There seems to be a connection between orbital revolution (one year) and distance from the star the body orbits. For example, Earth orbits the sun at a distance of 93 million miles and completes one revolution every 365 days.

This alternate Earth has a longer calendar--372 days per revolution. That is pretty much 31 days on all of the twelve months of the modern calendar, including February. How far would Earth orbit the sun at this calendar? And would it pose any difference on how bright daylight is?

  • $\begingroup$ brightness would not differ much at least not perceivable to humans $\endgroup$ Sep 13 '17 at 21:59
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    $\begingroup$ Please explain why this isn't a basic math problem. Also, you can achieve the same number of "days' per revolution by speeding up the spin of the Earth. $\endgroup$
    – Aify
    Sep 13 '17 at 22:06
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    $\begingroup$ The year had about 390 days during the Carboniferous period (359 to 299 million years ago); yet the Earth orbited the Sun at the same distance as it does today. How come? Is it really a mystery, or does the braking effect of the tidal forces acting on Earth because of the Sun and the Moon have anything to do with it? $\endgroup$
    – AlexP
    Sep 13 '17 at 22:59
  • $\begingroup$ Yep. That's pretty much it. $\endgroup$
    – Phiteros
    Sep 13 '17 at 23:07
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    $\begingroup$ Follow the link (section Climate). The Earth used to rotate faster in the past; in particular, in the Carboniferous a day was only about 22.5 hours long. During the long geological time, the tidal forces exerted by the Sun and the Moon have slowed down its rotation. They still continue to slow it down. The second was defined at the beginning of the 20th century so that a mean solar day was exactly 24 * 60 * 60 = 86400 seconds long at 1-Jan-1900; but today, 100 years later, we need to insert one leap second every 18 months (on the average). $\endgroup$
    – AlexP
    Sep 13 '17 at 23:44

There is in fact a relationship between a planet's distance from its star and its orbital period. This is known as Kepler's Third Law. Mathematically, this can be formulated as:

$$ \frac{P^2}{a^3} = \frac{4\pi^2}{G(M+m)}\approx\frac{4\pi^2}{GM} $$

Here, $P$ is the period, $a$ is the semi-major axis (distance), $M$ is the star's mass, and $m$ is the planet's mass.

We can simplify this by using the mass in solar masses, distance in AU, and period in years. This gives us:

$$ M=\frac{a^3}{P^2} $$

Rearranging to solve for distance:

$$ a=(MP^2)^{\frac{1}{3}} $$

372 days is 1.019 years, and assuming the star your planet is orbiting is the same mass as the Sun, you get a distance of ~1.01 AU. That is 1% further away from the sun than the Earth is now. Light intensity falls off as $1/d^2$, so this would reduce the light your planet receives by ~2%. This might be enough to make your planet a few degrees colder, but if your planet has a slightly thicker atmosphere, it might not. All in all, I'd say the effect would be rather negligible.

Now, this is changing your orbital period by changing the distance. The other way to change your period is by simply changing the length of your day. If you make your days slightly shorter, your planet will complete more rotations in the time it takes to do one full revolution.

  • $\begingroup$ Beat me by 5 seconds. +1 for math. $\endgroup$
    – Chris M.
    Sep 13 '17 at 22:08
  • $\begingroup$ @ChrisM. Well, I did just finish teaching a lab about Kepler's Laws. $\endgroup$
    – Phiteros
    Sep 13 '17 at 22:10
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    $\begingroup$ It's probably worth noting that the semi-major axis isn't distance exactly -- rather, it is the mean value of the minimum and maximum distances (perihelion and aphelion, respectively). Compare the table in Events in Earth's orbit and Semi-major and semi-minor axes [of an] Ellipse, both on Wikipedia because I'm lazy. Earth's orbit is pretty circular (perihelion 0.9833 AU, aphelion 1.0167 AU) so this works out okay for us, but that's not the general case. $\endgroup$
    – user
    Oct 3 '17 at 8:28

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