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My protagonist is stranded on an alien planet. Fortunately, along with ready supplies of food, shelter, water, electrical power, etc. she has a functioning high-end smartphone. She uses this smartphone as a sextant and after observations and a bunch of math she determines:

  • How long a day on the planet is
  • The orbital period of both moons and that of the planet itself (how long a year is)
  • The planet's approximate angle of inclination respective to the ecliptic
  • Rough latitude
  • Roughly what time of the year it is (eg. predicting that winter is coming or spring is coming)

The only major limitation is that the smartphone's accelerometer is only has a repeatable pointing accuracy of around 0.1 degrees and the time period of observation is limited (only around thirty days). Fortunately, due to local weather patterns, it can be assumed that the night sky is clear every night of observation. Also, she has no outside-context knowledge on local space. Before measurements, all she knows is that the planet she's on has two moons and a sun.

Is it feasible to determine all these facts using only the tools provided? If yes, what degree of inaccuracy could be expected?

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  • $\begingroup$ When you say "high-end smartphone", do you mean a future device with similar capabilities as a present-day smartphone? Or do you mean a literal macOS / android device? I see people making a bunch of assumptions that only hold in the latter case, but your premise (a life-sustaining world with available survival supplies) would seem to suggest the former (and also that this takes place quite some time in the future). $\endgroup$ – Matthew May 17 at 20:12
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    $\begingroup$ @Matthew the setting is present day-ish if maybe a tiny bit into the future. Regardless it can be assumed the phone has effectively unlimited computational power and has an IDE and compiler of some sort loaded on to it. There shouldn't be a "this is too difficult to calculate limit". Similarly, the clock on the phone is accurate, but not at atomic clock levels. I have yet to be convinced that a couple seconds of drift per day would meaningfully impact results. $\endgroup$ – Dragongeek May 17 at 20:42
  • $\begingroup$ @Matthew magic in this case, but it's out of the scope of the question. The survival aspects aren't important here. $\endgroup$ – Dragongeek May 17 at 20:47
  • $\begingroup$ (Gist of previous comment I deleted upon seeing an edit: how did the protagonist end up there and from whence come the supplies?) Okay, if your protagonist can write code, that makes things a LOT easier. Well, at least in terms of number crunching, in the sense that we can assume doing more complex math (calculus) isn't out of reach. As for the clock, I believe they tend to be consistent and I'm not sure it's actually of immediate importance if the phone thinks "one second" is actually 0.8 ISO seconds. $\endgroup$ – Matthew May 17 at 20:51
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    $\begingroup$ @Matthew It depends on exactly what the data is going to be used for. If it is to be integrated into calculations using any physics or astronomical constants, then the data would have to be consistent with the units that defined those constants in the first place. Calculating an interplanetary space journey, for instance. If they are for some religious purpose, or some archival record keeping local to only the planet, then accuracy would not be an issue. If they need to work locally, for purposes of science, that is a different scenario. Unless one recalculates the constants using local units. $\endgroup$ – Justin Thyme the Second May 18 at 3:00
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All info can be gathered in just 30 days, subject to some inaccuracy due to elliptical orbits.

Accuracy of tools:

  • Math: Any modern cellphone's calculator app gives good calculations to 16 digits, 13 displayed. This is more than enough.
    Default calculator also includes Sin, Cos, Roots, Exponent, ln, stored values for PI and e, etc.
  • Time: Any modern cellphone's time is accurate to less than 1/20th of a second per day. Usually much better than that. More than enough. Ironically older phones were more accurate!
  • Observation: Any (decent) modern cellphone can take photos of stars, with resolution of > 1000x1000. This is more than enough. Note that each photo includes a timestamp, down to the second. This is important
  • distance: Oh carp! I knew we were forgetting something. We have no means to measure distances more accurately than pacing off steps or touching elbows.

Measurements:

  • How long a day on the planet is:
    Just measure the time from first touch of sun to horizon, until the same horizon touch the next day, from the exact same location. If the local weather is the same, then you will have the time accurate to within about 10 seconds.
    Over more days, this can be fined down to about 2-3 seconds, if the sun is not too "fuzzy". Note that you expect both the precise location and time to drift between days, due to season progressing.

  • Year length: Do the exact same measurement for a specific star, that happens to enter the horizon at the same point as the sun.
    Due to the much star being a point light, you should be able to get the measurement accurate to within 1 second on the first try.

The difference between these gives you the number of days in a year, thus gives you the year length.
NB!! This works best with a planet on a perfectly circular orbit. Repeat every now and then, plot deviation over time to get orbit eccentricity. 30 days (rather 1/10th of a year) will not really be enough to determine this accurately, but should refine the possible range a lot.

