Timeline for If I toss a coin, vertically, on the surface of Mars, will it land back in my hand?
Current License: CC BY-SA 3.0
28 events
| when toggle format | what | by | license | comment | |
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| Sep 15, 2018 at 15:35 | vote | accept | Soham | ||
| Jan 8, 2016 at 22:06 | answer | added | Guill | timeline score: 0 | |
| Jan 4, 2016 at 15:51 | comment | added | user854 | Since your question explicitly mentions atmospheric resistance (air resistance), physics.stackexchange.com/q/227391 doesn't answer it, but might be worth reading. | |
| Jan 2, 2016 at 17:13 | history | edited | Qmechanic♦ |
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| Jan 1, 2016 at 0:20 | comment | added | emory | My understanding of the term vertical is probably tied to Earth's gravity. Is it well defined on other planets? | |
| Dec 31, 2015 at 22:02 | history | tweeted | twitter.com/StackPhysics/status/682683264142807040 | ||
| Dec 31, 2015 at 21:51 | answer | added | paparazzo | timeline score: 0 | |
| Dec 31, 2015 at 21:19 | answer | added | Floris | timeline score: 42 | |
| Dec 31, 2015 at 21:14 | answer | added | hobbs | timeline score: 11 | |
| Dec 31, 2015 at 19:34 | comment | added | garyp | @Sathyam Yes. The analysis needs to account for drag and the relative difference of tangential velocity between the coin and the earth's surface. For "normal" coin tosses, we can ignore both, but at some point (height of toss) both will become significant. I think the OP might be asking: what is that height? | |
| Dec 31, 2015 at 19:25 | comment | added | Sathyam | @garyp Agreed that parabola is an approximation. But the orbit is elliptical only in the absence of air drag. | |
| Dec 31, 2015 at 19:24 | comment | added | leftaroundabout | I, too, don't understand what you're getting at with this question, but at any rate the answer won't depend on whether you do this on Earth or Mars. | |
| Dec 31, 2015 at 19:23 | history | protected | Qmechanic♦ | ||
| Dec 31, 2015 at 19:22 | answer | added | amasics | timeline score: 10 | |
| Dec 31, 2015 at 19:18 | comment | added | garyp | @Sathyam The coin absolutely is in an elliptical orbit. It's approximately parabolic over the small distances involved, assuming the toss is not too high. But the height of the toss plays an essential role in the analysis. I'll wait for the OP to clarify his intent before commenting further. | |
| Dec 31, 2015 at 19:12 | comment | added | Sathyam | @garyp Once you toss a coin, it is no longer in any orbit. It traces a parabola if you haven't tossed it in the escape velocity. The questions you mentioned are the concerns of OP as far as I understood. | |
| Dec 31, 2015 at 19:11 | comment | added | HDE 226868 | Related: physics.stackexchange.com/q/166853 | |
| Dec 31, 2015 at 19:10 | answer | added | user98038 | timeline score: 0 | |
| Dec 31, 2015 at 19:07 | comment | added | HDE 226868 | @Sathyam All I'm saying is that you're not going to see any major effect. We can extrapolate from here that the effect will be extraordinarily small. A coin will travel much higher than a person can jump. But it won't go high enough to make the journey long enough for the forces at work to make it go any significant distance parallel to the ground. | |
| S Dec 31, 2015 at 19:07 | history | suggested | Sathyam | CC BY-SA 3.0 |
Removed redundant tags. Edited body for clarity.
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| Dec 31, 2015 at 19:04 | comment | added | garyp | I don't think I understand the question. Read this and let me know if I interpret your question correctly. Once I toss the coin it is in an elliptical orbit around the center of Mars, while my hand is constrained to be in a circular path that is not an orbit. Will the coin land in my hand in a perfect vacuum? What is the effect of atmospheric drag? Under what conditions will it land on my hand, or not land on my hand? | |
| Dec 31, 2015 at 19:02 | comment | added | Sathyam | @HDE226868 Well small as you can imagine. If you toss a point mass from a point on the surface, it will never come back to its original position. But since the coin is a physical entity with a definite shape and mass, we can consider small as small as we toss a coin on earth. I explained the point in my answer. | |
| Dec 31, 2015 at 18:59 | review | Suggested edits | |||
| S Dec 31, 2015 at 19:07 | |||||
| Dec 31, 2015 at 18:58 | comment | added | HDE 226868 | @Sathyam How small is small? I would assume that for the realm of this problem, the differences will be negligible. | |
| Dec 31, 2015 at 18:56 | answer | added | Sathyam | timeline score: 0 | |
| Dec 31, 2015 at 18:50 | review | Low quality answers | |||
| Dec 31, 2015 at 19:00 | |||||
| Dec 31, 2015 at 18:50 | comment | added | Sathyam | @HDE226868 The coin is moving with the planet, too This is incorrect. Once the coin is detached from the hand(contact) the centripetal acceleration stops. The coin would only come to your hand if it is tossed to small distances. | |
| Dec 31, 2015 at 18:31 | history | asked | Soham | CC BY-SA 3.0 |