The "Lagrange giant" in the title refers to a gas giant that orbits its star in the same distance as Earth does. The alternative description is "Trojan planet" or, to be most clear, "co-orbital". So in this alternate universe, the only difference in our solar system is that a gas giant (let's start with Jovian in size and Class I in Sudarsky's classification, as those are the most familiar factors) orbits the sun at a distance of 93 million miles, meaning that it must share its orbit with Earth.

How to do this without turning Earth into a moon? Physicist Sean Raymond proposed that such planets sharing that same orbital plane must be separated by an angular distance of 60 degrees. At such a distance, can anyone standing on the planet Earth even see this Trojan co-orbital giant? If yes, how would it look in the sky?

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    $\begingroup$ So the planet would be sitting at an L4 or L5 LaGrange point? en.wikipedia.org/wiki/Lagrange_point $\endgroup$
    – DWKraus
    Commented Mar 28, 2021 at 4:17
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    $\begingroup$ @DWKraus Does it matter which point it's in? $\endgroup$ Commented Mar 28, 2021 at 4:21
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    $\begingroup$ You can see Venus with the naked eye and it is even farther away but far smaller. It's that star that is always near the moon to the lower right, even when you can't see any other stars in the sky because it is too bright. $\endgroup$
    – DKNguyen
    Commented Mar 28, 2021 at 4:21
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    $\begingroup$ @JohnWDailey The 1,2, & 3 Lagrange points are wildly different than the 4 & 5. From the question, I assumed the 4 & 5. I just wanted to be sure. L3 would never see the gas giant. L4 & L5 make the most sense for trojan planets. $\endgroup$
    – DWKraus
    Commented Mar 28, 2021 at 4:25
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    $\begingroup$ @DKNguyen: Also Mercury, given good viewing conditions, and it's even further away, a good bit smaller. and isn't covered with highly reflective clouds. $\endgroup$
    – jamesqf
    Commented Mar 28, 2021 at 16:58

5 Answers 5


VERY visible

The giant planet would be at the L4 or L5 Lagrange "trojan" point of Earth around the sun.

Or, more accurately, Earth would be in the giant planet's L4 or L5 Trojan point.

For a L4 or L5 Trojan point to be stable, the mass ratio of the primary(the sun) to the Giant needs to be at least 25:1, and the mass ratio of the Giant planet to its Trojan companion needs to also be at least 25:1
Math here: https://wmap.gsfc.nasa.gov/media/ContentMedia/lagrange.pdf
For a Jupiter size giant planet, with Earth in the trojan point, these mass ratios are well satisfied.
The orbit will be stable, even in astronomical timescales. The Earth's moon would also be allowed in a similar configuration as present, although that orbit will be only somewhat stable, and will lead to loss of the Moon much sooner than for the current configuration.

Appearance: The giant will be visible at 60 degrees from the sun. Either leading before dawn (if Earth is at L5) or trailing after sunset (if Earth is at the L4 position).
Jupiter will display as a perfect half-circle gibbous orb, pointing towards the sun.
It will have an apparent diameter of 0.053476 degrees, almost exactly 1/10th as wide as the Moon.
Assuming that being nearer the Sun does not change its appearance, it will have an apparent brightness of 0.73% of the full Moon, or some 75 times brighter than Venus at its best.
It will definitely be visible even in full daylight, as a small pastel semi-circle.

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    $\begingroup$ How'd you get 5000x brighter than Venus? I'm reasonably happy with my calculations (which were originally 100x, and are now 25x brighter than Venus) but I couldn't say for certain that I'm right and you're wrong. $\endgroup$ Commented Mar 28, 2021 at 11:27
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    $\begingroup$ @StarfishPrime Did it the cheap way. compared to half moon. adjust for albedo, adjust for apparent diameter. Basically 1/100th the apparent surface of a half moon, 3.2-ish times the albedo. Half moon as 1/6th brightness of full moon. Seems i believed an incorrect figure for venus vs. fullmoon, factor 100 error. (thanx, would not have found if you didn't poke me) $\endgroup$
    – PcMan
    Commented Mar 28, 2021 at 11:57
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    $\begingroup$ Mystery solved! $\endgroup$ Commented Mar 28, 2021 at 12:14
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    $\begingroup$ I don't think it would be a perfect half-circle. That would only be the case if the line of sight to the giant planet and the line connecting the sun to the giant planet made a 90° angle. In this scenario, they would make a 60° angle, and you'd get a "gibbous" appearance instead. This would also affect the brightness somewhat. $\endgroup$ Commented Mar 28, 2021 at 13:43
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    $\begingroup$ @MichaelSeifert I was thinking the same thing...good point about it affecting the brightness, too. $\endgroup$
    – Qami
    Commented Mar 28, 2021 at 13:56

These are the Lagrangian points of the Earth-Sun system.

enter image description here

L1, L2, and L3 are on the line through the centers of the two large bodies, while L4 and L5 each act as the third vertex of an equilateral triangle formed with the centers of the two large bodies.