  • To get your latitude: Just do a rough guesstimate of where the south/north pole is. Take a photo of the stars there, with reference objects in the foreground. Repeat 1/4 day later, match up the two photos.
    This will give you the exact north/south celestial pole position.
    Angle between this and true horizon gives you your exact Latitude.

  • To get the planet's inclination: Just use your latitude and the sun's maximum elevation angle. Repeat over the next 30 days. You will have partial plot of 30/yeardays part of the year cycle, for the angle between the sun and the ecliptic. This should be enough to give you an exact planetary rotation axis inclination, if the orbit is circular.

  • To get season:
    From two consecutive sunsets, you can see whether the sun is heading towards the equator or towards a pole. From this and planetary inclination and year length, you can get exact season phase, thus know the season accurate to +- a very few days. (Actually, you can predict the equinoxes, but that's the same thing)

  • Orbital parameters of the moons:
    Just take two photos showing the moon, and background stars, on consecutive days. This will give you a good idea of angle covered per time, which gives orbital period.
    Again, elliptical orbits can skew the result, ideally you want to observe for one full orbit.
    This will also give you any orbital inclination of the moons and their rotation rates

  • Size of the planet, or the moons:
    Sorry, nope. Rough guesstimates only, unless you can travel a significant distance (some hundreds of km) over the surface is a very short time, to take two observations.
    Or, if slow movement only, you need to do the observations on identical orbit days, i.e. typically on the equinoxes or similar. Not doable within your timespan.

  • Your longitude on the planet:
    Arbitrarily declare your current position to be the Meridian, 0 east. Party!

Summary:

With only 30 days of observation, and assuming that this 30 days is < 10% of a year , similar to Earth, you can get:

  • Latitude, accurate to 500m. Seriously! It's only limited by your resolution of photo, and measurement of horizon. (and Oblateness of Planet, but day length and gravity will give you an order-of-magnitude estimate of that, and it is a very small error)
  • Day length, Accurate to about 1-2 seconds. Limited by: visual observation of star occultation by horizon.
  • Year length, accurate to well under one day. Accurate to under an hour if the orbit is circular, but it never is, quite.
  • Longitude: perfect, because its arbitrary
  • Moon orbit periods: Accurate to <10 minutes, IF your stay is longer than one orbit. Otherwise To the hour. Limited by: visual observation of occultation of a star(if there long enough), else visual of horizon touch.

Moon size, planet size, sun size/ orbit distance: rough guesswork based on how heavy you feel. Really just guesswork, unless your cellphone includes some form of accurate distance measurement.

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  • $\begingroup$ "Any modern cellphone's time is accurate to less than 1/20th of a second per day." Given drift I have actually observed, I am highly skeptical of this claim. (Although as I've said previously, I'm not sure how the difference between a "smart device second" and "ISO second" would actually matter.) OTOH, a device with a ToF camera actually can, at least in theory, measure distances reasonable well (yes, even without GPS, which is lousy for accuracy at around meter-resolution anyway). This would only work for relatively short distances, though (i.e. meters, maybe decameters). $\endgroup$ – Matthew May 17 at 20:38
  • $\begingroup$ @Matthew Here is a supporting reference. news.mit.edu/2018/… Smart phones depend on GPS signals for accuracy. Something about the cost of a $1,000 chip to make them accurate enough for GPS. Steve usually took the cheapest way out, to maximize profit, and used marketing to cover up the cheap. $\endgroup$ – Justin Thyme the Second May 18 at 2:16
  • $\begingroup$ Unless the planet wobbles. $\endgroup$ – Justin Thyme the Second May 18 at 2:40
  • $\begingroup$ @JustinThymetheSecond JUst for clarity, please give ONE example of a planet that "wobbles" $\endgroup$ – PcMan May 20 at 19:16
  • $\begingroup$ @JustinThymetheSecond cellphone time accuracy. I have an old samsung s3 mini. Battery is basically dead, but I keep it plugged in all the time. Phone is in airplane mode at all times, but more relevantly it has no sim, so none of the networks will ever try to talk to it. I use the thing as an mp3 player, and sometimes as alarm clock. It was last connected to a wifi or cellular signal in november 2019, at which time it should have been network synchronised. I checked it this morning, the clock is slow by 17 seconds. This means it lost one whole second every 33.35 days. $\endgroup$ – PcMan May 20 at 19:22
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It's a clock - and you can do this with a clock

The British Navy conquered much of the world because it developed a portable clock that allowed it to measure changes in longitude. You haven't even asked for that - the measurements you mention can all be done with simple pre-industrial observations, using the stars themselves as the clock.

