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So far I've designed two planets, and would like for them to orbit each other at a distance of roughly 1.25 million km (both planets would easily stay in the habitable zone), the smaller being tidally locked.

One has a radius of 6,371 km and a mass of $~4.776×10^{24} \ kg$ (the smaller, tidally locked planet), the other has a radius of 7,645 km and a mass of $~1.1002×10^{25} \ kg$ (the larger planet).

The estimated orbital period is 39 days 11 hours. The star is class KV, with 0.625 solar masses, and a luminosity of 0.16386 solar units.

I'm wondering what the tidal forces of each planet would be, if it would be possible for life to develop on the larger planet, and if the smaller planet would be habitable even though it would have a relative day/night length of almost 40 earth days. If any other factors are needed for clarification please tell me and I'll edit the post.

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  • $\begingroup$ Both planets seems plausible. Rocky density and razoable gravity aceleration. Depends of atmosphere and orbit around the star. The small one probably haven't one magnetosphere due slow rotation and will lose atmosphere quickly. $\endgroup$
    – Rodolfo Penteado
    Commented Mar 15, 2020 at 4:53
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    $\begingroup$ This is known to be possible. The Earth and moon is a real life example. $\endgroup$
    – Kilisi
    Commented Mar 15, 2020 at 5:29
  • $\begingroup$ Kilisi, while it is possible for the larger of the two planets to sustain life, the question was moreso centered on whether or not the smaller of the two, which would have a day/night cycle equal to that of its orbital period, would be habitable. As well as what the tides compared to earth would be. $\endgroup$
    – Rebel110
    Commented Mar 15, 2020 at 5:38
  • $\begingroup$ @Rebel110 I have another question. The smaller planet has a radius similar to Earth but lower mass, and the larger planet has roughly twice the mass of Earth. The size of the Hill Sphere of Earth is about 1.5 million kilometers, though I think that a moon will have a stable orbit only within the inner third of the Hill Sphere radius. The larger planet's Hill Sphere will have to be calculated from its mass and the mass and distance of its star. It is possible that the smaller planet won't have a stable orbit at 1.25 million km. $\endgroup$
    – M. A. Golding
    Commented Mar 15, 2020 at 20:26
  • $\begingroup$ @M. A. Golding I hadn't considered this fact, I've found, due to the calculator, that the two planets could orbit each other at roughly 3× the distance between earth and its moon. I didnt put the hill sphere into account, as I was greatly oversimplifying the orbital calculations. Thank you for mentioning this however. $\endgroup$
    – Rebel110
    Commented Mar 15, 2020 at 21:26

2 Answers 2

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What seems to be a bit off is the ratio of the masses of the two bodies. Earth is about 100 times more massive than Moon, while here the big one is just 10 times more massive than the little one.

This means that the co-orbiting around their center of mass would be more noticeable, the COM being at about 70% of their mutual distance, measuring from the center of the smaller planet: 880 thousand kilometers, outside of both bodies.

I am not sure that the above would allow tidal locking with the main star alone. With tidal locking out of the picture, developing life is "just" a matter of having the right conditions: temperature, water and so on.

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I found a calculator that suits my needs exactly! Neither body actually needs to be tidally locked. Tides on the major planet are 2.56 solar tide magnitude (1.4 m), and 1.24 lunar tide magnitude (1.3 m). For the smaller planet, the solar tide magnitude is equal, but the lunar tide magnitude increases to 1.29 (1.4 m). I am also sufficiently satisfied with the tidal forces that I will be removing the tidally locked stipulation from the smaller planet. Thanks everyone! Edit; link to the calculator :https://docs.google.com/spreadsheets/u/0/d/1uSjlohnk_dR_WNqFaqebrd2myOw8HrBMupr-5-WBWhU/htmlview

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    $\begingroup$ If you want to self answer your question, at least refer to the calculator you have found. $\endgroup$
    – L.Dutch ♦
    Commented Mar 15, 2020 at 6:40

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