Timeline for Why is the Moon considered the major cause of tides, even though it is weaker than the Sun?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 30, 2016 at 14:41 | comment | added | Wouter | I've already written an answer to a closely related question that shows a different, more explicit comparison between the Lunar and Solar Tides. Since it might be of interest to readers of this answer, I'll link it here: physics.stackexchange.com/a/111695/16660. | |
| Jan 2, 2015 at 23:36 | history | edited | user65081 | CC BY-SA 3.0 |
deleted 55 characters in body
|
| Dec 2, 2014 at 13:03 | comment | added | user65081 | @MarcvanLeeuwen Thanks! I agree with you, the grammar is not good, and/or the reasoning is not straight. But I think people understood what I meant. I am going to improve the writting, due to the large number of votes. | |
| Dec 2, 2014 at 12:57 | comment | added | Marc van Leeuwen | I think there is a gap in the reasoning in this answer. You start by saying the absolute gravity is not the relevant quantity, and then in the second paragraph you nonetheless talk about how the force (itself) grows with $r$. I think what is missing is that the differential of a $r^{-2}$ force fields is proportional to $r^{-3}$ and that this precisely cancels the factor $r^3$ by which mass increases under constant apparent size and density assumptions. Therefore such objects should have tidal effect independent of $r$. Then finally it is their density ratio that distinguishes Sun and Moon. | |
| Dec 2, 2014 at 4:23 | comment | added | user65081 | $Floris @G.Bach I am gonna try to edit it to make it less confussing (at least the number, who were wrong) and put it back. | |
| Dec 2, 2014 at 3:25 | comment | added | G. Bach | @Floris Thanks for mentioning it, it still shows up in the edit history. | |
| Dec 2, 2014 at 0:07 | comment | added | Floris | I liked the picture and I'm sorry it was edited out again. | |
| Nov 30, 2014 at 23:19 | comment | added | JiK | This explanation about having the same apparent size seems very confusing. I'd say "[...] the volume $V$ of the object will grow as $r^3$ - - the force grows linearly with $r\rho$", so there'd be no need to first assume (without saying it explicitely before making the calculations) that the Sun and the Moon had the same density. | |
| Nov 30, 2014 at 12:00 | vote | accept | Moctava Farzán | ||
| Nov 30, 2014 at 6:26 | history | edited | user65081 | CC BY-SA 3.0 |
deleted 71 characters in body
|
| S Nov 29, 2014 at 19:57 | history | suggested | Christopher King | CC BY-SA 3.0 |
Added drawing
|
| Nov 29, 2014 at 19:40 | review | Suggested edits | |||
| S Nov 29, 2014 at 19:57 | |||||
| Nov 29, 2014 at 16:42 | history | edited | user65081 | CC BY-SA 3.0 |
deleted 5 characters in body
|
| S Nov 29, 2014 at 16:20 | history | suggested | user36790 | CC BY-SA 3.0 |
Modified the spellings and grammer
|
| Nov 29, 2014 at 16:04 | review | Suggested edits | |||
| S Nov 29, 2014 at 16:20 | |||||
| Nov 29, 2014 at 15:59 | history | answered | user65081 | CC BY-SA 3.0 |