How long does it take to get to the Sun/Earth L4 Point from Earth?

Adam Hibberd

Let’s say we eject a spacecraft with a small velocity decrement in the opposite direction to Earth’s own velocity around the Sun, what will happen to it exactly? It will stay around Earth’s heliocentric orbit since its velocity is close to Earth’s, however we shall assume that the spacecraft flies outside of Earth’s theoretical gravitational sphere of Influence (SoI) and also that it is only affected by the Sun’s gravitational field.

Well the first thing to observe is that the body will gradually drift IN ADVANCE of Earth’s orbit around the Sun. That may seem counter-intuitive but look at it this way. By having a lower velocity relative to the Sun than Earth, that means it will reach a perihelion (closest point to the Sun) slightly lower than Earth’s circular orbit and so the time period, T_C of the craft will be shorter than the time period of Earth’s orbit, T_E (1 year). Since the time-period of the spacecraft’s orbit is shorter, then that means it will gradually get ahead of Earth. The reasoning is that for every 1 period the spacecraft takes, the Earth will be slightly behind it since its period is longer.

So let’s assume that the time period of the spacecraft is indeed lower than that of Earth’s and allow the spacecraft to creep ahead of Earth until it reaches the so-called Sun/Earth L₄ point. For the uninitiated, that sits at exactly 60° (π/3 radians) in advance of the Earth along Earth’s orbit around the Sun. How long would that take exactly?

The answer to this turns out to depend on the velocity with which the object leaves the Earth’s SoI, which we shall equate here to the object’s hyperbolic excess speed relative to Earth.

I have done precisely this research and I generated the following two plots.

In these plots, the red-dashed line represents the 2-body problem we are addressing in this blog post, and for information the dark blue solid line indicates the 3-body model, with the Earth included.

When you look at the 2-body model in the first of these plots you can observe some ridges which are even more evident in the second plot. These turn out to be harmonics. They are also present to a lesser extent in the 3-body simulation.

Let’s look at what is happening a little more deeply.

The synodic period between the craft and Earth,T_S, is given by the following expression:

If we wish to reach a point at θ in advance of the Earth in time t, then it follows that approximately:

Let us say we wish to reach the S/E L4 point, thus θ=π/3 , and so:

Furthermore we wish elapsed time to be some integer multiple, n, of the craft’s time period T_C, whence:

Inserting this into (1) we get:

Now rearranging to get T_C , we find:

$latex \displaystyle T_C = T_E\frac{6n}{6n-1}$

Thus:

From (3) & (4), this leads to a time required to reach L₄ as:

Let us now determine the theoretical hyperbolic excess speed, V_∞, needed at Earth to allow a passage to the L₄. We know the time-period of the orbit of the spacecraft, T_C, how does that translate to, V_∞ ?

First we find from Kepler’s third law that, given the ratio T_C/T_E, then the ratio of semi-major axes, a_c/a_E , is given by:

The heliocentric velocity, V_C , of the spacecraft when it returns to 1 au is described in the following equation from the well-known orbital energy relationship:

The hyperbolic excess needed at Earth is then:

Using equation (8) and with the identity, T =2 π √(a³/μ) , then we can restate equation (10) in terms of time periods as follows:

The L₅ Point.

For this we note that the RHS of (1) & (5) must be changed in sign, like so:

We then follow a similar line of reasoning to arrive at:

Also note that equation (10) now changes to:

From which we eventually obtain: