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If object one is moving $0\frac{m}{s}$ relative to the Earth and object two is moving $1\frac{m}{s}$ and they both have $1kg$ of mass, then object one has no kinetic translational energy (relative to the Earth?) but object two has $\frac12$ joules of kinetic translational energy.

Which should mean that if I add in their kinetic energies due to the speed at which the Earth is rotating, object two should half of a joule more kinetic energy? But upon doing the calculations, the fact that kinetic translational energy is given as $K=\frac12mV^2$ means that by moving $1\frac{m}{s}$ faster than object one, object two has much more kinetic energy.

Could someone explain what happened?

Simply Beautiful Art
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    what happens is that kinetic energy does depend on your chosen reference frame (same with their difference). There is no law that says that the difference in kinetic energy is the same across reference frames –  Dec 14 '15 at 22:20
  • Check out the answers to this question https://physics.stackexchange.com/questions/535/why-does-kinetic-energy-increase-quadratically-not-linearly-with-speed/14752#14752 – jim Jun 25 '19 at 08:21

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The initial calculations of kinetic energy are correct in the local rest frame (the Earth). However the relative energy differences are not the same in other rest frames.

One way of thinking about this is to remember that work is "force times distance". Imagine you want to "push" two objects to accelerate them each by 1 $\frac{m}{s}$, but one is at rest and the other is already travelling 1 $\frac{m}{s}$. You will have to exert the same force on the two objects - but that force will be exerted over a longer distance for the object which is already moving - thus it will accumulate more kinetic energy going from 1 $\frac{m}{s}$ to 2 $\frac{m}{s}$ than the object which goes from 0 $\frac{m}{s}$ to 1 $\frac{m}{s}$.

Paul Young
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