Timeline for Why don't we consider centrifugal force on a mass placed on earth?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Sep 2, 2017 at 1:00 | comment | added | Floris | @MartinBeckett presumably if you want to buy enough gold that it matters, you would want to confirm the calibration of the supplier's scale - and if your measurements disagree, walk away. See also this | |
| Sep 1, 2017 at 20:52 | comment | added | Martin Beckett | @Floris - if you know a gold seller that will accept my calibration weight I can make money without 'g' ;-) | |
| Sep 1, 2017 at 20:04 | comment | added | Floris | If you buy gold, you bring a calibration weight with you and use a balance. That removes the effect of $g$ from your purchase... | |
| S Nov 23, 2014 at 15:47 | history | suggested | Piper McCorkle | CC BY-SA 3.0 |
MathJaX applicable here?
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| Nov 23, 2014 at 15:16 | review | Suggested edits | |||
| S Nov 23, 2014 at 15:47 | |||||
| May 12, 2011 at 19:46 | comment | added | Martin Beckett | @Luboš - or my new diet that just moves people form Oslo to Mexico city | |
| May 12, 2011 at 12:49 | comment | added | Luboš Motl | Yes, sure, unless you of course buy gold for a million of dollars and using a different weight, they steal $3000 from you. ;-) | |
| May 12, 2011 at 12:32 | comment | added | Martin Beckett | @Lubos - yes, I worked it out. But it's still small enough that if you are using 9.8 for'g' you don't need to care | |
| May 12, 2011 at 7:41 | comment | added | Luboš Motl | Dear @Martin, you're wrong that the variation of $g$ due to the centrifugal force is smaller than the variation of $g$ because of the bulge, which is also caused by the same centrifugal force. Up to a factor of at most 2, they're the same. See physics.stackexchange.com/q/8074 | |
| May 12, 2011 at 5:28 | vote | accept | claws | ||
| May 12, 2011 at 5:26 | comment | added | Martin Beckett | ω is small - it's 2pi/(24*60*60) rad/s. It might help to think of it in F=mv^2/R terms, remember R is big | |
| May 12, 2011 at 5:18 | comment | added | Martin Beckett | Interestingly doing the numbers it's larger than I guessed | |
| May 12, 2011 at 5:15 | history | edited | Martin Beckett | CC BY-SA 3.0 |
added 143 characters in body; added 13 characters in body
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| May 12, 2011 at 5:12 | comment | added | Martin Beckett | The magnitude of the force is smaller than the variation in 'g' - the centrifugal force is about 0.1 - 0.2% of 'g' | |
| May 12, 2011 at 5:09 | comment | added | claws |
How is it smaller? $\omega$ is angular velocity of earth which is large and R is radius of earth which is also large.
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| May 12, 2011 at 5:06 | history | answered | Martin Beckett | CC BY-SA 3.0 |