Potential energy in a moving reference frame

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Consider an object dropped from a height h above the earth's surface. Observe the motion from the following two reference frames :-

1. The frame fixed to the earth's surface :-
Initially,
Potential energy of the object = [tex]mgh[/tex]
Kinetic energy of the object = [tex]0[/tex]
Finally,
Potential energy of the object = [tex]0[/tex]
Kinetic energy of the object = [tex]mgh[/tex]

2. A frame moving towards the earth's surface at a constant speed of [tex]\sqrt{2gh}[/tex] :-
Initially,
Potential energy of the object = [tex]mgh[/tex]
Kinetic energy of the object = [tex]\frac{1}{2} m(\sqrt{2gh})^2[/tex] = [tex]mgh[/tex]
Finally,
Potential energy of the object = [tex]0[/tex]
Kinetic energy of the object = [tex]0[/tex]


It appears that in the second case, the energy is not conserved. What's the flaw in the above reasoning?
 

Answers and Replies

  • #2
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Vinter, according to the moving frame of reference in the second case, not only the object is initially moving, but also the Earth.
So I think the flaw is this: If you measure the potential energy relative to the Earth, then the speed of an object should also be measured relative to the Earth.
 
  • #3
Doc Al
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The potential energy is a property of the earth-object system, not just a property of the object. Energy is conserved in both frames.

In both frames, the change in PE of the system is -mgh.

In the earth frame:
change in KE of earth = 0
change in KE of object = mgh

In the moving frame:
change in KE of earth = 2mgh
change in KE of object = -mgh

In both frames, energy is conserved.

Note: It is perfectly OK to treat the object as an isolated system in a gravitational field, if done properly. The "PE" of the isolated object does depend on the frame, as it reflects the work done against gravity. Say the earth is on the left, the object on the right. In the earth frame, the object moves to the left a distance h, so the change in PE is -mgh. But in the moving frame the object is seen to move to the right (opposite to the field) a distance h, thus the change in PE is +mgh. Again, energy is conserved in both frames.
 
  • #4
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OK, forget about the earth. Suppose you are in space occupied by a uniform gravitational field in the positive x direction. You are in a frame moving towards the negative x direction at a speed of v. You have a ball of mass m in your hand. To define the potential energy, you take the origin as the reference point, i.e., the zero potential point. Now, in your reference frame, the ball is throughout stationary, so no change in kinetic energy. However, the distance between the ball and the potential energy reference point (the origin) is continuously changing and so is the potential energy of the ball. What's wrong?
 
  • #5
Andrew Mason
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vinter said:
2. A frame moving towards the earth's surface at a constant speed of [tex]\sqrt{2gh}[/tex] :-
Initially,
Potential energy of the object = [tex]mgh[/tex]
Kinetic energy of the object = [tex]\frac{1}{2} m(\sqrt{2gh})^2[/tex] = [tex]mgh[/tex]
Finally,
Potential energy of the object = [tex]0[/tex]
Kinetic energy of the object = [tex]0[/tex]

It appears that in the second case, the energy is not conserved. What's the flaw in the above reasoning?
Measurement of distance and speed depends upon the frame of reference. Since they are frame dependent quantities, potential and kinetic energy must be measured in the same frame. The initial and final potential energy of the object is measured in the earth frame but you are measuring its kinetic energy in the moving frame. It is no wonder they aren't conserved.

AM
 
  • #6
russ_watters
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vinter said:
What's wrong?
Nothing is wrong. You're calling the ball "stationary" and "moving" at the same time. You can do that, you just have to remember that you're dealing with two different rest frames at the same time.
 
Last edited:
  • #7
Doc Al
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vinter said:
OK, forget about the earth. Suppose you are in space occupied by a uniform gravitational field in the positive x direction. You are in a frame moving towards the negative x direction at a speed of v. You have a ball of mass m in your hand. To define the potential energy, you take the origin as the reference point, i.e., the zero potential point. Now, in your reference frame, the ball is throughout stationary, so no change in kinetic energy. However, the distance between the ball and the potential energy reference point (the origin) is continuously changing and so is the potential energy of the ball. What's wrong?
First off, if you are in a uniform gravitational field, yet are moving with a constant speed v, then there is an external force acting on you! (You could be in a spaceship with thrusters on to balance the gravitational force.)

Potential energy is a way to bookkeep work done against/by gravity. The reference point for PE must be in the same frame that measures the KE, not a different one!

In the "moving" frame, gravity does no work on the ball -- the ball doesn't move. Pick any point in that frame as your "zero potential point", it doesn't matter. No change in PE. Net work done on ball = 0, change in KE = 0.

In the "stationary" frame, gravity does do (negative) work on the ball. So PE does increase. But the external force is also doing work on the ball. Work done by external force = change in PE. Change in KE = 0.
 
  • #8
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I think I am getting it now...
Should I say, that the reference point for measuring the potential energy should be stationary in the reference frame you are measuring it?
 

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