As I was thinking about space flight for my world, I thought about how they would need to know a planet's gravitational force before they could land, and found that humans figured out the moon's gravity using its orbital period, tides, etc.

However...the situation that native astronomers find themselves in is an interesting one, as the "moon" in question is actually the home planet's binary twin. Since the two planets are similar in mass, both planets are tidally locked to each other which means that, from a stationary point on land, the moon would not move in the sky and there would be no tides, thus our human methods are useless.

Would it be possible for pre-space age astronomers to figure out the "moon's" gravity?

In my mind, a really easy way would be to send a satellite to the planet and find how long it takes to orbit at a distance, but this question assumes no such technology is possible.

Also, if anything here is important, the data of the planets are as follows:

Home Planet:

  • ~0.7x Earth mass
  • ~0.9x Earth radius & gravity
  • **density of both planets are the same as Earth

Moon Planet:

  • ~1.84x Earth mass (technically this makes Home planet the moon, but unimportant)

  • ~1.26x Earth radius & gravity

  • Distance between planets is about 129,000km

  • Day length (both planets) is 79.9 hours

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    $\begingroup$ See in Quora: How is the mass of the moon calculated?. Actually, can be copypasted here for an answer. $\endgroup$ – Alexander Jul 31 '19 at 23:06
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    $\begingroup$ I see that you used the hard-science tag so I'm adding the special notice for those. Please read it. If that's not what you meant, please flag for a moderator to remove it. Thanks. $\endgroup$ – Monica Cellio Aug 1 '19 at 3:07
  • $\begingroup$ @Alexander Those methods were effective for calculating the real Moon's mass (and thereby gravity) due to the fact that the Earth was not tidally locked to it. In my scenario, the planet stays mostly stationary in the sky, aside from slight wobbling in place. This would also mean that although tides still do technically exist, it would not be observed as moving up and down like on Earth. Although, natives would easily be able to determine their "moon's" orbit period due to the fact that it's exactly the same as their own day length, if that helps them at all. $\endgroup$ – Foosic17 Aug 1 '19 at 3:25
  • $\begingroup$ Jules Verne, From the Earth to the Moon, 1865. $\endgroup$ – AlexP Aug 1 '19 at 3:56
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    $\begingroup$ @Foosic17 In Quora answer, methods starting from "ii) Airy (1849)" do not rely on tides. $\endgroup$ – Alexander Aug 1 '19 at 7:17

1. Tides

the moon would not move in the sky and there would be no tides

Wrong. The sun provides tides too (albeit smaller ones than our moon), so the sea still rises and falls on a regular cycle. You'll even get variations in tide height depending on the relative angles of the sun and your "moon", though not to the same extent as spring tides on earth. This means that it may still be possible to use tides on your planet to calculate the mass of the moon in the way that was done historically here (see the quora answer linked by AlexP above) though as I am not a mathematician I couldn't tell you the exact technique or how precise it may be.

2. Parallax.

Establish the distance to the "moon". Our moon's distance can be judged using lunar parallax. You'll have to observe your moon from two different places on the surface to see the effect, of course. This'll also get you the size of the moon, so you can see that it is about the same size as your own planet, another big clue.

Determine where the barycenter of your "moon"-planet system lies. I believe you can do this using diurnal parallax, only in this case your planet doesn't rotate itself, but it does rotate around the two bodys' barycenter which will cause visible parallax effects. Not big enough to see the stars wobble, I suspect, but it should be enough to cause other planets in the same solar system to show some movement.

Now you've determined that the barycenter of your system lies at more-or-less the midpoint of the two worlds, that's a massive clue that they mass about the same, assuming you've got some idea of the nature of gravity.

3. Tidal locking

There's a big clue in the tidal locking, of course... I'm not certain on the history of understanding of tidal forces on solid a objects, or how the idea came about, but you'd still get tides on your world so it wouldn't require a huge leap of logic. If tidal forces were understood, it may well be seen that bodies will become tidally locked if they have similar masses, or if they have existed long enough to lose their rotation to tidal effects.

