^2}{2}\) where \(\mathbf{c}\) is the acceleration of our car, we get after some algebra:

$$ \mathbf{F} = m\frac{\mathrm{d}^2 \mathbf{x}}{\mathrm{d} t^2} = m(\frac{\mathrm{d}^2 \mathbf{x'}}{\mathrm{d} t^2} + \mathbf{c}). $$

Reorganising, we get:

$$ \mathbf{F} - m\mathbf{c} = m\frac{\mathrm{d}^2 \mathbf{x'}}{\mathrm{d} t^2}. $$

This is our new equation of motion in the non-inertial uniformly accelerating frame. In fact, it almost looks the same as our old one except for the new \(-m\mathbf{c}\) term. This extra term behaves just like a force except that it disappears in the inertial frame (where \(\mathbf{c}\) = 0). Thus, we call it a fictitious force \(\mathbf{F}_{fictitious} = -m\mathbf{c}\). This fictitious force is the one that pushes you into your seat when accelerating in a car.

Just like the uniformly accelerating car, a uniformly rotating frame with angular velocity \(\mathbf{\omega}\) is also a non-inertial frame. In this case, we can also calculate the fictitious forces and end up with:

$$ \mathbf{F}_{fictitious} = -2m \mathbf{\omega} \times \mathbf{v} - m \mathbf{\omega} \times (\mathbf{\omega} \times \mathbf{x}). $$

The first term of this equation is the Coriolis force and is responsible for large scale weather patterns such as hurricanes. The second term of this equation is the centrifugal force and is the force you feel when making a sharp turn around a corner.

Integrating the centrifugal force term, we get a “centrifugal potential”:

$$ U_c = -\frac{1}{2} \omega^2 r^2. $$

This is plotted as the purple surface.

Intuitively, this makes some sense since the further we are away from the axis of rotation, the more we are flung outward by the centrifugal force. If we think of this potential function as a “hill”, we experience more force the steeper the slope. We can now add these two potentials into an effective potential.

Our effective potential \(U_{eff} = U_g + U_c\) is plotted here in red.

What about the Coriolis force? The Coriolis force is always perpendicular to the motion, so it doesn’t do any work. Although the Coriolis force definitely affects the motion of our rocket, it does not enter into any energy considerations, so we can safely ignore it for now. Let’s draw a contour plot of our effective potential.

From this diagram, we see that there are certain points of equilibrium.

In fact, there are exactly five, which are called the Lagrange points Here, they are labelled \(L_1\) to \(L_5\). These points are special because if we were to sit there without moving, we could in principle stay there forever. But give a small nudge and we would fall right off.

But one of the quite surprising things about Lagrange points is that you can in fact orbit around them. Continue to the next slide to find out more.

Halo orbits

These orbits exist around the Lagrange points \(L_1\) and \(L_2\) and are called Halo orbits. They are unlike the usual orbits around celestial bodies because there is, in fact, no mass at all at the Lagrange point. We are in outer space after all! So how can we have a closed orbit around nothing?

Remember when we decided to transform into a frame where the Moon was stationary and introduce fictitious forces, in particular, the centrifugal and Coriolis forces? So far, we have only dealt with the centrifugal force by incorporating it into our effective potential. But as it turns out, it is the Coriolis force that is responsible for these weird Halo orbits.

Let’s start off a little to the side of the Lagrange point \(L_1\) with a velocity purely in the vertical direction. As we fall away from the Lagrange point, we are pushed sideways by the Coriolis force.

This gives us a bunch of different trajectories for slightly different velocities. To find the trajectory that results in a closed orbit, simply select the one that cross the horizontal again at a perpendicular angle. This one!

Now to get a full orbit, all we need to do is take the mirror image on the other side.

This is not the only Halo orbit. There are, in fact, an infinite family of Halo orbits for different distances away from the \(L_1\) point. We can find them by repeating the same method as we have described above.

This particular family of orbits are called the Lyapunov orbits. There are many other kinds of orbits around Lagrange points, but they extend up into the third dimension, such as the Lissajous orbits. Since we are only dealing with two dimensions, we’ll keep our lives simple but just considering these Halo orbits.

Phase space, stable and unstable manifolds

Here we have the potential function \(U\) as a function of spatial position \(x\) and the corresponding phase space.

Phase space

The “space” in phase space does not refer to our usual 3-dimensional space. Instead, it refers to the more abstract notion of the space of all possible states of the system.

A state of the system must contain everything we can possibly know about our system at a given moment. For example, a point particle needs three parameters to locate it in space. But that alone is not enough; we need three additional parameters for specifying the velocity. This gives us a total of 6 parameters per state or a 6 dimensional phase space.

For a more complicated example, say, a rigid cube, 6 parameters is not enough. We need an additional 3 parameters to describe the orientation and then another 3 parameters to describe its angular velocity. Hence for a rigid body, we need a total of 12 parameters or a 12 dimensional phase space.

In our case, we will just be considering point particles in 1 dimensional or 2 dimensional “real” space so we only need to be concerned about 2 or 4 dimensional phase space. In the former case, this is convenient because 2 dimensions can easily be visualised which we shall do here.