  • The main advantage of the cell phone is that you can say the day is X hours long, rather than "a day" long, or 360 degrees of right ascension.

  • Observing the orbital period of a moon is another simple timing.

  • The latitude can be measured, for example, by identifying a constant pole star (or a way to identify the pole's location - even on Earth we're not always so lucky as to have a single good pole star). Excluding some simple thinking about the definition of north vs south, the angle of the pole star above the ground is your latitude.

  • Observing the angle of the sun, knowing your latitude, and observing how it changes will yield the season and approximate angle of inclination.

For the last two, it may be convenient if the device has a sine function to work out the angle from a slope measurement, but I suspect you would find only a simulated first-generation calculator mostly intended to give smart phone users a way to multiply by 100.

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    $\begingroup$ There are plenty of "scientific calculator" apps available, and most devices will have a way to calculate [co]sines and tangents. If they do any "3D", they absolutely will be capable of these functions. The question becomes whether they are user-accessible. (That said, my default calculator app can do sines.) $\endgroup$ – Matthew May 16 at 13:04
  • $\begingroup$ Assume the actual number-crunching isn't difficult. The smartphone has Excel. $\endgroup$ – Dragongeek May 16 at 13:23
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    $\begingroup$ @Dragongeek Actually, it doesn't. Excel is in the cloud. It needs an internet connection. A lot of functions in a smart phone are actually in the cloud. Everything is moot if she can just ask Alexa or Siri or Google. $\endgroup$ – Justin Thyme the Second May 16 at 16:38
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    $\begingroup$ @JustinThymetheSecond The Excel smartphone apps work offline. $\endgroup$ – notovny May 16 at 17:14
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    $\begingroup$ @JustinThymetheSecond I can't test on an iPhone, but as far as Android goes, with both Wifi and mobile service turned off, and the application force-stopped, Excel will launch functionally enough to open files that are saved on the phone, do calculations, and save files. Don't know how long that authorization will last, admittedly. $\endgroup$ – notovny May 16 at 18:55
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Disclaimer: this is an incomplete answer and may contain errors. Some of these procedures may also require the sun to actually set, i.e. may not work too close to a pole in the middle of summer. (This also assumes you aren't on a tidally locked planet at just the right distance so that the hemisphere of eternal sun is livable.)

How long a day on the planet is

Day length AFAIK can be determined over at most a few days using nothing more complicated than a crude sun dial and an accurate clock. The real catch is the accurate clock; because they are expected to have external time sources available, even a "high end" smart device may not have a very accurate clock.

The procedure is to set up your sun dial (which can be as simple as a relatively straight stick stuck in a relatively flat patch of dirt) and watch it for a day or several days to determine when the shortest shadow is cast. This gives you "noon", and also "north". Once you've marked this, just measure the time between noon and noon for several days to get an average. (or just once if you don't need much accuracy).

If your smart device has enough storage, and you can afford to do without it for some hours each day, you may be able to set it to take a photo every minute or so around noon and use the photos to determine noon. This could also give you a result in a little over a local day.

(Another option, as Mike noted, is to take pictures of the night sky with the device in an absolutely fixed position and find which two are most similar. The above is essentially the same procedure, but using the position of the sun for your comparisons.)

The planet's approximate angle of inclination respective to the ecliptic

Assuming the planet isn't wobbling or doing something similarly crazy (which might not preclude this method giving you an answer, it just might be wrong)... I think this may actually be quite easy, if you can either take or somehow fake a long duration photograph. Assuming the stars are fairly stationary relative to the planet's rotation, you just need to take a long enough exposure to get some good star trails (you may be able to fake this with many separate shots, but you might then need to manually track stars), then use that to determine the planet's axis of rotation relative to your current position. Then just take several readings of the sun to determine the ecliptic and compare notes.

Rough latitude

Should be trivial if you can solve the previous problem. Ignore the sun, as its position is seasonally dependent. Instead, use your star trails to find a true (rotational) pole and compare the angle to that to the angle to the sun at noon.

Roughly what time of the year it is (e.g. predicting that winter is coming or spring is coming)

With the prior disclaimer about accurate timekeeping, if you are near an equinox, this should be fairly trivial to accomplish by making two marks on your sun dial (toward morning and evening, but measuring between two marks may be more accurate than trying to judge sunrise and sunset) and comparing the length of time it takes to go between these over a period of time. (Bonus: you don't necessarily need daily readings for this, but you do need readings separated by several planetary days; the more separated, the better.)

