I'm not a physics student but I'm interested in this topic. One thing I don't understand is kinetic energy. We know that if an object moves, its kinetic energy increases. Consider a rocket accelerated to 99.9999999% of the speed of light. Its kinetic energy is huge. Now let us imagine that there are no other objects in the universe except this rocket. Then what will be the kinetic energy of the rocket and how can it be calculated?
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$\begingroup$ Why do you think the other stuff in the universe matters for the kinetic energy of the rocket alone? $\endgroup$– Marius Ladegård MeyerCommented May 1 at 8:04
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$\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$– Community BotCommented May 1 at 8:49
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$\begingroup$ Voting to reopen. This is a perfectly clear (and a very good) question - how to define velocity and kinetic energy in an empty universe. There are good answers below. $\endgroup$– gandalf61Commented May 1 at 12:49
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4 Answers
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The velocity and kinetic energy of an object depend on the frame of reference. Special relativity states that there is no preferred inertial frame. It is meaningless to specify a velocity without specifying what it is relative to. The formula for the kinetic energy is $(\gamma-1)mc^2$.
answered May 1 at 8:21
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Well,I Don't understand what do you mean by ''Now let us consider all the other objects in the universes except this rocket removed''. Because in newtonian mechanics, it doesn’t matter whether you remove anything from surrounding or not.If they do not become a obstacle for the Rocket,they do not impact the kinetic energy of the rocket.If you go by Einstein relativity then It's also depend on which frame of reference we are considering.So Either your question is uncleared or i simply Don't understand what you are trying to say.But in general you use K.E=$1/2 m \cdot v^2$ or K.E=$m \cdot c^2 [\sqrt{1-(v^2/c^2)}-1]$ Since you've mention that the rockets velocity is 99.99% of the Light speed,i prefer you to use the second formula.There are many derived form of that equation.I've just given you the simple one.Hope this helps you.And advance sorry if i've made any mistake.
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Kinetic energy has no measurable effect on the object itself, in its own rest frame. An observer inside the rocket will see no change in its clock rates etc. Kinetic energy is only relevant in the context of an objects velocity relative to some reference frame. In a universe where the rocket is the only object there is no notion of velocity as it is relative. The concept of kinetic energy of the rocket is meaningless in this context. The kinetic energy of an object is only relevant if the object interacts with another object, e.g. in a collision or in a comparison of two objects with relative motion e.g. in determining the difference in clock rates of the two objects, but if there are no other objects in the universe it has no relevance at all.
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I am not sure if I correctly understand your question, but I am going to make a couple of comments:
We know that any object If it moves, its kinetic energy (KE) increases.
This is technically wrong. If there is movement, then there is a kinetic energy. The "more movement there is" (this is, how heavy and/or fast the object is), the more KE there is.
Now let us consider all the other objects in the universes except this rocket removed. Then what will be the kinetic energy of the rocket and that How can it be calculated?
Well, if now there is no rocket, then it can't have an energy.
Now, if your question is something like "what is the KE of everything BUT the rocket", then the answer might depend on the observer. If you are moving with the rocket, so that for you the rocket is not moving, then you will have to sum the KE of all the objects of the universe with the speed measured from your point of view (as if you were in the rocket).
If you are an external observer, and you see the rocket moving at $99,9999\%$ then, again, you just simply sum the KE of all the objects of the universe from this point of view. Note that now the speed of the objects will be different than when seen from the rocket, so the total KE will be different from the last case!
This is as when you are traveling in your car (so the car wouldn't be moving for you), you see the road signs travelling $towards you$ at 100 km/h, but if you were standing on the road close to the sign, now you would see the car moving $towards you$ and the sign wouldn't be moving at all.
answered May 1 at 8:46