A Milankovitch cycle is defined in Universe Today as

a cyclical movement related to the Earth’s orbit around the Sun.

There are three elements to a Milankovitch cycle that affects the amount of solar heat and with it, the Earth's climate:

  • Eccentricity (Orbital Shape)--The elliptical shape of the Earth's orbit varies from 0.000055 to 0.0679 over a 100,000 year period. Eccentricity affects the lengths of a season.
  • Obliquity (Axial Tilt)--The Earth's axis shifts from 22.1 to 24.5 degrees over a 41,000 year period. Obliquity affects the amount of heat gathered on either hemisphere.
  • Precession (Axial Wobbling)--Currently, Polaris is the North Star, but it hadn't always been the case. In 3000 BCE, Thuban was the North Star. in 14,000 CE, Vega will be the North Star. Each wobble has a 26,000-year cycle. The tidal relationship between Earth and the moon plays a part in precession. Also...

Precession as well as tilting are the reasons why regions near and at the poles experience very long nights and very long days at certain times of the year. For example, in Norway, the Sun never completely descends beneath the horizon between late May to late July.

In an alternate universe, orbiting a yellow G-type main-sequence star are not four Inner Planets, but eleven.

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Note that the farthest of TRAPPIST-1's planets, TRAPPIST-1h, is only 0.06 AUs (5,577,348.436046 miles) from the star. By comparison, the closest planet, Mercury, orbits the sun from a distance of 0.387 AUs (35.98 million miles). So it seems ideal to mesh the two systems without problem.

In this same universe, one natural satellite still orbits Earth, except that it is bigger--3,273 miles wide and a mass of 148,148,148,148,148,148,148.15 tons, as big as Ganymede--and orbits the planet from a distance of almost 475,000 miles.

How will the changes listed above affect Earth's nightscape and its Milankovitch cycles?

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    $\begingroup$ That is an incredibly specific mass for the satellite. $\endgroup$ Feb 27, 2017 at 20:14
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    $\begingroup$ I suspect that the inner planets would simply get vaporized if they were that close to the sun. Mercury gets to 430 degrees C on the daylight surface; the furthest of these planets is 1/6th the distance and would therefore get something like 36 times the incoming radiation. It would evaporate in short order. $\endgroup$ Feb 28, 2017 at 14:18
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    $\begingroup$ seems that since the only real difference ( once the inner planets vaporize in a few minutes) would be the mass and distance of Earths' new , hypothetical moon... are there other differences in consideration? $\endgroup$
    – Joe
    Feb 28, 2017 at 18:31
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    $\begingroup$ @JohnWDailey you need to learn to use significant digits. You started with 81, not 81.0000000000000 exactly. The magnitude of this approximation needs to be preserved after you divide it. $\endgroup$
    – JDługosz
    Mar 6, 2017 at 19:35
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    $\begingroup$ @JohnWDailey how precise a value did you get for earth's mass? Let the moon's mass have identical precision. $\endgroup$ Oct 15, 2017 at 6:22

2 Answers 2


You are missing at least one important feature. Size. "Oh Be A Fine Girl, Kiss Me" is one mnemonic for main sequence stars. O stars are the largest. Then comes B, A, F, G (our star), K, and M. The Trappist star is an "ultra-cool dwarf." Dwarf stars are smaller than M. There is a nice Wikipedia article: https://en.wikipedia.org/wiki/Star

You also need to consider if you have a binary or trinary system. However, working out reasonable orbits for even a binary is well past my abilities.

Stars change size with age: In about 5 billion years (give or take a few million), our star will run out of hydrogen to burn. It will then expand. Mercury, and Venus will be vaporized and Earth might suffer the same fate. Even if Earth survives, everything on it will not.

While Earth orbit is well within a class O star, it would be possible to move planets to get them into the Goldilocks zone for a particular star. The sky will be influenced by the gaseous composition of the atmosphere and the color of the star.

  • $\begingroup$ This does not answer the question. $\endgroup$ Mar 4, 2017 at 4:54
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    $\begingroup$ 1) If you restrict your answer to a G start, then the information about Trappist is irrelevant. 2) You have already answered your own question. Based on atmosphere, eccentricity, obliquity, and precession the ice ages can be anything you want. The bigger problem will be the orbital mechanics to fit 11 planets in stable orbits within the Goldilocks zone. We can currently work out a 3-body problem with difficulty en.wikipedia.org/wiki/Orbit. You may be in line for a Nobel Prize if you can work out the equations for 11 bodies. Strictly speaking, your question was unanswerable. $\endgroup$ Mar 5, 2017 at 14:55
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    $\begingroup$ @JohnWDailey "This sentence is false." Is the previous statement true or false? If you think that question has an answer then it is you who is being hubristic. $\endgroup$
    – AngelPray
    Mar 6, 2017 at 7:15
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    $\begingroup$ This answer is perfectly fine under the reality-check tag. $\endgroup$
    – JDługosz
    Mar 6, 2017 at 19:38
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    $\begingroup$ @F1Krazy who is OP? Determining the length or severity of ice ages is not possible knowing only that there are 11 planets. At the very least the mass of the star and planets will be needed along with orbital velocity. Also, the planets are not unchanging objects. The moon is moving away from the Earth, and the Earth rotation is slowing down. forbes.com/sites/brucedorminey/2017/01/31/…. If you think that planetary and solar mass is not relevant then I don't know where to start. How could the type of star not be relevant? $\endgroup$ Feb 19, 2019 at 4:40

The answer is not any more complex than an identical question involving only a change in the Earths' moons' changing to a) more massive and b) more distant.

The insertion of Trappist-1 worlds as "Inner Planets" is Moot. They simply could not persist as solid bodies for more than a few minutes around our own sun at the distances stated. Gasses, whether atmospheric, or as a super dense mantle as our own gas giants have, would boil off almost instantly. Rocky, denser worlds would last longer, but as stated above, with more than 36x the energy that Mercury receives, a solid guess of a few minutes seems reasonable.

A no point would any of this affect any part of Earths' pre-existing Climate.

As far as your actual Milankovich Cycle question, it is limited to the changes that would occur with the difference in gravitational force, exerted on the Earth. Our own Moon exerts a force of N= 1.982E+20 whereas the new satellite would have a reduced force of N= 1.01E+20. While no change is without a "ripple effect", this small change in lunar gravitation would not affect the eccentricity, axial tilt or precession of Earth, and would have zero effect on the amount of solar energy at Earth's surface. Despite the snazzy graphic attached, there would likely be zero changes to Earths' Milankovitch cycle.

  • $\begingroup$ not a clever dude, but how would the increased mass and retained velocity of the earth-lunar system not affect it's orbit around the sun? //well, ok, the orbital duration is a stated parameter, nvm. $\endgroup$
    – Giu Piete
    Mar 13, 2019 at 7:53

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