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Change in Potential Energy levels of a body in Gravitational field.

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Now , I Will Begin by stating the calculation and i will post the query at the end.

    so here we begin,
    we know that If W is the work done by a conservative force,
    potential energy change is given by:
    ΔW= - ΔU .....(1)
    ie when work done by conservative force is positive , Potential energy decreases ..right?

    now im going to derive the work done in lifting a body up from say earth's surface (r1) to some height(r2) let small body be of mass 'm' and the Huge body (earth ,say) be 'M".

    now F(conservative force) ,points downwards ,and the displacement is upwards
    so, dw = -Fdr
    integrating over the limit , r1 to r2,(Using Newtons Law F = GMm/r^2)
    ............r2.....................................r2
    ∫dw = -∫fdr → w = - GmMv ∫(1/r^2)dr → w= GmM(1/r2 - 1/r1)
    ...........r1..................................... r1

    now since r2>r1 (Its Going up) , W is Negative , and Hence by (1) Potential Energy Change is Positive (Which is true),
    but Suppose we consider a case where a body 'Falls' from r1 to r2

    now here F(gravitational force) and dr(displacement) both Point along the same direction.


    so , dw = fdr

    ..........r2 ............................. r2
    ∫dw = ∫fdr → w= GMm(-1/r) → w= GMm(1/r1-1/r2)
    ..........r1................................r1


    Now Since r1>r2,
    w is coming negative
    and then by ....(1) ΔU is coming Positive .... I mean How is THis ? THe Change in Potential energy levels should be negative.

    Now i do know that potential energy at any point is by convention considered negative , but this here is the 'change' and it should come negative too , but it doesnt ... i dont know why this is happening ? can anyone explain ,where i am going wrong ?

    if i just solve it using change in energy levels,im getting it right , but why not this way ?

    NOTE : Here i am thinking of the work done by the gravitational force , which is pointing downwards
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 15, 2012 #2

    Doc Al

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    Staff: Mentor

    The work done by gravity is always W = ∫Fdr. Using 'r' as your variable of integration, the Force of gravity is always negative (in the direction of -r), F = -GMm/r^2.

    The sign of the work is determined by the direction of integration. Let r1 < r2. So, when the body moves up the work is negative:
    [tex]W = \int_{r_1}^{r_2} (-GMm/r^2)dr = GMm(1/r_2 - 1/r_1)[/tex]
    And when going down, the work is positive:
    [tex]W = \int_{r_2}^{r_1} (-GMm/r^2)dr = GMm(1/r_1 - 1/r_2)[/tex]
     
  4. Sep 16, 2012 #3
    So does it mean that when the body is coming down, the gravity acts along -r? ie, in upwards direction? why? if you can explain
     
  5. Sep 16, 2012 #4

    Doc Al

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    Staff: Mentor

    Gravity acts down, of course, which is always in the -r direction. That's true regardless of whether the body is moving up or down.
     
  6. Sep 16, 2012 #5
    I'm afraid Al I can't understand it properly, can u please be more specific?
    ie, while going down
    dw =f.dr (dot product ) = fdr (angle is zero right?) = -GMm /r^2.dr ..

    while going up ,
    dw =f.dr = -fdr (since angle is 180° ,so cos180°= -1)
    = -(-GmM /r^2.dr )

    is that it?
     
  7. Sep 16, 2012 #6

    Doc Al

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    Staff: Mentor

    I thought my previous post was quite specific. :smile:

    In any case: W = ∫F.dr

    When going up, F acts down and the displacement is up, so the work done by gravity is negative.

    When going down, F acts down and the displacement is down, so the work done by gravity is positive.

    But realize that the integration limits take care of the sign for you. Note that 'dr' is positive along the +r direction. So when you integrate from r1 to r2, which represents going up, the displacement is in the direction of positive r; when integrating from r2 to r1, which represents going down, the displacement is in the negative direction, opposite to dr.
     
  8. Sep 16, 2012 #7

    so sir, that whenever I express work done by gravity dw = fdr ,and I don't need to express the sign by taking into account the angle? I mean why does this happen? because of the integration is it?

    I'm sorry if I am irritating you, but I just want to understand
     
  9. Sep 17, 2012 #8
    I'm sorry for this previous ,now I clearly understand the logic behind this. thanks a lot

    :)
     
  10. Sep 17, 2012 #9

    Doc Al

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    Staff: Mentor

    Excellent. (I was just about to respond to your previous post. Sorry for the delay. I'm glad you have it all straight now.)
     
  11. Sep 23, 2012 #10
    indeed ,thanks for your help , i guess i was ignorant about the change in sign of small displacement 'dr' by the change in direction of integration itself , so i in my ignorance was actually jumbiling it both together

    W =F.dr and dw=F*dr
    now i realise that the signs of the forces and displacement automatically decide the sign and by using the dot product again on top of that , i was messing up the result .
    THANK YOU!
     
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