Timeline for If a massive object like Jupiter flew past the Earth how close would it need to come to pull people off of the surface?
Current License: CC BY-SA 4.0
18 events
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| May 30, 2019 at 6:11 | history | edited | uhoh | CC BY-SA 4.0 |
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| May 30, 2019 at 6:10 | comment | added | uhoh | @LoganPickup yep I think you're right, changing now. Thank you! | |
| May 30, 2019 at 6:03 | comment | added | Logan Pickup | "On the close side, being 6,378 kilometers closer, would feel an acceleration 1.2 m/s^2 less" <- 1.2 m/s^2 more? | |
| May 29, 2019 at 21:48 | vote | accept | Yevgeny Simkin | ||
| May 28, 2019 at 5:24 | comment | added | uhoh | @mckenzm yes. Earth's center of mass is closer to Jupiter than the people on the far side, so Earth experiences a greater acceleration than the people do. That's why the diagram showing outward-pointing arrows on both sides of the Earth is always confusing at first. | |
| May 28, 2019 at 5:04 | comment | added | mckenzm | Lighter on the far side? | |
| May 28, 2019 at 4:07 | comment | added | Peter Cordes | Glad I could help, thanks for doing the math and writing it up, this is an interesting Q&A. :) I thought about adding in the phrasing "having the Earth pulled out from under them (even faster than the extra pull of Earth + Jupiter)" for the people on the far side, but I don't see a place to put it without being redundant or rewriting a whole chunk. | |
| May 28, 2019 at 0:06 | comment | added | uhoh | @PeterCordes that's so much better than I could have done I've just quoted you, thank you. Please feel free to edit the answer further! | |
| May 28, 2019 at 0:05 | history | edited | uhoh | CC BY-SA 4.0 |
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| May 27, 2019 at 22:56 | comment | added | Peter Cordes | TL:DR Jupiter isn't dense enough for its gravity gradient over Earth's radius to produce a 1g tidal acceleration, even right at Jupiter's surface. | |
| May 27, 2019 at 18:49 | comment | added | uhoh | @Barmar double check the second paragraph that begins with: "It's not like a vacuum cleaner that selectively lifts small and light objects..." The Earth's momentum does not affect the Earth's acceleration, and the acceleration of objects of different sizes near Jupiter will be (almost) the same because the force depends on the mass; $F_E=M_EM_J/r^2$ so the acceleration $a_E=F/M_E=M_J/r^2$ is independent of the mass. I say "almost" because the Earth is so big that it's gravitational effect on Jupiter is very small but can't be completely neglected. | |
| May 27, 2019 at 18:35 | comment | added | Barmar | Isn't the earth's momentum a zillion times larger, so its motion won't be perturbed as much as a person? | |
| May 27, 2019 at 17:34 | history | edited | uhoh | CC BY-SA 4.0 |
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| May 27, 2019 at 17:33 | comment | added | uhoh | @ShakesBeerCH it looks like your edit was rejected, but there was indeed an error in the arithmetic. $GM_J/(R_E+R_J)^2=20.9$ m/s^2, etc. Can you check again, thanks! | |
| May 27, 2019 at 17:29 | history | edited | uhoh | CC BY-SA 4.0 |
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| May 27, 2019 at 11:54 | review | Suggested edits | |||
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| May 27, 2019 at 9:56 | history | edited | Glorfindel | CC BY-SA 4.0 |
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| May 27, 2019 at 1:06 | history | answered | uhoh | CC BY-SA 4.0 |