First, determine the mass of your planet (there are various ways to do this, see this earth science SE answer for example, but I won't go into them here). Next, establish the age of your planet (again, left as an exercise to the reader). It should be possible to determine that insufficient time has passed for your planet to become locked to a much less massive body, therefore the mass of your moon must be pretty similar to the mass of your planet.

I think. There are a lot more leaps of logic here, but it wouldn't do to underestimate people's ability to figure stuff out!

4ish. Meteorite Impacts

Meteorites will sometimes hit your moon. They sometimes hit the earth. There's good reason to believe the meteorites may be made of the same material. Given that you know the size of the moon and how far away it is (see parallax above), it is possible to estimate the height of the craters. Assuming your "moon" isn't made of something particularly bizarre and scifi-y, it'll be possible to make estimates of the density of its surface given the sizes of the craters and the spread of ejecta, given some understanding of gravity and ballistics.

I'm not totally certain that this will provide a useful estimate of the mass of your moon (because you have to make assumptions about density of subsurface layers), but it is certainly a contributing piece of evidence, especially when combined with assumptions about a common origin of the "moon" and the observer's planet.

(note that this does make some assumptions about the nature of the "moon". One with a thick atmosphere, or a liquid surface or whatever else may make this bit of analysis impossible)

(note 2: there's a related useful thing here if the "moon" is a rocky world, because you can use impact cratering to help estimate its age, which is a pointer towards a common age and common origin of the observer's planet and the "moon")

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  • $\begingroup$ Re: tides - how can tides created by the sun help calculate the mass of the moon? $\endgroup$ – G0BLiN Aug 1 '19 at 12:23
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    $\begingroup$ Since the "moon" is our case is bigger than the planet, it would be likely that it has considerable atmosphere, water and maybe even plate tectonic, so #4 is not going to be very useful. $\endgroup$ – Alexander Aug 1 '19 at 16:31
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    $\begingroup$ Yeah, I kinda skipped over the whole thing about solar tides existing, but you're correct that they are rather small. I used Artifexian's tide calculator a while back and I believe the solar tide was somewhere around 0.15 m. Definitely not a lot, but it's noticeable if someone was trying to notice. As the other guy said, I don't know if solar tides would help with the moon's mass, but 2 and 3 should certainly be effective. (and the "moon" does in fact have an atmosphere and plate tectonics, so 4 might not work) $\endgroup$ – Foosic17 Aug 2 '19 at 4:36

"Can early astronomers determine the gravity of their planet's “moon” without ever going there?" Of course they can. Our Earth-bound astronomers did, and we know how they did it.

  1. The distance from the Earth to the Moon and the size of the Moon were known since the Antiquity. The Moon is close enough that the parallax method works well enough with the naked eye instruments of the ancients.

    For example, Ptolemy in his Almagest gives the average distance between the Earth and the Moon as 59 Earth radii; the correct value is 60.06 Earth radii. I'd say that Ptolemy's measurement is quite good for the 2nd century CE.

  2. The mass of the Earth was measured in the 18th century; by 1798, the famous Cavendish experiment (performed by Henry Cavendish applying an idea by John Michell) came within 1% of the true value. (The same experiment measures the gravitational constant.)

  3. Because the Earth-Moon system rotates around their common barycenter, astronomical objects seen from Earth show a monthly parallax. Towards the end of the 19th century, by measuring the monthly parallax of the asteroid 12 Victoria, British astronomer David Gill determined the position of the barycenter and thus the ratio between the mass of the Earth and the mass of the Moon was determined with great accuracy.

  4. Since the mass of the Moon was known and its size was known, the gravitational acceleration on the surface of the Moon could be easily computed.

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    $\begingroup$ @G0BLiN tidally locked bodies have a movement called libration, which helps a little. $\endgroup$ – Renan Aug 1 '19 at 12:43
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    $\begingroup$ @G0BLiN if you can track a feture on the moon, you can use parallax to measure the distance. And though there might not be diurnal libration, the other two modes are enough to use parallax. $\endgroup$ – Renan Aug 1 '19 at 13:07
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    $\begingroup$ @G0BLiN: You measure the position of the Moon in the sky (with respect to the fixed stars) at the same time from two faraway places. That's how they did it. It works just like a rangefinder. $\endgroup$ – AlexP Aug 1 '19 at 14:28
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    $\begingroup$ @G0BLiN, tidally locked and fixed in the sky doesn't make a difference. As one method, measure the angular height of the other planet above, say, the eastern horizon, then go directly east or west and make the same measurement. The height above the horizon will change, and from that and knowledge of the size of the home planet, you can calculate distances. $\endgroup$ – Keith Morrison Aug 2 '19 at 3:42
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    $\begingroup$ @G0BLiN, that's why tidally locked and fixed makes it easier. You're not limited to a given direction; I just used east and west arbitrarily. You can go pretty much in any direction, so long as you know what it is, and that means you can pick the ideal places to make your measurement. $\endgroup$ – Keith Morrison Aug 5 '19 at 0:57

