In this question, it's pointed out that the L3 Lagrange point, where a "true" Counter-Earth would lie, is in fact unstable, and over time any object there would drift into a different orbit.

Obviously this means that if we're going to put a habitable planet there, we need to give it some means of station-keeping, to adjust itself as it drifts away from the desired point; this is how the titular planet of the Gor series remains in its orbit.

My question is, how much energy would be required to keep a habitable world in Earth's L3 point? Assume a world that's roughly 85% the mass of Earth, though an answer where any mass could be plugged in easily would be stellar. And let's not worry ourselves with the means just yet (let alone questions of efficiency in converting a power source into thrust), and instead just focus on the energy output required: How much energy do we need to expend to maintain our orbit?

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    $\begingroup$ Tangential: If a planet were in Earth's L3 point, wouldn't Earth then be in Earth Prime's L3 point and, thus, suffer the same consequences? $\endgroup$
    – Frostfyre
    Commented Jul 29, 2015 at 16:55
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    $\begingroup$ If a planet is in the L3 point, it's not an L3 point anymore. $\endgroup$
    – Samuel
    Commented Jul 29, 2015 at 16:57
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    $\begingroup$ @Samuel . . . nor is it a planet anymore. Neither have cleared their orbits, technically. One other thing to worry about is that now Earth is at the L<sub>3<\sub> point of this Earth. $\endgroup$
    – HDE 226868
    Commented Jul 29, 2015 at 16:59
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    $\begingroup$ The L3 point is not in the location of a counter earth (in earth's orbit 180 degrees around) it is slightly farther out. $\endgroup$
    – Oldcat
    Commented Jul 29, 2015 at 17:29

2 Answers 2


The scenario you have proposed has more problems than just the L3 point's instability. You also have issues with mass, interaction with other planet's gravity, and the shape of Earth's orbit.

Starting with the mass problem: In order for a Counter-Earth to exist, it must have precisely the same mass as Earth does. This is because orbital equations mandate that orbital velocity is a factor of mass and distance from the gravitational body. Assuming a circular orbit to simplify things...

$$ v = \sqrt{\frac{G(m_1 + m_2)}{r}} $$

...where $m_1$ is the mass of the body in the center (i.e. the Sun) and $m_2$ the mass of the body orbiting it (i.e. the Earth or the Counter-Earth).

If you wish to maintain velocity with a smaller mass, you must move it closer to the body it is orbiting. The L3 point has forces that try to hold you at a similar orbital velocity, but they wouldn't be strong enough to hold an entire planet there. If your planet had a lower mass, and thus higher velocity than Earth, then it is going to have a wildly elliptical orbit with a different orbital period (meaning we'd have probably crashed into each other by now if we are in the same plane)

Orbital shape is also an issue. Since Earth sits in an elliptical orbit, in order to sit constantly in Earth's L3, Counter-Earth would need a matching elliptical orbit that is precisely the opposite of Earth's. Your energy needs for station keeping would vary depending on the time of year.

For a lower mass planet, with Earth's same orbital velocity, your challenge would be keeping the planet from sliding further out on Earth's orbital path because of its lower mass.

Now, on to the Lagrange Point issue. The L3 point is the spot at which Earth and the Sun's gravitational pulls line up, so there is no angular force tugging you off-station. Beyond that, it is your centrifugal force from orbital velocity that holds you on-station with respect to distance from the sun. With a mass smaller than the Earth, you would tend to drift further from the sun, so station keeping would mandate that you exert an force propelling you inward towards the sun. Further complications will be introduced by the orbits of other planets, particularly Venus, which will tug inwards.

Now...on to the math. Ignoring Venus for a moment, lets calculate the gravitational force generated by the Earth/Sun system that you are in the L3 point of. Note that I am also ignoring the in/out drift of Earth and Counter-Earth for this in the name of simplicity and am using the Average distance. If you want to calculate minimum and maximum energy requirements, swap out the distances for the Earth's aphelion distance and perihelion distance. (Remember to double the distance for the gravitational pull between Earth and Counter-Earth)

Force of Gravity can be calculated like so...

$$ Fg = \frac{G*m_1*m_2}{r^2} $$

So, to get total Force of Gravity for the system...

$$ F = \frac{G*m*1.989\times 10^{30}}{149,597,870,700^2} + \frac{G*m*5.972\times 10^{24}}{299,195,741,400^2} $$

Where $m$ is the mass of your planet and $G$ is the Gravitational Constant, $6.6726\times 10^{-11}$

Plugging in the '85% Earth Mass' value that you gave, we get a force of gravity equal to $3.0119\times 10^{22}\text{ N}$. In order for our planet to neither fly away, nor fall into the sun, we need for its centrifugal force (+ generated force) to be equal to the force of gravity.