The L4 and L5 points are stable gravity wells and have a tendency to pull objects into them. The points L1, L2, and L3 are positions of unstable equilibrium. Any object orbiting at L1, L2, or L3 will tend to fall out of orbit.

The L4 and L5 points are stable provided that the mass of the primary body (e.g. the Earth) is at least 25 times the mass of the secondary body (e.g. the Moon)

L4 and L5, for an observer on the terminator, will be 60 degrees above the horizon, and will therefore be visible for about 4 hours after sunset or 4 hours before sunrise.

Being the body a giant, it will surely reflect enough light to be visible.

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    $\begingroup$ I'm curious. Since 2 bodies at Earth and L4/L5 are interchangeable, a body 25 times less massive than earth would also be stable? But not one comparable in mass? $\endgroup$
    – Stilez
    Commented Mar 28, 2021 at 10:12
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    $\begingroup$ @Stilez, yes. with similar masses the mess starts. See Theia. $\endgroup$
    – L.Dutch
    Commented Mar 28, 2021 at 10:14
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    $\begingroup$ @Stilez actual math behind it.. detail here.wmap.gsfc.nasa.gov/media/ContentMedia/lagrange.pdf With large enough mass ratios, and no other external influences, L4 and L5 are stable even on astronomical timescales. $\endgroup$
    – PcMan
    Commented Mar 28, 2021 at 12:27
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    $\begingroup$ @Stilez: Just to be clear, L4 & L5 are only equilibrium points if the body there is much smaller than either of the two primary bodies. They are stable equilbrium points for this small body if the two primary bodies have a mass ratio of about 25x or greater. If you put a body with 1/25 of Earth's mass at the Earth-Sun L4 point, it's not clear to me that this would be stable in the long term; 4% of Earth's mass might not be sufficiently small for the approximations that are made to calculate L4/L5's positions. $\endgroup$ Commented Mar 28, 2021 at 13:39
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    $\begingroup$ Colour me impressed - I really enjoyed that paper. Thank you! $\endgroup$
    – Stilez
    Commented Mar 28, 2021 at 13:39

I think the title of trojan planet would go to the smaller body in this case.

The L4 and L5 points sit at the apices of conveniently equilateral triangles, assuming the orbit in question is basically circular. That means that if Earth were a trojan, the body it co-orbits with would be ~1AU away from us (or about 150 million kilometres). If the body were Jupiter sized (ie. ~143000km diameter) its apparent angular diameter would be about 3'17" of arc... that's about a tenth of the diameter of the moon or sun as seen from Earth, and much larger than any of the other planets. Given that the average human eye (whatever that is) can a resolution of approximately one arc-minute, the planet being three times larger than that should be obviously a small round blob in the sky, not just a bright point of light. (I did originally say "circle" here, but it'll be only a partial circle on account of being at an angle to the sun... it may or may not appear to be a circle to the naked eye, but even with low-powered binoculars or other assistance it'll be obviously non-circular. It'll always have the same size and shape, regardless of time of day or year)

For a rough scale example, here's a picture of the moon with an approximately 3 arc-minute diameter circle drawn on the Mare Serenetatis, which is about 6 arc-minutes across and you can pop outside the next time you have a clear night with a full moon and see the relative size for yourself.

3 arc-minute circle drawn on the mare serenetatis

Jupiter's absolute magnitude $H$ can be calculated from its diameter and its geometric albedo (0.538), giving approximately -9.5. The apparent magnitude of the same planet in Earth's orbit but 60 degrees away would be:

$$m = H + 5\log_{10}\left({D_{BS}D_{BO} \over D_0^2}\right) - 2.5\log_{10}\left(q(\alpha)\right)$$

where $H$ is the absolute magnitude, $D_{BS}$ is the distance from the body to the sun, $D_{BO}$ is the distance from the body to the observer, $D_0$ is the distance between Earth and the Sun and $q(\alpha)$ is something called the phase integral which for a diffuse reflecting sphere (which is a reasonable model for a planet) at 60° is about .406, and represents the portion of sunlight scattered from the planet that is bounced towards us. All of the distances are conveniently 1AU, giving the gas giant an apparent magnitude of -8.5. That's bright, by the way... brighter than any other star or planet in the sky and exceeded only by the Moon and the Sun. It would be 25x brighter than Venus at its brightest (and Venus can be seen by the naked eyes at dawn and dusk), 10x brighter than the ISS, and equivalent to the brightest iridium flares. You can't reliably see things that are as bright as those flares these days now that the original iridium satellites have been deorbited, but they were apparently possible to see during daylight hours, even outside of the normal Venus-viewing timescale.

