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Possibly incorrect professor statement

  1. Aug 28, 2014 #1

    RMZ

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    My professor today stated that if a ball was to fall a certain height, then the work done by earth's gravity on the ball would equal the work done on the earth by the ball's equal and opposite gravitational pull. W=F*d means that the earth must travel the same distance as the ball. Upon someone asking him about it, he responded that it would take longer for the earth to travel the same distance. So if the ball falls 300 meters, the earth would eventually rise 300 meters in the direction of the ball's gravitational pull? Has it been awhile for him or am I missing something? can you please give me your input on this particular matter?

    I think, say in an elastic collision perpendicular to Fg, W=ΔK could be used to find the amount of work done on the ball if it hits something. By Newton's third law, the force on the ball will equal the force on the object it is colliding with, and while they are exerting a push on eachother they will be in contact and therefore will move together. So the work done on the object by the ball=the negative of the work done on the ball by the object=-ΔK of the ball. Therefore the amount of kinetic energy that was lost by the ball can be said to have been used to do work on the object. right? So I can kind of see where a relationship such as the one my professor described might be seen as possible at first, but after some investigation I cannot see how what he said can be true.

    Also, I remember him insinuating that the declining potential energy of a falling ball=increasing potential energy of the earth. But that makes absolutely no sense to me, as the ball's potential energy is being converted into kinetic energy. Even if I view the earth as falling towards the ball, the earth's capacity to do work due to its position in the ball's gravitational field is being turned into kinetic energy of the earth as they come closer together. I cannot see how in the world it would increase.

    Please help! If I can just either know that he is wrong, or see mathematically that the work on the ball by the earth =the work on the earth by the ball, I can rest easy.
     
  2. jcsd
  3. Aug 28, 2014 #2

    Matterwave

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    Your professor...seems to have completely confused himself...

    The potential energy between two objects can not be said to be stored in either object. It is "stored" in the configuration itself if anything. The potential energy of a system of two masses separated by a distance d is given by ##U=-Gm_1 m_2/R##. One does not say that either mass 1 or mass 2 "posses" a portion of this energy, the energy is the energy of this particular configuration of the 2 masses.

    In the case of the Earth-ball system, we make the approximation that the Earth is so massive that it doesn't move (note this approximation) and so we can regard the ball only, and in that case it is easy to ascribe it a potential energy like mgh (which is also an approximation of course of the real formula for small heights in comparison to the radius of the Earth).

    In this approximation, the Earth does not move. That is the only way you can make sense of ascribing to the ball all of the potential energy. Using this approximation, it makes no sense to talk about the work done by the ball on the Earth, since the Earth literally does not move. Indeed, this assumption is the one allowing us to stay in the Earth's frame of reference since it is only inertial if it doesn't move!

    Of course, in real life, the Earth does move, but only by a tiny amount. In that case, one would, strictly speaking, have to switch to using the center of mass frame, as that is now the only inertial reference frame around (the Earth being no longer inertial since it moves). In the center of mass frame, the ball does almost all the moving, and the Earth does basically none of it. The work done by the Earth on the ball is much more than the work done by the ball on the Earth. The Earth gains almost no kinetic energy, while the ball gains almost all of it.

    One can do some math to figure it out. Say the ball has mass ##m=1kg##. The COM of the system obviously lies along the line which connects the two centers. Along this line, the distance from the center of the Earth to the center of mass is given by:

    $$d=\frac{m}{M}(h+R)$$

    Where h is the distance from the surface of the Earth to the center of the ball, R is the radius of the Earth and M is the mass of the Earth. Now, we are dropping this ball from some height (let us approximate the ball as a point so we don't have to worry about its radius) and then it hits the surface of the Earth. This means that the CoM, as measured as a distance from the Earth's center changes:

    $$\Delta d = \frac{m}{M}(h+R)-\frac{m}{M}(R)=\frac{m}{M}h$$

    This ##\Delta d## is actually how much the Earth's moves in the CoM frame, while the ball moves the rest of the height h (the separation between the Earth and the ball decreases by h, and so the CoM is regarded as stationary while the Earth moves ##\Delta d## and the ball moves ##h-\Delta d##.

    Let us use some round numbers. Let's estimate ##M_E \approx 6\times 10^{24}kg## and ##R\approx 6000km##. Using these numbers, we find:

    $$\Delta d\approx 2\times 10^{-24}m$$

    Which is miniscule, so that our original approximation was pretty good.
     
  4. Aug 29, 2014 #3

    Andrew Mason

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    To follow up on Matterwave's comments, the forces are equal and opposite. So the work done on each is proportional to the distance moved relative to the centre of mass. The earth stops accelerating when the ball hits the earth and stops. The work done on the earth (in the inertial centre of mass frame of reference) is a tiny, tiny fraction (mball/Mearth) or roughly 1/1025) of the work done on the ball.

    AM
     
  5. Aug 29, 2014 #4

    RMZ

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    Thank you guys so much. I can't stand the guy. i had a pretty bad headache for a while because of that guy. I just have a few questions before committing this to my brain:


    1. You are both using the COM reference frame, but i feel weird using a COM reference frame since gravity is accelerating them. Is the COM going to stay the same even if the ball bounces back and both several times? The calculation further down seems to suggest so, but i'd like to verify it.

    2. And if it doesn't move... Are we calling the COM frame reference inertial since the center of mass doesn't move (I hadn't used that term for a reference frame before)? I did a rough calculation using the assumption that the earth and the ball experienced uniform (although unique, of course) acceleration and came up with:

    MEΔVE=ME*aE*Δtfrom fall to collision
    =ME(-Fg/ME)*Δvball/aball
    =-mballΔvball.

    Which means that,
    MEVE+MballVball=0
    which suggests that VCOM=0 since VCOM=(MEVE+MballVball)/(ME+mball

    But I feel OCD worried that it might be different if i were to do this using integrals in the proof.

    If VCOM=0, then i would see that the normal forces of the ball and the earth would mean that both don't move anymore after settling (once the ball doesn't bounce anymore).
     
  6. Aug 29, 2014 #5

    nrqed

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    By definition, the center of mass of a system moves at constant velocity along a straight line if there are no external forces (in this case, this would mean ignoring the force of gravity of the Sun, of the Moon, etc. )
     
  7. Sep 10, 2014 #6

    RMZ

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    Thank you guys so so so much.
     
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