For a 2 dimensional phase space, it is useful to plot our this space in which we have our physical space “\(x\)” on the horizontal axis, and the associated velocity \(v\) on the vertical axis. Every state of our system can then be associated with a point on this diagram.

We want to use this formalism to gain insight into how motion evolves in this system. To do so, let’s shoot some rockets into the potential hill and trace their corresponding paths in phase space.

What we obtain is a phase-space diagram of our system. This diagram, much like the diagram of a fluid flow, shows where a test object located at a certain point in phase-space evolves with time. That means that an object located at some point on one of the green curves will follow that green curve as time goes on.

Furthermore, since the potential energy only depends on position, all the forces involved are conservative and hence total mechanical energy is conserved. Therefore, each curve in our phase space diagram is also a line of constant total energy. The diagram thus shows that all states with an energy under a certain amount fail to go over the bump, whereas those states with an energy higher than that certain amount do go over.

What if we had just the exactly right amount of energy? One possible state would be that of sitting exactly at the top of the hill with 0 velocity. On our phase-space diagram, this would correspond to the point right in the middle.

Manifolds!

Once we have identified our equilibrium point, we can draw the contour lines of constant energy that pass through this point. These lines are called the stable and unstable manifolds.

What do “stable” and “unstable” mean in this context? They don’t refer to stable and unstable equilibrium, since this point is clearly an unstable equilibrium. Instead, they are the curves that lead to and away from our equilibrium. That is, any point on the blue “stable” curve eventually tends to the equilibrium point (although it would take an infinite amount of time to reach it). Similarly, any point on the red “unstable” curve is where particles would go if they were ever so slightly pushed off from the equilibrium (or equivalently, one can imagine that it would take an “infinite” amount of time to fall off).

And not only do these stable and unstable manifolds tell us how to get to and get away from the equilibrium, they also tell us the general structure of motion in our system.

Any state that is located between the two manifolds do not have enough energy to reach the equilibrium and fall back down the same side. On the other hand, any state that is above or below the manifolds all have enough energy to pass over the top.

If we shoot a rocket that has almost, but not enough energy to go over the top, it will follow the stable manifold towards the equilibrium and then fall away along the unstable manifold. If we shoot a rocket that has enough, but just enough energy to go over the top, it will follow the stable manifold (but this time on the other side) and then fall away again along the unstable manifold in the other direction.

Multiple Manifolds

Let’s get back to our restricted 3-body problem. Here, we have the Sun-Earth system and our equilibrium point is the periodic orbit around the \(L_1\) Lagrange point.

In our previous example, the equilibrium was simply a point but now the equilibrium is an entire orbit? This probably deserves some justification.

Since the periodic orbit around the Lagrange point is unstable (meaning that a small perturbation would cause bodies in this orbit to fall off), we can create analogous definitions of stable and unstable manifolds, this time not for a point but a whole set of points!

We say that the stable manifold is the set of all points, which, when evolved in time, approaches a point in the periodic orbit. This means that a body on the stable manifold will spiral into the periodic orbit, albeit never completely reaching it. Likewise, the unstable manifold is the set of all points that will be reached when we fall off our periodic orbit.

This means we can have stable and unstable manifolds leading to and from this orbit. And these manifolds, which look like tubes, go all the way down to Earth. We therefore fire from Low Earth Orbit following the inside of the stable manifold tube and going back in with the unstable manifold tube.

Well, this tube is only for the Sun-Earth system. We can also consider the restricted Earth-Moon system, which has its own set of manifolds. And the stable manifold tube for the Moon reaches all the way out to intersect with the Earth stable manifold tube. We therefore hop manifolds to find our way to the Moon!

In fact we can think of the Solar System as a collection of many restricted-3-body systems, each with its own set of manifold tubes. We can use a similar method to travel to virtually anywhere else in the Solar System forming what is called the Interplanetary Transport Network.

Ballistic Capture

Going back to our ballistic capture trajectory, we see that this is indeed what happens.

We follow the stable manifold tube leading to (the periodic orbit around) the \(L_1\) Lagrange point and falling back via the unstable manifold. We are caught however by the stable manifold from the Earth-Moon system which lets us get captured by the Moon.

If this is such a great way of travelling in space, why do we still use Hohmann transfers? Well, the biggest reason is time. A Hohmann transfer to get to the Moon takes a few days. The low-energy trajectory takes a few months. A Hohmann transfer from Earth to Mars takes a few months. A low-energy trajectory could take over a thousand years!

This would clearly be unacceptable for a manned mission, not only due to the psychological challenges but also because of all the resources it would take for life support. However, the Interplanetary Transport Network is a great way for probes where it is not so important to get to the destination as fast as possible.

In fact, this has already been done quite a few times in real life, such as salvaging the lunar probe Hiten in 1991 by performing one such low energy transfer as we have seen. Other examples include the Genesis mission (2001 - 2004) or ESA’s SMART-1 (2003 - 2006) probe.