So, calculating the natural centrifugal force of our planet, moving at Earth's orbital velocity renders this equation:

$$ F = \frac{mv^2}{r} $$

For the sake of simplicity, we are going to ignore the fact that the orbital center is not the center of the sun...it's a little bit off, but not by much. So, plugging in values gives us this finalized equation...

$$ F = \frac{m*30,000^2}{149597870700} $$

Again plugging in our 85% mass, we get a Centrifugal force of: $3.0539\times 10^{22}\text{ N}$

This leaves us having to make up for $4.2015\times 10^{20}\text{ N}$ of Force on a constant basis. These forces give the planet an outward acceleration of $.0000828\text{ m/s}^2$ (using $a=\frac{F}{m}$). Calculating for displacement using...

$$ x = vt+.5at^2 $$

Shows that our planet will attempt to slip off station by $.0000414\text{ m}$ every second. This is a quite small number, but we have to keep it perfectly balanced. The further we slip from the sun, the weaker the cumulative gravity of the sun and Earth is, so the faster we'll fall away. And if we fall away, our orbit will restabilize with a higher eccentricity and a different orbital period...likely ending in gravitational interactions with Earth that will either cause a collision or fling one of us out of our orbit (chaos theory, yay)

So, calculating our energy requirements, we convert over to joules.

$$ J = F * d $$

Plugging in values, we need a constant feed of $1.739\times 10^{16}$ joules (17.3 petajoules) per second ($1.739\times 10^{16}$ watts) to maintain our position.

This works out marvelously for a basic 3-body problem...as long as we are ignoring the effects that this other planet has on Earth...and every other planet in the Solar System. Venus would be the biggest troublemaker, applying anything between $2.421\times 10^{16}\text{ N}$ and $2.863\times 10^{18}\text{ N}$ and is usually not pulling in-line with the Earth/Sun arrangement. This is going to require further adjustments as well to keep it all lined up. And again, you can't drift in or out...if you do, your instability will grow exponentially. To roughly ballpark your energy demands, I'd suggest tacking on a few extra petajoules as your upper end, and understand that you are likely to have to vary your output and direction constantly. Also note that this is joules of kinetic energy...I'm ignoring energy waste here and assuming perfect conversion to kinetic.

You said not to worry about 'how,' but 1 petajoule is about equal to the biggest boom ever made by humans: The Tsar Bomba. We need that much energy every second.

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    $\begingroup$ Why not assume that the counter-Earth's orbit is also elliptical, but simply has perihelion and aphelion at opposite points around the Sun but at the same time? Also, Lissajous orbits might be needed, adding some more parameters to consider in the $z$-axis. $\endgroup$
    – HDE 226868
    Commented Jul 29, 2015 at 18:49
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    $\begingroup$ The problem is that then you are drifting in and out of your opposing planet's L3 point, which 1: defeats the request of this question and 2: Introduces exponential instability into the system due to the drift. Lissajous orbits may be useful, but their usefulness of a massive body, like a planet, has not really been studied. $\endgroup$ Commented Jul 29, 2015 at 19:00
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    $\begingroup$ Good point about Lissajous orbits, but regarding the eccentricity, isn't Earth already moving out of the counter-Earth's L$_3$ point, causing issues? $\endgroup$
    – HDE 226868
    Commented Jul 29, 2015 at 19:02
  • $\begingroup$ @HDE226868 huh. Now that I think about it...the L3 point of the system will drift relative to where Earth is in its elliptical orbit. So, actually, you are completely right. In order to stay in the L3 point, it would need to have a reverse Elliptical Orbit relative to ours. I'll tweak my answer to reflect this. Thanks! $\endgroup$ Commented Jul 29, 2015 at 19:03
  • $\begingroup$ I took the liberty of editing your orbital velocity equation to use the form that includes both bodies' masses, to make your point that a smaller body has to be moved closer to maintain the same orbital velocity. Feel free to revert if you feel it wasn't warranted, but it was the only quibble I could find in this excellent post -- not exactly the answer I wanted to hear, but from everything I can tell it's the right one! $\endgroup$
    – Kromey
    Commented Jul 30, 2015 at 16:48

As I understand this problem, the amount of energy needed would be dependent on the accuracy and responsiveness of your guidance system. If you keep it exactly on coarse then the amount of energy to keep it on course would be very small because the gravity pulling it in either direction would be equal. It's only once it deviates significantly that it becomes a massive job to move it back into position.



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