Conservation of KE in a moving frame

  • #1
Muu9
129
84
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).
 
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes sophiecentaur
  • #3
Muu9 said:
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.
 
Last edited:
  • Like
Likes sophiecentaur
  • #4
Muu9 said:
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.
 

Similar threads

Replies
13
Views
1K
Replies
25
Views
1K
  • Mechanics
Replies
33
Views
2K
  • Mechanics
2
Replies
53
Views
2K
Replies
9
Views
2K
Replies
7
Views
860
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
254
Replies
12
Views
1K
  • Mechanics
Replies
11
Views
980
Back
Top