Timeline for Almost tidally locked to moon and the tides it would create
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 10, 2017 at 15:34 | comment | added | pablodf76 | Tidal force is proportional to the cube of the distance. With the Moon at GSO (an order of magnitude closer than IRL) the tides would be three orders of magnitude stronger (i.e. ~1000 times as strong as IRL). | |
| Jul 10, 2017 at 15:31 | comment | added | pablodf76 | @JamesK Yes of course. If the planet is Earth-sized and the mass of the moon is negligible compared to it and you want the lock to happen at 24 hours (which is not specified in the question), then yes, the moon should be at today's GSO. | |
| Jul 10, 2017 at 15:15 | comment | added | James K | @pablodf I agree entirely. But if we make the moon and the planet mutually locked, and the Planet is the same mass as the Earth, then the moon has to be closer. The question isn't about the Earth. | |
| Jul 10, 2017 at 11:41 | comment | added | G0BLiN | Earth - Moon distance is around 384,400km, this is an order of magnitude larger than the 36,000km of this scenario. Since gravity's force diminishes in a square proportion to distance, you'll need to re-scale the moon's mass by two orders of magnitude (diminish it to 0.0088 of it's current mass). If you assume homogeneous density for simplicity, it means diminishing the moon's radius by around 0.2 (compared to the distance, which was diminished by around 0.1) - all this means that you'll get a much closer, much much less massive but quite bigger (around twice it's angular size) moon... | |
| Jul 10, 2017 at 10:15 | comment | added | pablodf76 | @JamesK The Earth cannot become tidally locked to the Moon and remain rotating at the same speed as today. Indeed that's why the Earth's spin is slowing down right now while the Moon recedes at about an inch per year (I think). | |
| Jul 10, 2017 at 5:45 | comment | added | James K | @pablodf76 If Earth and Moon one day end up in a mutual lock they will be much further apart than today, THe rotation period of the the earth will be much slower, and the height of a gso will be at the distance of the moon, since in a mutual lock the orbit period of the moon = rotation period of Earth = rotation period of moon. If the Earth has a 24 hour rotation, and in mutual lock, then the moon would be 36000km high. | |
| Jul 10, 2017 at 1:39 | comment | added | pablodf76 | A geosynchronous orbit has nothing to do with a tidal lock. If Earth and Moon one day end up in a mutual lock they will be much further apart than today, and the Moon is already ten times more distant from Earth than a GSO. | |
| Jul 9, 2017 at 22:33 | history | edited | James K | CC BY-SA 3.0 |
deleted 9 characters in body
|
| Jul 9, 2017 at 22:33 | comment | added | James K | @JayLemmon 1m is bigger than I'd remembered. corrected. | |
| Jul 9, 2017 at 22:21 | comment | added | Jay Lemmon | Er, sorry, I should have said "cubicly proportional to apparent size". | |
| Jul 9, 2017 at 22:14 | comment | added | Jay Lemmon | Also, a note on your numbers, from lhup.edu/~dsimanek/scenario/tides.htm, the tidal bulge in the mid ocean is ~ 1 meter. | |
| Jul 9, 2017 at 22:11 | comment | added | Jay Lemmon | Thanks, I hadn't thought about it in terms of where that necessarily places the moon. From npl.washington.edu/av/altvw63.html it sounds like the tidal effect of a celestial body is more or less proportional to its apparent size in the sky, so even if the distance is fixed I can still play with the size of the actual moon to get reasonable tidal forces. | |
| Jul 9, 2017 at 17:56 | history | answered | James K | CC BY-SA 3.0 |