Timeline for Is it possible for a moon to have a higher surface gravity than the planet it is attached to?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Dec 3, 2019 at 21:56 | comment | added | CJ Dennis | @Chronocidal That is what the answer should have said originally. I understand how it works, I was just pointing out it was badly written. | |
| Dec 3, 2019 at 16:47 | comment | added | Pete Kirkham | @FooBar if you hold density constant, yes, if you hold mass constant, then density is inversely proportional to the cube of radius, and so gravity becomes inversely proportional to the square; the two are the same. Though I'd agree that it's might be more practical physically to hold density more-or-less constant when constructing your moon. | |
| Dec 3, 2019 at 16:03 | comment | added | Foo Bar | Fun fact: Surface gravity of a sphere is linearly proportional to radius (and density). Gravity is based on mass divided by distance squared. Mass equals density * volume. Volume of a sphere is based on radius cubed. Therefore surface gravity is proportional to radius cubed divided by radius squared = radius^1. | |
| Dec 3, 2019 at 13:18 | comment | added | Chronocidal | @CJDennis g=mG/r² : So, gravity is directly proportional to both Mass and the Gravitational Constant, but inversely proportional to the square of the radius (or distance from the centre of mass - gravity is lower at the top of Mt Everest, but by less than 0.5%) The question you forgot to ask yourself is "If it is proportional to the square of the radius, then what is that a proportion of?" | |
| Dec 3, 2019 at 8:24 | comment | added | Starfish Prime | @CJDennis this is why I said proportional not equal. | |
| S Dec 3, 2019 at 5:30 | history | edited | Mark Omo | CC BY-SA 4.0 |
As comments point out, readers could easily assume constant density, not constant mass. Or forget about mass entirely, apparently.
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| S Dec 3, 2019 at 5:30 | history | suggested | Peter Cordes | CC BY-SA 4.0 |
As comments point out, readers could easily assume constant density, not constant mass. Or forget about mass entirely, apparently.
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| Dec 3, 2019 at 4:59 | comment | added | Peter Cordes | @CJDennis: I made a suggested-edit that adds (for constant mass) to address your comment; fair point that readers might be thinking of constant density until they get to (or miss while skimming) the word "compressed" in the next sentence about the Moon. | |
| Dec 3, 2019 at 4:57 | review | Suggested edits | |||
| S Dec 3, 2019 at 5:30 | |||||
| Dec 3, 2019 at 4:49 | comment | added | CJ Dennis | @Draco18s I would not have commented if the statement mentioned both mass and radius. | |
| Dec 3, 2019 at 4:19 | comment | added | Draco18s no longer trusts SE | @CJDennis The Earth also has 83 times more mass. The moon has 1.2% the mass, yet manages 16.7% the surface gravity. | |
| Dec 3, 2019 at 3:37 | comment | added | CJ Dennis | "Surface gravity of a body is inversely proportional to the square of its radius." This doesn't make sense. If that was true, the surface gravity of Earth would be much less than the Moon's. | |
| Dec 2, 2019 at 16:39 | history | edited | Starfish Prime | CC BY-SA 4.0 |
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| Dec 2, 2019 at 16:27 | history | edited | Starfish Prime | CC BY-SA 4.0 |
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| Dec 2, 2019 at 15:57 | history | answered | Starfish Prime | CC BY-SA 4.0 |