Timeline for How can kinetic energy be proportional to the square of velocity, when velocity is relative?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 9, 2018 at 19:52 | comment | added | LSpice | Your edit has the answer end mid-sentence ("But not in general, because if they start off with a different"). | |
| Sep 10, 2013 at 18:01 | comment | added | Manishearth | @babou It's perfectly OK for total energy to be different in different frames. When talking about energy differences, yes, the energy can distribute differently but the net difference has to be the same, i.e, zero. | |
| Sep 10, 2013 at 17:56 | comment | added | babou | I was bothered by that too. I think the answer is that though the total energy posted is the same, the energy ratio between ship and reaction mass is different, so that the ship gets the same velocity increase. I have not done the actual computation, but I do not see any other explanation. You accelerate twice, each time getting the same increment in a different inertial frame, since it was done twice the same way. But speed of inertial frames adds vectorially, or am I mistaken. | |
| Sep 10, 2013 at 17:37 | comment | added | Manishearth | @babou The speed increase will not be the same if there is initially a nonzero relative velocity between the two ships. That was my point, you can't generalize it too much. | |
| Sep 10, 2013 at 17:32 | comment | added | babou | "He's talking about relative velocity between two ships". Right, but if the energy posted is the total energy spent on both craft and reaction mass, the speed increase will be the same for each burst of acceleration. I think that is correct, and his question is to understand why it will not work on Earth. It is really a good question. He could not answer it because he did not consider the reaction mass needed for momentum conservation. | |
| Sep 10, 2013 at 14:19 | comment | added | babou | Actually, I answered this question (hopefully correctly). Initially it just looked interesting to sort out why the OP was confused. I find it fruitful to write answers that attempt to give the intuition for what is going on, trying to avoid maths while being precise. I also believe it is useful to users, but it is considerable work if one attempts to be clear and unambiguous. This confirmed what I said on meta: answering an old question is a waste of time even if you are really contributing. The system needs a way to give a chance to new answers. | |
| Sep 10, 2013 at 13:15 | history | edited | Manishearth | CC BY-SA 3.0 |
added 28 characters in body
|
| Sep 10, 2013 at 13:02 | comment | added | Manishearth | @babou I took it as the energy released by the burnt fuel. (I'm not so sure if that makes sense anymore, I'll fix the answer) Yes, the speed change of the ship in its inertial frame before acceleration is the same. But that's not what's asked. He's talking about relative velocity between two ships. | |
| Sep 10, 2013 at 12:52 | comment | added | babou | For me the screen is posting the energy spent on accelerating both the ship and the reaction mass in opposite directions, not just the part going to the ship. Then if the same energy is spent twice with the same mass ratio, the speed change for the ship in its inertial frame before each acceleration is the same. That is what I understood. And I think that was meant by the OP except for the fact that he missed the issue of reaction mass. Did I say anything wrong ? If I did, then I do not understand how the laws of physics can be the same in all inertial frames. | |
| Sep 10, 2013 at 10:26 | comment | added | Manishearth | @babou No, it won't. It will produce the same $\Delta v^2$, not $\Delta v$. Re:energy: Of course in the end everyone agrees on the total energy expended. However, the values on the screen won't match with the values for their frame as the distribution is different. (Like I said, it is consistent in different frames). I may have interpreted the question differently, see Alfred's answer -- the OP has not been very clear here. | |
| Sep 10, 2013 at 10:17 | comment | added | babou | Don't you think you are misleading the OP with your last sentence. Actually, the same amount of energy spent in the same way with respect to ship frame will always produce the same velocity increment (depending on ship orientation) in any inertial frame (though less than OP says). I think that is what the OP was saying. - Also, regarding the screen posting of energy in first sentence, all inertial frame will agree on total energy expended by the ship. What they may disagree on is the way it is distributed between ship and reaction mass expelled. | |
| Nov 23, 2012 at 1:00 | history | answered | Manishearth | CC BY-SA 3.0 |