Timeline for Why are these periods the same: a low earth orbit and oscillations through the center of the earth?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
replaced http://physics.stackexchange.com/ with https://physics.stackexchange.com/
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| Jul 13, 2014 at 2:44 | comment | added | user2338816 | Dropping an object into a hole through the Earth has an implicit assumption that the drop takes place from ground level. Perhaps a better assumption would be that the drop was done from an altitude of 160 km. Even better might be using the Moon instead of Earth so that air resistance is minimized from consideration (though uniform density, etc., are still issues). Regardless, such "oscillations" are simply variations of elliptical orbits; one is just "flatter" than the other. | |
| Jul 12, 2014 at 13:28 | comment | added | David Hammen | @User58220 - That calculation assumes the Earth is of a uniform density Earth. Density is not constant, so that 84 minute period for any length tunnel isn't the case. | |
| Jul 12, 2014 at 10:16 | answer | added | Qmechanic♦ | timeline score: 2 | |
| Jul 12, 2014 at 10:13 | answer | added | ticster | timeline score: 7 | |
| Jul 12, 2014 at 7:47 | history | edited | Johannes | CC BY-SA 3.0 |
Replace "fine-tuning" with "zero altitude"
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| Jul 12, 2014 at 2:37 | history | tweeted | twitter.com/#!/StackPhysics/status/487787810456629248 | ||
| Jul 12, 2014 at 2:31 | answer | added | David Hammen | timeline score: 2 | |
| Jul 11, 2014 at 20:30 | answer | added | Thomas Pornin | timeline score: 5 | |
| Jul 11, 2014 at 17:48 | comment | added | DJohnM | The "coincidence" is even more far-reaching. If you drill any straight tunnel of any length between two points at the same elevation, a frictionless body will take the same 84 minutes to oscillate back and forth...10 km or 10,000 km... | |
| Jul 11, 2014 at 16:22 | comment | added | Cruncher | I don't think this qualifies as an answer so: I think the intuition here is that, objects in orbit are free falling toward the centre of the earth. Much the same way the dropped ball is free falling toward the centre of the earth. So the time it takes an object to do a quarter turn around the earth, should be about as long as it takes the ball to reach the centre of the earth. Come to think of it, maybe the ball should be slower. Since half way to the core, it has the earth above it pulling it back up. The orbit is always free falling. | |
| Jul 11, 2014 at 16:16 | comment | added | Steve Jessop | Note that you aren't just ignoring air drag, you're ignoring resistance from the trees and stuff that the satellite would be crashing through, since it's orbiting at surface level ;-) | |
| Jul 11, 2014 at 16:08 | vote | accept | Carlos | ||
| Jul 11, 2014 at 15:43 | comment | added | ACuriousMind♦ | Not really. You are comparing two oscillations, one of which you adjusted to be precisely of the same period as the other. I really don't get the question. | |
| Jul 11, 2014 at 15:40 | answer | added | Phil Frost | timeline score: 42 | |
| Jul 11, 2014 at 15:40 | history | edited | Carlos | CC BY-SA 3.0 |
added 46 characters in body
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| Jul 11, 2014 at 15:38 | comment | added | Carlos | @ACuriousMind: I think my question is comparing two different oscillations: (1) the period of the person oscillating through a hole in the earth and (2) the period of a LEO orbit fine-tuned to an altitude that resulted in a period of 84.5 min? Is this clearer? I modified to question to reflect your comments. | |
| Jul 11, 2014 at 15:33 | comment | added | ACuriousMind♦ | Wait a moment. You say "If I fine-tuned the LEO orbit to be 84.5 min" and then you wonder why it would be exactly 84.5 min? | |
| Jul 11, 2014 at 15:30 | history | asked | Carlos | CC BY-SA 3.0 |