  • $\begingroup$ The shape (if you can discern it with the naked eye) should be like a gibbous moon, though a little more towards the half than towards full. The moon on day 9 or 10 after a new moon is about the right shape. $\endgroup$
    – Blckknght
    Commented Mar 29, 2021 at 8:29
  • $\begingroup$ @Blckknght i suspect you wouldn't be able to discern it with the naked eye... it should be obviously not a point source like all the other stars and planets, but it would still be quite small and comparatively bright. $\endgroup$ Commented Mar 29, 2021 at 8:47

Yes, a Jupiter sized planet in the same orbit as Earth would definitley be visible. The planet Jupiter is visible to the unaided eye in the night Sky even when about 400 million to 600 million mils from Earth, so if it is moved to about 93 million miles from Earth it will be much brighter and more visiple.

If the solar system has a much dimmer star, the analogs of Earth and Jupiter would have to share an orbit which is much closer to the star than in our solar system for the analog of Earth to be warm enough.

Because the Moon's orbit is elliptical and it gets closer to and farther from Earth, the angular diameter of the Moon as seen from Earth varies between 29.3 and 34.1 arcminutes. Since there are 60 arcminutes in a degree of arc, the angular diameter of the Moon as seen from Earth is about half a degree.

Jupiter has a equatorial radius of 71,492 kilometers, and thus an equtorial diameter of 142,984 kilometers. To have an angular diameter of 0.5 degrees, Jupiter would have to be at such a distance that the circumference of the circle around the observation point would be 720 times 142,984 kilometers, or 102,948,480 kilometers. So the radius of that circle would be 16,384,773.32 kilometers.

So the Jupiter analog and the Earth analog would orbit the Sun analog at a distance of 16,384,773.32 kilometers, and they would be spaced 16,384,773.32 kilometers apart along the orbit. 16,384,773.32 kilometers is about 0.109524447 Astronomical Units (AU), or almost 11 percent of the distance between Earth and the Sun.

I note that the potentially habitable exoplanet Gliese 180 b orbits the red dwarf star Gliese 180 in its circumstellar habitable zone at a distance of 0.103 AU with a period of about 17.38 days. Gliese 180 is a specral type M2V or M3V star.

And a planet orbiting its star that closely would become tidally locked to that star, having its rotation slowed until one side always faced the star and the other side always faced away. That might make life impossible on the planet. On the other hand a sufficiently dense atmosphere and oceans might transport heat from the day side to the night side of the planet. And the gravity of the nearby Jupiter analog planet might prevent the Earth Analog planet from becoming tidally locked to its star.

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    $\begingroup$ This does not answer the question. $\endgroup$ Commented Mar 28, 2021 at 18:21
  • $\begingroup$ @Draft I naturally assumed that the "Earth" would be in the L4 or L5 point relative to the "Jupiter" since the other Lagrangian points are too unstable. $\endgroup$ Commented Mar 29, 2021 at 13:17
  • $\begingroup$ @JOhnWDailey Yes it answers the question, jus tlike several other answers do. I pointed out that Jupiter is usually visible despite being a few times farther away than it would be in your scenario. I also pointed out a way to make "Jupiter" appear larger and brighter in the sky of "Earth", by moving both planets closer to the "Sun", and also pointed out aproblem, that the "Earth" might become tidally locked if too close to the "Sun". $\endgroup$ Commented Mar 29, 2021 at 13:23

Suppose Planet X is located at the Earth-Sun L3 point - would it be hidden from us all the time? No. Since Earth's orbit is elliptical, Planet X would be visible shortly after Earth passes its perigee. This problem is found in Goldstein's Classical Mechanics.

  • $\begingroup$ Whilst the question is not particularly focused as we like them to be, it does ask how X would look from Earth, adding that detail would complete your answer. We invite you to take our tour and when you have the time, read-up in the help center about how we work, welcome to worldbuilding. $\endgroup$ Commented Mar 29, 2021 at 22:18

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