You can also compare your ecliptic to your poles to refine your guess. This can't give you the answer outright, however, as, unless you are near one of the solstices, on its own it will give two solutions. Both together, however, should give you a good guess, and may also be able to suggest the planet's orbital period (i.e. year). However, you may need to be able to do some calculus for that.


p.s. I really hope this "survival kit" is intended for use on sparsely populated worlds and includes a powerful distress beacon. (Or maybe it's only intended for use on something like a derelict ship?) Not because it will help, but because if the designers considered unexplored planets in their possible scenarios, they darned well ought to have included tools designed to figure this stuff out.

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  • $\begingroup$ Unless the planet wobbles. Two moons, not necessarily in orbits related to each other. $\endgroup$ – Justin Thyme the Second May 16 at 16:41
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Not all info can be gathered in just 30 days.

  • How long a day on the planet is

To determine this one just needs to measure the time interval between two sunsets/sunrises. Measuring the passage of a distant star at a certain celestial meridian is going to take some more infrastructure. If I would do the same in my present spot between today and tomorrow, I would measure a day length of 24 hours and 1 minute.

  • The orbital period of both moons and that of the planet itself (how long a year is)

If they make the assumption that the moons orbit is rather circular, they need to measure the time between for example new moon and first quarter, and multiply that by 4. If they want to be really sure, they need to measure the time between two identical phases. This will have an error of a few hours.

  • The planet's approximate angle of inclination respective to the ecliptic

For this they would need to wait both solstices and measure the height of the sun at noon. The axial tilt would be the complement to 90 of the average of the two.

  • Rough latitude

Once they know the axial tilt and the local time, they can calculate the latitude based on sun height and local time.

  • Roughly what time of the year it is (eg. predicting that winter is coming or spring is coming)

For this they just need to monitor the length of the day, from sunrise to sunset, across a couple of days. If the daylight is getting longer, they are in the local spring/summer, if it is getting shorter they are in the local fall/winter.

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  • $\begingroup$ "To determine this one just needs to measure the time interval between two sunsets/sunrises." Assuming you can be reasonably accurate about what constitutes "sunrise" or "sunset", this will still give you a slightly dubious answer. Measuring noon to noon is going to be more accurate. $\endgroup$ – Matthew May 16 at 13:08
  • $\begingroup$ How do you determine the axial tilt? That takes a LOT of sequential observations, and the ability to VERY accurately plot the movement of the stars over a considerable time frame. Unless you already knew the 'sky map'. The first thing would be to find the stars that do not rotate, to find the axis of rotation of the planet. And if the planet wobbles - there are TWO moons, not necessarily in the same orbits, not necessarily even with the same period of rotation, and a sun, maybe not in the planar of the axis of rotation of the planet, giving conflicting 'tidal lock' configurations. $\endgroup$ – Justin Thyme the Second May 16 at 16:26
  • $\begingroup$ "For this they just need to monitor..." In this case, there is no such thing as 'just need to...' The first thing that' needs to be done' is that she 'just needs to' make and validate all of the correct assumptions, and that is a bit more complex than 'just needs to'. 'Assume a perfect sphere in a perfect orbit with a perfect rotation', and everything else falls into place. $\endgroup$ – Justin Thyme the Second May 16 at 16:33
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    $\begingroup$ You're going to need to know the orbital period of the planet to accurately calculate the orbital period of any moons; To take the example of Earth, because the planet moves roughly 1/12 of its orbit around the Sun between full Moons, measuring the time between full moons will give you a value a shade over two days longer than the Moon's sidereal orbital period. $\endgroup$ – notovny May 16 at 17:05
  • $\begingroup$ The time from sunrise to sunrise or even from noon to noon can vary, but a recognizable star should rise at once-daily intervals. From the planet's perspective, the sun "moves" all the way around the sky once a year, but the sphere of the fixed stars revolves (rotates?) around the planet at a constant rate. $\endgroup$ – Mike Serfas May 17 at 8:12
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Why would she go to all of that trouble? Why wouldn't she just enquire of the local population?

They would obviously have to be advanced, given that they have GPS-type satellites in orbit and cell towers all over the place that enables the apps in the smart phone to work.

A smart phone itself is pretty useless. It needs constant signals form outside sources to enable it to function. Without them, at best you can use it as a calculator.

Even the compass functions would be useless unless she had a firm understanding of the magnetic field, if any, around the planet.

EDIT Let me TL:DR this

Impossible without her knowing MUCH more about the planet. Shape? Orbital wobble? Magnetic pole? Gravity? Axis of rotation vs the sun? Far too much has to be assumed, without her having an external intelligent source of data.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – L.Dutch May 18 at 3:52

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