There are many questions on WBSE that cover most aspects of building a planetary system. (some examples below):

Creating a realistic world(s) map - planetary systems

How many planets should I have in my planetary system?

What distances would be involved in this planetary system?

But it occurred to me that while a sci-fi story setting might need the complete details of the system, a story set in a medieval or any similar society with pre-telescope or primitive telescope technology would only need to build the parts of the system that would be observable by the inhabitants of participants of the story, or the well known and widespread knowledge of the society that those participants are members of.

So ... How far away (maximum distance) can humans detect planets with the naked eye?

Assume best observable conditions for naturally occurring planets. As far as I can tell from my limited research, this should limit the albedo of the observed planet to about .8 (unless you can give a reasonable explanation of why it should be more or less than that), and the radius of the observed planet should be no more than about that of Jupiter (again, unless a reasonable reason for an exception is given). Assume ideal observing conditions, such as no light pollution, good (perfect specimen human) eyesight, ideal alignment of star and observing planet and observed planet for best lighting of observed planet, etc. Also assume a sun-like star, and earth-like planet as far as atmosphere and other observation characteristics, though human habitability is not a requirement except as it directly relates to human-like observation (a fictional atmosphere is allowed, as long as there is an explanation of how it improves observation, while not entirely preventing complex life in general). Earth-like orbit is NOT required for either planet.

This question has applications not only for general world-building, but can also be used as a basis for calculating orbit times which then apply to things like creating mythologoes, calendars, religious influences, cultural iconography, and much more.

  • $\begingroup$ Are multiple start systems allowed? Any size of planet up to the brown dwarf limit (i.e. about 13 Jupiter masses?) $\endgroup$ – Rafael Dec 18 '18 at 21:50
  • $\begingroup$ @Rafael Yes, but in order to show that it would be better observation than what's possible in a system with a single star, the observation characteristics of the single-star system would also be needed, for comparison. $\endgroup$ – Dalila Dec 18 '18 at 21:52
  • $\begingroup$ Do they have to know that the bright point in the sky is not a star, but a planet? $\endgroup$ – DarthDonut Dec 19 '18 at 8:23
  • $\begingroup$ @DarthDonut It should be possible to tell that it moves differently across the sky, over time, yes. Though I place no restrictions on how long it takes to compile the observations required to reach that conclusion (as this could be applied to fantasy settings, immortal or near immortal races could observe for hundreds or thousands of years to accomplish it, or even multiple generations of actual humans could do it as well) $\endgroup$ – Dalila Dec 19 '18 at 16:22
  • $\begingroup$ I tweaked the title of your question. Feel free to roll back or to Edit further if you disagree with the edit. $\endgroup$ – a CVn Dec 19 '18 at 18:24

With the naked eye, humans can see approximately 6th-magnitude objects. We can compute the apparent magnitude of a planet at a given distance, and find the distance corresponding to an apparent magnitude of +6. The formula is $$m_p=M_p+5\log\left(\frac{d}{10\text{ pc}}\right)$$ where $m_p$ and $M_p$ are the apparent and absolute magnitudes and $d$ is the distance from Earth to the planet. $M_p$ can be calculated if we know $M_S$, the absolute magnitude of the Sun: $$M_p=M_S-2.5\log\left(a\frac{R_p^2}{4d_s^2}\right)$$ where $d_s$ is the distance from the planet to the Sun and $r_p$ is its radius. Let's say that $d_s\approx d$, by assuming that the planet is much further from the Sun than Earth is. Finally, we get $$m_p=M_S-2.5\log\left(a\frac{R_p^2}{4d^2}\right)+5\log\left(\frac{d}{10\text{ pc}}\right)$$ This has the solution $$d=\sqrt{\frac{a^{1/2}R_J\cdot10\text{ pc}}{2}}10^{\cfrac{m_p-M_S}{10}}$$ Let's say $R_p\approx R_J$, the radius of Jupiter, and $a=0.5$ - also like Jupiter. Then, given that $M_S=4.83$, I get 24 astronomical units - about halfway between Uranus and Neptune. We can't see Neptune with our naked eyes, but we can see Uranus under good conditions, which matches our calculations (although note that the ice giants have different albedos). Let's say we choose an even higher albedo - say, $a=0.8$, as you suggested. This gets me up to 27, even closer to Neptune.

  • $\begingroup$ I think this answer defines every variable used in the formulae, except actually for $d$ (so you don't really tell us what you're calculating). We can see that from the question, but it'd be a small change to make the answer stand more on its own. Also, you might consider \cfrac for the exponential term in the last formula, but that may require a little bit of rearranging to make it read nicely. $\endgroup$ – a CVn Dec 19 '18 at 18:21
  • $\begingroup$ @aCVn Thanks for the suggestions; I've implemented them. The final equation's still a bit bulky, but that's the best I can do for now. $\endgroup$ – HDE 226868 Dec 19 '18 at 18:36
  • $\begingroup$ We can see Uranus with our naked eyes, under good conditions. It was seen in the pre-telescope age, but nobody realized it was a planet. To be classified as a planet, it has to attract enough attention so someone looks for it more than once, with enough elapsed time to get noticeable motion. There's lots and lots of sixth-magnitude stars, and nobody's going to track all of them. The minimum brightness for a visible planet as detected is 1.86 for Mars, and the lowest mean brightness is 0.71 for Mars. There's very few stars that bright, so it would be easy to note one moving. $\endgroup$ – David Thornley Dec 19 '18 at 22:58
  • $\begingroup$ @DavidThornley Thank you for the correction. $\endgroup$ – HDE 226868 Dec 19 '18 at 23:17

Uranus was detected by naked eye in the classical age but because it is too slow it wasn't recoginzed as a planet. In the "discovery" section. So the classical astrological manuals that are aware of its existence classify the planet as a fixed star.

So it's not only visibility that matters but also the speed because if it is too slow it will be classified as fixed star.

  • $\begingroup$ Very good points, but not actually an answer to the question ... A Jupiter sized planet (significantly larger than Uranus) that is more reflective than any in our current solar system should (I suspect) be visible at a significantly greater distance, and while speed would make it hard to differentiate from fixed stars, it should still be possible with enough long term observations. $\endgroup$ – Dalila Dec 18 '18 at 21:55
  • 1
    $\begingroup$ An addendun - since the question deals with ""primitive"" societies it also deals with astrology, specifically something like the classical, pagan, astrology. In this system what matters is what you see, the human perception of the world. So, invisible objects are irrelevant, even if they exist. Also an object that is too slow won't be classified as a planet but as a fixed star. Also the ancients were aware that the fixed stars werent that fixed in a long enough time, like the crab nebula supernova shown. $\endgroup$ – Geronimo Dec 20 '18 at 21:40
  • $\begingroup$ Yes, astrology in stories is one purpose I intended for this question. And the fact that invisible objects are irrelevant in that sense is also a basis for the question. Determining a full planetary system isn't useful in that type of setting, because anything outside the range of observation couldn't possibly have an effect on the society in that way. However, even if said society classified it as a star, a distant planet could still "behave" noticeably different ("special/important star") from actual stars, and could be given preferential status in their astrology/mythology as a result. $\endgroup$ – Dalila Dec 21 '18 at 22:41

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.