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If I had a vehicle in a completely frictionless environment, say a rocket mass 2kg, which had a battery with 100j of energy, how fast could it go?

The obvious answer is $ke = \frac{1}{2}mv^2 = \frac{1}{2}2v^2 = v^2$ so $100 = v^2, v = 10$ .At this point all energy from the battery is in the form of kinetic energy, so the battery has no energy. Final state: battery has 0j, vehicle at $10m/s$

An alternate and seemingly equally correct explanation would be: The object accelerates to $5m/s$, $ke = \frac{1}{2}mv^2 = \frac{1}{2}\cdot2\cdot5^2 = 25$. Now the object moves at $5m/s$ and the battery has $75j (100-25)$. Then we switch reference frame to one at equal speed to the object. The object is now moving at $0 m/s$, and the battery still has 75j of energy. Now accelerate again in our new frame of reference to $5m/s$ from $0$. $\frac{1}{2}mv^2 = \frac{1}{2}2\cdot5^2 = 25j$. The energy in the battery should now be $50j (75-25)$, and the object is at $5m/s$ in this frame of reference. Moving back into the original reference frame, the energy of the battery is still 50j, but the vehicle is at $10m/s$. How can this be - where does the 50j of energy come from?

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    $\begingroup$ Kinetic energy is not the same in different frames of reference! $\endgroup$
    – Sid
    Commented Sep 7, 2023 at 10:37
  • $\begingroup$ Yes, but the energy in a battery is the same in all frames of reference, and changing kinetic energy within a frame of reference will always deduct that quantity from the battery. $\endgroup$
    – Mercury
    Commented Sep 7, 2023 at 10:41
  • $\begingroup$ There will be a frame where the change in kinetic energy will equal the battery energy and a frame where there will be no change in kinetic energy (the frame of the rocket) $\endgroup$
    – Bob D
    Commented Sep 7, 2023 at 10:53
  • $\begingroup$ Does this answer your question? Why is the work done by a rocket engine greater at higher speeds? $\endgroup$
    – Dale
    Commented Sep 8, 2023 at 15:05

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The amount of energy in the battery available to alter the masses speed IS frame dependent. A joule applied to a slower moving mass has a greater effect on the speed of that mass than when it’s at a greater speed. In your boosted frame the apparent effect of 1 joule on the masses speed is greater than that in the original frame.

In your boosted frame when you go from 0 to 5m/s this costs 25J to achieve in that frame. However in the original frame the mass went from 5 to 10m/s the energy cost would have been 75J in the original frame. There IS a disagreement on the energy in that battery between frames.

This doesn’t violate any energy conservation however as the amount of energy in the battery that is available to change the masses speed is determined by what speed the mass is already at relative to something. That 75J you refer to on the battery is the energy available to do work on the rocket in the original frame not your boosted frame. The remaining energy in the battery for the boosted frame would have to be recalibrated to 25J as to conserve the total amount of work the fuel can do on the rocket in the initial frame.

The other 50J will now have appeared to be transferred to the exhaust gas instead of being transferred to the rocket, as in the new frame the exhaust gas is moving faster relative to you with an extra 5m/s than in the original frame.

Some kind of exhaust/opposite force is required as per newtons 2nd law. There must have been a reaction force to the force that accelerated the rocket.

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  • $\begingroup$ Ok, but this seems to cause a contradiction; If instead of a battery we view this as say fuel which we burn, and at the start we have 10g of this fuel which when burnt gives us 100j. when we burn 2.5g to accelerate to 5 m/s we have 7.5g of fuel left, which doesnt change in any reference frame. This also should provide 75j of energy, as the energy density of a substance should also be frame independent otherwise we run into loads of contradictions. Recalibrating to 25j is equivalent to recalibrating 7.5g to 2.5g, which seems very clearly false. $\endgroup$
    – Mercury
    Commented Sep 7, 2023 at 22:07
  • $\begingroup$ It’s not the mass of the fuel that is being recalibrated. It’s the ability that mass of fuel has to provide a change in kinetic energy that is being recalibrated. Remember the 100J of energy that fuel had was measured in the original frame. $\endgroup$
    – Blue5000
    Commented Sep 8, 2023 at 7:00
  • $\begingroup$ Yeah, but the energy of the fuel is mass * energy density, and the energy density of a substance is just constant in all reference frames. So in the original frame, 100j = m*density, m = 100/density. If the energy the fuel has to change the kinetic energy changes, m = 25/density, implying mass changes. Either that or despite having 100j of energy it can only supply 25j, which also doesnt make sense. $\endgroup$
    – Mercury
    Commented Sep 8, 2023 at 7:53
  • $\begingroup$ All the maths you have just done was based on a value of energy that was measured in a specific frame. Where do you think this value of 100J came from? $\endgroup$
    – Blue5000
    Commented Sep 8, 2023 at 8:36
  • $\begingroup$ I mean it depends on how it is stored. If its stored in heat for example, energy = Mass times specific heat capacity times temperature, which would then be calculated to be 100j. If its in a fuel which is burnt, it would be mass times energy density, if it were in pure mass it would be mass times c^2. In all situations this measurement is frame independent. $\endgroup$
    – Mercury
    Commented Sep 8, 2023 at 8:39
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You have explained what is observed in the original frame, all the fuel is burnt and the velocity of the rocket relative to that frame, starting from rest, is $10 \,\rm m/s$ to the right.

