Free falling object violating conservation of energy?

[Nanosuit]

Homework Statement

This is elementary level stuff and I am pretty much past this, and yet I can't seem to find a suitable way of explaining this.I was thinking about Potential Energy of a particle of mass m falling freely under gravity(ignoring air resistance, again, beginner stuff :P) from a point A 10m above a point B which is on horizontal ground, vertically below A.GPE at A is 10mg.At ground level all GPE=KE so KE=10mg.That was scenario 1, but the 2nd scenario is a bit different;lets say that the instant the particle reaches point B(just before actually touching but theoretically there-sounds weird I admit :/) the ground(or in this case, point B) drops by 10 meters.Now, GPE at A is again 10mg(remember the ground was there when it was first let go)again all GPE=KE so KE=10mg but since it drops further 10 meters so KE actually becomes 20mg!

Homework Equations

K.E=1/2mv²

P.E=mgh

The Attempt at a Solution

I tried saying that it all depends on the point from which we take the height as standard and that this isn't always at sea level.I have even included a diagram to help u better understand the situation if I happen to sound confusing.

 

[paisiello2]

When comparing two different cases for conservation of energy you always have to use the same datum points. It looks like you moved point B.

 

[dauto]

Kinetic energy will be K = 20mg and potential energy will be U = -10mg for a total energy of E = 10mg

 

[flatmaster]

I think you may be thinking that potential energy has an absolute scale and that there is some place where height=0. The exact value for the height isn't important. It is the change in height that is important. Immagine you were to perform a similar experiment in a very tall building with many floors. If you call the ground h=0, the ball may actually fall from h=174m to h=164 meters. It doesn't matter what the actual heights were, its the change in height that was important.

 

[goraemon]

As others have mentioned above, potential energy can be less than zero - it just depends on what coordinate system you use. Kinetic energy, however, can never be less than zero (because speed, as a scalar quantity, can never be less than zero).

 

[haruspex]

Kinetic energy, however, can never be less than zero (because speed, as a scalar quantity, can never be less than zero).

Or perhaps because the speed gets squared to calculate KE?

 

[Nanosuit]

I always thought GPE was absolute since GPE is a scalar...I never really thought about it that way silly me :P

Thanks a lot for the replies guys :D

 

[ian_dsouza]

Yeah. I think we call it "potential" energy because there is a potential for the associated force (in this case, gravitational force) to do work. When you drop the particle, it falls through a distance - Work = Force X Distance, for a constant force (Gravity is almost constant through the distance the particle travels in this case)

We usually measure the height from the ground because we assume that the particle does not fall below ground level and thus the GPE calculated with this height is an accurate representation of the work that can be derived from the particle - in this case it shows up as KE when the particle reaches the ground. For eg, that KE is used by hydroelectric paddles to produce electricity, with some loss in efficiency of conversion.

Strictly speaking the GPE is not zero either at ground level or 10 m below it. I think it is zero at the center of the Earth - assuming the Earth and the particle are the only two objects in the universe. If you'd place the particle at the center of the Earth, it would just sit there.

Hope this makes sense! Feel free to post constructive criticism.

 

[adjacent]

I think it is zero at the center of the Earth

Not center of the Earth. Center of mass of the Earth.

 

[Nathanael]

I think it is zero at the center of the Earth - assuming the Earth and the particle are the only two objects in the universe.

And it would be zero at an "infinite distance"! (With the same assumption.) Zero and infinity often seem to be related

 

[haruspex]

Not center of the Earth. Center of mass of the Earth.

If you're making that distinction to allow for arbitrary density distributions, that won't do it either. A system consisting of a particle mass 10kg at x=-1 and a 1 kg particle at x=10 has a mass centre at x=0, but that won't be the lowest potential position for a test particle.

 

[adjacent]

If you're making that distinction to allow for arbitrary density distributions, that won't do it either. A system consisting of a particle mass 10kg at x=-1 and a 1 kg particle at x=10 has a mass centre at x=0, but that won't be the lowest potential position for a test particle.

I assumed that the Earth and the particle were the only two objects in the Universe.

 

[ian_dsouza]

If you're making that distinction to allow for arbitrary density distributions, that won't do it either. A system consisting of a particle mass 10kg at x=-1 and a 1 kg particle at x=10 has a mass centre at x=0, but that won't be the lowest potential position for a test particle.

Where would you say is the point of lowest potential in this case and how do you arrive at the conclusion?

 

[voko]

Strictly speaking the GPE is not zero either at ground level or 10 m below it. I think it is zero at the center of the Earth - assuming the Earth and the particle are the only two objects in the universe.

You need to understand that the law that says $mgh$ is only an approximation of the true law. It becomes inaccurate as go to any significant distance above or below the surface of the Earth. It is definitely wrong near the centre of the Earth, no matter how the "centre of the Earth" is defined.

 

[voko]

Where would you say is the point of lowest potential in this case and how do you arrive at the conclusion?

You would need to start with the correct equation for potential energy of a test particle with respect to a massive particle; then recall how you obtain potential energy in a compound system.

 

[adjacent]

\It is definitely wrong near the centre of the Earth, no matter how the "centre of the Earth" is defined.

Why? Can you elaborate?

 

[voko]

Why? Can you elaborate?

The approximate law assumes that the force of gravity remains the same everywhere. That is why it is simply linear in the distance. But as you go both up or down from the surface of the Earth, the force of gravity decreases and becomes zero at both the centre of the Earth and infinitely far away from it.

The above assumes that the Earth is perfectly symmetrical and its density also is centrally-symmetric. It is not really the case, so the above is also an approximation, albeit a much more accurate one.

 

[adjacent]

the force of gravity decreases and becomes zero at the centre of the Earth

So Gravitational potential energy at the centre is $0J$?

 

[voko]

So Gravitational potential energy at the centre is $0J$?

Not necessarily. What we can say is that because the force of gravity is zero at the centre, the potential energy has a minimum or a maximum there. Analysing this further, we can conclude it has a minimum there. Whether this minimum is zero or not is entirely up to us, because we can always add any constant to potential energy, which won't change anything physically.

 

[haruspex]

I assumed that the Earth and the particle were the only two objects in the Universe.

The two particle example is to illustrate that when mass is arbitrarily distributed the lowest potential point in the field it generates is usually not at its mass centre. This applies also to a sphere with arbitrary density distribution.
Consider a solid sphere radius r with uniform density, except for a spherical inclusion radius s at distance a from its centre, O (centre to centre distance). Let the larger sphere have mass M, and the inclusion have 'extra' mass m. Consider a point distance x from O towards the centre of the inclusion, but not inside the inclusion (say). The least potential will be where the two attractions balance: $\frac{Mx}{r^3} = \frac{m}{(a-x)^2}$. But the common mass centre is given by $ma = (M+m)y$, where y is its distance from O.