Timeline for Does distance matter for gravitational U in non-earth/ non-celestial-body systems?
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Apr 8, 2020 at 22:54 | comment | added | Alexander Ye | @PM2Ring I am talking not about the height but the distance Btwn the carts | |
| Apr 8, 2020 at 22:50 | comment | added | Alexander Ye | @PM2Ring the problem does not give any horizontal distances | |
| Apr 8, 2020 at 22:49 | comment | added | PM 2Ring | @Alexander Sorry, I don't understand your last comment. Why wouldn't you know $r$? Maybe you will find this article helpful: en.wikipedia.org/wiki/Gravitational_energy | |
| Apr 8, 2020 at 22:33 | comment | added | Alexander Ye | @PM2Ring that still does not solve the problem unless when you get closer/farther to infinite the PE is still 0... otherwise it will just create the same problem of not knowing what r is | |
| Apr 8, 2020 at 21:34 | comment | added | PM 2Ring | @Alexander Yes, you can set the PE zero point anywhere you like, because it's a torsor. So you might as well set it somewhere convenient for calculation. | |
| Apr 8, 2020 at 21:03 | comment | added | Alexander Ye | @PM2Ring wait so I’m assuming we are allowed to do that even if it is not an orbital mechanics set up right? To be sure ... | |
| Apr 8, 2020 at 20:57 | comment | added | Alexander Ye | @JoeIddon does that make sense? | |
| Apr 8, 2020 at 20:57 | comment | added | Alexander Ye | @JoeIddon For your example, the system includes the earth. For the question though, it specific excludes it. Think about the concept of gravitational PE, it only exists Between two objects, and not within the object itself. So your question, you already assumed that the system was between the object and the earth, which is different from the question. | |
| Apr 8, 2020 at 20:55 | comment | added | Alexander Ye | @PM2Ring I’ll try using the barycentre! | |
| Apr 8, 2020 at 20:49 | comment | added | PM 2Ring | @AlexanderYe It's convenient when calculating stuff on Earth to set the zero point for gravitational PE at $h=0$. But in orbital mechanics the usual convention is to set the gravitational PE zero point at infinite distance from the centre of the main body, or from the barycentre of a system of bodies. | |
| Apr 8, 2020 at 20:48 | comment | added | Joe Iddon | Hmm, I'm not sure that makes sense... I'm struggling to understand your justification for why ME is not conserved: "the external force of the earth’s gravitational force acting on the objects". Consider a simple example: you drop a ball from a height of $1m$. Is ME conserved? Yes, it is under the Earth's gravitational force and it is conserved as the potential is converted to kinetic. | |
| Apr 8, 2020 at 20:44 | comment | added | Alexander Ye | I would answer the question directly By writing that ME is not conserved or rather there is no conservation of energy bc of the external force of the earth’s gravitational force acting on the objects. The one problem here though is that how would you prove ME Increases? (Without being given a distance that the two objects are apart by) Because it intuitively does increase. | |
| Apr 8, 2020 at 20:42 | comment | added | Alexander Ye | I would say that too, but my teacher specifically made a point of Ug being separate from things outside the system and also, there was another sub question above this that was the same exact question, except just with “the two cart-AND earth “ system. So it would be weird to have the same question twice conceptually. I think I would answer the question by writing ... (next post) | |
| Apr 8, 2020 at 20:40 | comment | added | Joe Iddon | True. In this question though, it is clear that they only want you to consider the potential energy an object gains from being lifted away from the surface of the earth: $U = mgh$. This "approximate" potential energy formula assumes that the only objects in the universe is the object, of mass $m$, and the Earth (or any planet), which has a gravitational field strength, $g$ at its surface. It breaks down when the height lifted, $h$, is very large since the gravitational potential field can no longer be approximated as linear. | |
| Apr 8, 2020 at 20:38 | comment | added | Alexander Ye | Earth is in the set up, but when we define Gravitational PE, we have to limit/ specify a system, because PE is based on the Point of View or system you take it... if not, then there is no way you can calculate PE, there’s just too much objects in the universe who act on an object . | |
| Apr 8, 2020 at 20:36 | comment | added | Alexander Ye | So would you just say that the system starts off with ... “negligible U”? And then increases to A large quantity of K? (Thus There is no conservation of energy?) | |
| Apr 8, 2020 at 20:36 | comment | added | Joe Iddon | If there was no Earth in this setup (which raises questions like what is the ramp sitting on!), then what incentives would there by for the carts to even roll down the slopes at all? | |
| Apr 8, 2020 at 20:34 | history | edited | Joe Iddon | CC BY-SA 4.0 |
added 523 characters in body
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| Apr 8, 2020 at 20:33 | vote | accept | Alexander Ye | ||
| Apr 8, 2020 at 20:33 | |||||
| Apr 8, 2020 at 20:32 | comment | added | Alexander Ye | Note though that it says the system of ONLY the two carts. There is another part to this question that includes the system of the two carts AND the earth specifically (for which I chose to include the minute U between the carts)... I am pretty sure it did canceled out though. So basically this question is ONLY the two objects’ potential energy, even though Earth’s gravity is involved. | |
| Apr 8, 2020 at 20:28 | history | answered | Joe Iddon | CC BY-SA 4.0 |