Other answers have discussed the methods used by ancient, medieval, and modern pre-spaceflight astronomers to discover the distances, sizes, and masses of various astronomical bodies, from which the surface gravity and escape velocity of those bodies could be calculated once Newton published his laws of physics.

Here is another type of answer.

Any large library should contain astronomy, space flight, & science fiction books written before the space age.

I have read enough of them to know that popular books said that the surface gravity of the Moon was about one sixth that of Earth, while textbooks would give the actual scientific value of the Moon's surface gravity and of it's escape velocity.

Because of the Moon's low surface gravity, it was common for artistic depictions of the Moon's surface before the Space Age to show very steep mountains, unlike the ones actually photographed there during the Space Age.

The first Earthlings landed on the Moon in 1969, and the first space probe from Earth was sent to the orbit of the Moon in 1959. And if the astronomers of that era could not use data from Earthbound observations to provide the space agencies with reasonably accurate figures for the distance, orbital speed, size, mass, density, and gravitational attraction of the Moon, how could space flights to the Moon have been planned well enough for space probes to reach the Moon, and sometimes go into orbit around the Moon, and sometimes make soft landings on the Moon?

The first soft landing on the Moon was Luna 9, and the first lunar orbiter was Luna 10, both in 1966.

And I have read enough non fiction and science fiction from before the Space Age to know that the low surface gravity on the Moon was well known by amateur and professional astronomers, spaceflight enthusiasts, and science fiction fans, even in the remote and prehistoric era before 1959.

The lower surface gravity of the planet Mars was well enough known that Edgar Rice Burroughs, in A Princess of Mars (1912) had his hero from Earth, John Carter, discover that his Earthly muscles grave him a great advantage on Mars:

Carter finds that he has great strength and superhuman agility in this new environment as a result of its lesser gravity and lower atmospheric pressure.


Added August 4, 2019. The answers to this question should be interesting:


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Distance and size to our own moon were calculated with a surprising accuracy by the old greeks. That should make it fairly simple to calculate gravity relative to your own planet assuming the same density.

It should be noted that the moon is tidally locked to us because it has a bulge on one side, that caused it to slow it's rotation (by interaction with earths gravity) untill the bulge is permanently pointed towards earth. A similar formation should probably exist on both your planets if they are tidal locked to each other.

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  • $\begingroup$ An uneven distribution of mass helps tidal locking, but I don't think it is in any way essential. $\endgroup$ – Starfish Prime Aug 1 '19 at 8:17
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    $\begingroup$ Also, the question is flagged hard science, so you should probably back up your comments with a reference or two! $\endgroup$ – Starfish Prime Aug 1 '19 at 8:27

The planet's inhabitants are going to need, at minimum, to understand Newton's law of gravitation. Once they have that, there are many viable ways to calculate the mass of the moon. Here's one involving pendulums:

First, measure the distance to the moon using lunar parallax. Then, set up identical pendulums at 2 points on the planet's surface and record the number of ticks that occur over the span of a day (either via a clockwork mechanism or by tedious counting, depending on how early you want people to have figured this out). This, effectively, measures the strength of gravity at 2 points on the planet's surface.

If you picked 2 points with the same elevation, almost all of the differences in gravity that you record will be due to the different distances from the Earth-moon barycenter. There will be a term from centrifugal force and another from the gravity of the moon. Both depend only on the mass of the moon and known parameters (the mass of the Earth, radius of the Earth, distance to the moon, and the rotation period / length of a day), and so the moon's mass can be readily calculated from there.

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