Here's the background. I'd like the planetary year to be just under 800 days, with a longer number of hours per day-cycle. If the settlers live in the Mediterranean type climate area only and use greenhouses in addition to outdoor farming, could there be enough atmosphere and gravity to make scientific sense?

The planet was selected to allow human life. There is drinkable water, breathable air and enough gravity to make sense for long-term human occupation -- births and so on. It took over 90000 years to get there, so the settlement planet has to make sense. They were not pressured to land.

I can adjust the length of anything, but would still like my planet to be as different as possible and still make real sense from a hard-science pov.

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    $\begingroup$ Sure, just change the star. $\endgroup$
    – Frostfyre
    Commented Nov 30, 2016 at 13:36
  • $\begingroup$ could you explain 'change the star'? Distance, strength, both? $\endgroup$
    – WRX
    Commented Nov 30, 2016 at 13:38
  • $\begingroup$ As a star gets hotter, the habitability zone moves farther away. I'm not good with stellar physics, so someone else would need to do the math to figure out the numbers needed. $\endgroup$
    – Frostfyre
    Commented Nov 30, 2016 at 13:52
  • $\begingroup$ my problem IS with maths, so I understand. Everyone is giving thoughtful and helpful answers. I don't yet understand them, but I will. $\endgroup$
    – WRX
    Commented Nov 30, 2016 at 13:54
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    $\begingroup$ The Martian year is 687 days, and if Earth were magically transported to the same orbit, it might remain habitable. $\endgroup$
    – jamesqf
    Commented Nov 30, 2016 at 18:42

3 Answers 3


Having a longer orbital period of 800 days (assuming Earth days) and keeping a terrestrial climate will require you balance a number of variables. First, you have the generalised form of Kepler's law:

$$p^2 = \frac{4\pi^2 r^3}{G M}$$

Where $p$ is the period, $r$ is the distance from the centre of gravity, $G$ is the gravitational constant, and to simplify from $M_1 + M_2$, I will use $M$ because planetary masses are negligible compared to stars. Basically, the period is itself dependant on the orbital radius.

Then, I have the equation for extraterrestrial solar irradiance at different radii, holding the Earth value of 1367 $\textrm{Wm}^{-2}$ constant.

$$1367 = \frac{L}{4 \pi r^2}$$

Where $L$ is the luminosity of the light source. This is what is important for keeping the climate terrestrial. Now, because I can relate luminosity to mass, I can then plug the following equation into this one. I assume a mass-luminosity relation with the general approximate value for low-mass main sequence stars of around 4:

$$\frac{L}{L_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^{4}$$

This is where $L_{\odot}$ is the Sun's luminosity, $M_{\odot}$ is the Sun's mass, and $M$ is the mass of this hypothetical star. Substituting all of these equations together, starting with the solar radiation equation, then substituting the mass-luminosity relation for $L$ and then Kepler's law, solved for $r$, I get:

$$1367 = \frac{\left(\frac{M}{M_{\odot}}\right)^{4} L_{\odot}}{4\pi \left(\sqrt[3]{\frac{p^2 G M}{4\pi ^2}}\right)^2}$$

I then set $p$ seconds to the appropriate value for your 800 day year and come up with some mass value $M$ necessary for that star. Using Wolfram Alpha (because I really don't want to solve that by hand), with this query,

1367 = {(\frac{M }{1.988e30})^{4} * 3.848e26}/{4 * \pi * ({(6.912e+7)^2 * 6.67e-11 * M}/{4 * \pi^2})^(2/3)}, solve for M

I get a value for the mass around $2.719 \cdot 10^{30}$ kg, or around 135% the mass of the Sun. The distance from the star, therefore, would be somewhere around (using the expression substituted for $r$ and this query string),

(((6.912e+7)^2 * 6.67e-11 *(2.719e+30) )/ (4pi^2))^(1/3)

1.87 astronomical units from the star, which would be very much like our Sun, just brighter and a third more massive. Thus, because the orbital parameters fit and such a star has very similar properties to the Sun, life can certainly exist.

Note: For those who are familiar with Kepler's law, the generalised Newtonian form of the equation shows that the mass of the central object is inversely proportional with the square of the period, which is why the doubled period has a not-so-large effect on the mass of the central object. This means that the increased time leads to a higher mass. I wanted to constrain the distance variable by providing some energy per square metre requirements, which therefore, could also have been solved by substitution of $r$ from the second equation here.

  • $\begingroup$ wow -- phew! Way over my head! I suppose I should edit the question, but the answers continue to be so interesting. I decided that the planet had a more Earth-normal year and day length. Not identical, but close enough. My settlers are born there, so it is not an issue for them. $\endgroup$
    – WRX
    Commented Dec 28, 2016 at 18:13
  • 1
    $\begingroup$ It's because there are mathematical relationships between pretty much all of these variables, many of which are overlapping. Luminosity is dependant on mass. So is orbital period and radius, which also partly determine the solar irradiance. Because everything is dependent on each other, I have to substitute so many different equations. $\endgroup$
    – ifly6
    Commented Dec 28, 2016 at 18:52
  • $\begingroup$ 1.87 AU is also pretty close to what we know as Mars. Its 1.52 AU semi-major axis and 687 Earth days orbital period are pretty close to what the doctor ordered. $\endgroup$
    – user
    Commented Dec 28, 2016 at 19:02
  • $\begingroup$ The star, whose mass is calculated to be 1.35 solar masses, would output 3.3 times the energy output of the Sun, per the mass-luminosity relation above, and therefore, allow for a terrestrial climate and not one like that of Mars which is cold enough to support actual frozen carbon dioxide. $\endgroup$
    – ifly6
    Commented Dec 28, 2016 at 19:13


Of course that would change some things and needs to be taken into account when designing life that could live there.


Has analysis on the rotation periods. Radius of the planet would be the best variable to achieve what you are looking for.

  • $\begingroup$ I am not math smart. My settlers do not have to understand the math or even know how large the planet is, but you are saying that it is possible? I just need it to make sense but do not need to explain it to the reader. I can make the radius be as big or as small as necessary, but do not have a clue how to figure that out. If you think I do have to understand, I could bother my maths-genius hubby, but you know how that goes -- this isn't his bailiwick. $\endgroup$
    – WRX
    Commented Nov 30, 2016 at 13:49
  • $\begingroup$ Yes, it would probably make the planet also stormier. $\endgroup$ Commented Nov 30, 2016 at 13:56

To change the ammout of time in a planets year one must adjust the height of the orbit via keplers law. If you want that planet to remain in the green zone of a star you would have to increce the heat output of the star. Changing the heat/size would change the colour of the star which would change the evolution of flora and fauna to maximise use of this diffrent light.


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