Conservation of KE in a moving frame

[Muu9]

TL;DR Summary

Accelerating an object requires more energy from a moving reference frame, but the energy changed from changing height or temperature seems independent of the observer. Why is kinetic energy special?

Suppose I accelerate a mass from rest to 1 m/s using n J of energy. From the sun's perspective, I've just accelerated it from (say) 29,785 m/s to 29,786 m/s, which would require 59571n J of energy. Where is the extra 59570n J coming from?

If the answer is "nowhere, changes in KE are relative to the frame of reference", then why is this unique to speed/kinetic energy? With height/gravtational PE, regardless of the height of my frame of reference, a change in height of 1 meter always leads to a change in gravitational PE of mg J. With temperature, an increase in a material of 1 C always leads to the same change in thermal energy regardless of the temperature I set to be 0 (my frame of reference).

 

[Orodruin]

When you accelerated the mass to $v_1 =$ 1 m/s you also conserved momentum, giving the Earth a small amount of velocity in the opposite direction. The Earth velocity change was $v_2 = - mv_1/M_\oplus$. The change in the Earth's kinetic energy was negligible in the original frame.

In the moving frame, the Earth velocity changed from $v_0 =$ 30 km/s to $v = v_0 + v_2 = v_0 - mv_1/M_\oplus$ and so the difference in the Earth's kinetic energy is
$$
\frac{M_\oplus}{2} \left[(v_0 - mv_1/M_\oplus)^2 - v_0^2 \right]
\simeq - m v_1 v_0
$$
to leading order (the $v_1^2$ term is the same as the term neglected in the Earth frame so I have neglected it here as well).

The difference in the mass' kinetic energy is
$$
\frac{m}{2}[(v_0 + v_1)^2 - v_0^2] \simeq mv_0v_1 + \frac{mv_1^2}{2}.
$$
The total difference in energy is therefore
$$
mv_0v_1 + \frac{mv_1^2}{2} - m v_1 v_0 = \frac{mv_1^2}{2}
$$
just as in the Earth frame.

 

[Ibix]

Where is the extra 59570n J coming from?

You can't just accelerate something - something else must have been pushed in the opposite direction in order to conserve momentum. Account for the energy change there and you'll find the discrepancy goes away.

 

[sophiecentaur]

TL;DR Summary: Accelerating an object requires more energy from a moving reference frame, but the energy changed from changing height or temperature seems independent of the observer. Why is kinetic energy special?

Where is the extra 59570n J coming from?

When you use 'numbers' you lose the pattern of what's going on (it was a blur to me and I eventually found that difference). @Orodruin 's post has a full description but I suspect you may also have find it a blur - but it is correct and sufficient.
Here's a limited version which may be easier to get hold of:
The formula for KE is Mv²/2. So the change in KE by an increase in v of 1m/s will be
M(v+1)²/2 - Mv²/2
Take out the M/2
Algebra shows that the difference between (v+1)² and the v² terms is
v²+2v+1 -v²
=2v+1
And bringing M/2 back in, the KE increases by
M(2v+1)/2
which depends on the v you started with. So increasing from 0 to 1 is much less than increasing from 100 to 101
How can that be? I do see your problem but it's all a matter of definition and our intuition tells us wrong. KE is Frame Dependent and the result may not make sense at first but the sums don't lie.