Now consider the whole event from the point of view of the frame moving at $5\,\rm m/s$ to the right.
Initially the rocket is moving to the left at $5\,\rm m/s$.
$100\,\rm J$ of fuel is used up and the rocket is moving at $5\,\rm m/s$ to the right relative to the frame that is moving at $5\,\rm m/s$ to the right.
In this frame there is no change in the kinetic energy of the rocket.

You may thinks this is rather strange?
Consider that the rocket is the system and in some way there is a constant external force $F$ acting on the rocket, You are dealing with two inertial frames and so certain parameters are the same in each frame: mass $m$, the displacement $s$, external force $F$, acceleration $a$.

In the original frame for the whole of the motion the external force $F$ and the displacement $s$ are in the same direction with $Fs = 100\,\rm J$ and this produces a change in the kinetic energy of $100\,\rm J$ which leads to a final speed of $10\,\rm m/s$.

In the moving frame the motion can be thought of in two parts.
In the first part the force is to the right but the displacement of the rocket is to the left so the work done by the force is $\color{red} -F\frac s2$ and for the second part of the motion the work done by the force is $\color{red}+F\frac s2$ because force and displacement are in the same direction.
Thus the net work done by the force is zero and so the change in kinetic energy of the rocket is also zero.
Thus in the moving frame the final velocity of the rocket is $5\,\rm m/s$ to the right and so in the original frame it is $5+5=10\,\rm m/s$ to the right.

The (change in) kinetic energy is not a conserved quantity when moving between reference frames.

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  • $\begingroup$ I get that kinetic energy is not a conserved quantity when moving between reference frames, but the energy in the battery should be. The problem is that in the situation where we move the frame, we end with 50j of energy in the battery, but when we do not we have none. Furthermore, when we shift from the moving frame back, we have a situation where the velocity and position of the rocket is the same as in the none moving frame, but we suddenly have 50j of energy in the battery. Despite the exact same initial conditions, one ends with more energy than the other in the same initial ref frame. $\endgroup$
    – Mercury
    Commented Sep 7, 2023 at 12:54
  • $\begingroup$ If we end in the same reference frame which we started in, energy should be conserved, but we see that it is not. $\endgroup$
    – Mercury
    Commented Sep 7, 2023 at 12:55
  • $\begingroup$ If the change in kinetic energy is not the same then how can the external work done by the battery be the same when jumping between frames. So the work done to get the rocket from $0\,\rm m/s$ to $5\,\rm m/s$ in the original frame is not the same as the work done to get the rocket from $-5\,\rm m/s$ to $0\,\rm m/s$ in the moving frame. $\endgroup$
    – Farcher
    Commented Sep 7, 2023 at 13:29
  • $\begingroup$ well because switching reference frames is not a physical action. By starting in frame a, switching to frame b, doing some actions, then switching back to frame a we should have conserved all energy compared to the start and the end. And I show 2 situations, both starting in the same conditions and reference frame, and ending in the same reference frame, but energy is not conserved in one. In other words, one can only go to 10m/s, but another is able to go even faster for no reason. $\endgroup$
    – Mercury
    Commented Sep 7, 2023 at 13:40
  • $\begingroup$ You use frame changing to easy. take your 100J batterie, and your rocket, than seen from your rocket earth has the velocity 5m/s relativ to you so its kinetic energy is M/2v^2 with the mass M of earth! a lot of energy for 100J! $\endgroup$
    – trula
    Commented Sep 8, 2023 at 15:20
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It appears you are questioning what happens to the decrease in battery chemical potential energy in those frames where there is not an equal increase in kinetic energy of the rocket, an apparent violation of the law of conservation of energy. You appear to already know kinetic energy is not a conserved quantity. In order to account for all the energy forms for all the components of a given system in different reference frames, we need to consider how the rocket is given thrust by the battery, which is missing from your analysis.

Consider your rocket to be in outer space (a frictionless environment, as you stipulated). Instead of a battery as the energy source, consider rocket fuel as the energy source (both the battery and fuel are sources of chemical potential energy). The products of the combustion of the fuel (gases, particles) are pushed out of the rocket giving the rocket thrust.

In the reference frame of the rocket the change in kinetic energy of the rocket is zero. However, in that frame there is an increase in the kinetic energy (KE) of the center of mass (COM) of the products of combustion moving away from the rocket, plus an increase in the internal kinetic energy of the fuel (molecular kinetic energy as reflected by an increase in temperature of the fuel (commonly, but technically incorrectly, referred to as "heat").

In the frame of the center of mass (COM) of the products of combustion of the fuel the change in kinetic energy of the COM of the fuel is zero. However in that reference there is an increase in the kinetic energy of the rocket as well as an increase in internal energy of the fuel. The reduction in chemical potential energy of the fuel thus equals the increase in kinetic energy of the rocket plus the increase in internal kinetic energy of the fuel.

Governing all of this is conservation of momentum, which applies in all reference frames. If we can ignore any gravitational forces, the rocket plus fuel is an isolated system (there are no external forces acting on it) which means the total change in momentum (rocket plus fuel) is zero. That means the change in momentum of the COM of the rocket plus fuel is zero.

Hope this helps.

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