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Gravitational Potential Energy

  1. Jul 23, 2016 #1
    1. The problem statement, all variables and given/known data
    A body of mass m is taken at a constant speed from the surface of the earth (radius = Re) to infinity.
    (a) What is the work W1 done on the body in the process?
    (b) If m is taken from a distance r > Re to infinity, how much work W2 is required?
    (c)Which quantity is larger, W1 or W2? Why?
    (d) How much work is required to take the body from Re to r?
    (e) The body is dropped from r with no initial velocity. Find an expression for the velocity with which it hits the earth's surface. (Neglect all frictional forces)
    (f) With what minimum velocity must m be thrown upward in order to reach an altitude h = r-Re
    (g) If m is thrown with the above velocity upward but not exactly vertically, will it reach the above altitude? Explain your answer.
    (h) What is the potential energy U of the body at altitude h?
    (i) What is the potential energy U of the body at the surface of the earth?
    (j) Which is more, the potential energy at the surface or that at altitude h?How much more?
    (k) Is the work done on m taking it from the surface to altitude h equal to the increase in potential energy of m?

    2. Relevant equations
    U(r) = -GMm/r

    3. The attempt at a solution
    For a) I did an integral from Re to infinity and got W1=GMm/R
    For b) I did the same from r to infinity and got W2= GMm/r
    For c) I took did GMm/r - GMm/Re = GMm (1/r - 1/Re) so W1>W2
    For d) I took the integral from Re to r and got GMm ( -1/r + 1/Re)
    I'm not sure of (e) onwards
    For e) I have an idea that K.E = P.E so 1/2 (mv^2) = U but I'm not sure what U in this case will be. I know it can't be negative.
     
    Last edited by a moderator: Jul 23, 2016
  2. jcsd
  3. Jul 23, 2016 #2
    Addendum to my attempted solution:
    I think for part h) it should just be U = -GMm/ (Re +h)
    and part i) should just be either U=mgh or -GMm/Re
    for j) we subtract the 2 and see I presume

    for e) I believe the equation we'll be using is K1 + U1 = K2 + U2
    K1 will be 0 because and U1 will be -GMm/r and K2 is 1/2 mv^2 and U2 is -GMm/Re?

    And we do the same for f but K2 would be zero in this case?
     
  4. Jul 23, 2016 #3

    SammyS

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    You don't actually state what quantities you are integrating

    Your answer for part c) is correct, but not very convincing. What you have there is actually W2 - W1 which is negative if W1 > W2 . To show that 1/r - 1/Re is negative, write it as a single fraction with a common denominator.

    Your answers to parts a) through d) are correct.

    For e): It's more proper to say that the increase in KE is equal to the decrease in PE .
    You say that U (potential energy) can't be negative. However, in this case we usually consider U to be zero at infinite distance and that is the case where U is maximum. Therefore, U is negative at any closer distance. Also, notice your formula for U.
    U(r) = -GMm/r​

    h) is incorrect . That's not how to subtract fractions - if that's what you intended.

    Part i): U = mgh is good for potential energy near Earth's surface with U = 0 at the surface. - not good for this problem.

    Part j): Use the basic idea behind potential energy .See if subtracting the two values is consistent with this.
     
  5. Jul 23, 2016 #4
    Thank you for your reply.

    Why do we have to subtract fractions for part h? It just asks for the potential and so substituting for r in
    U(r)=-GM/r as (Re+h)? Or no, actually I think it's the potential at r - potential at Re.

    And how would you do part f? Similar way to part e?

    For part e) I get (-GMm/r) = 1/2 mv^2 but that doesn't seem right because you can't take the root of a negative number? Or would I need to take into account the potential at Re as well
    So -GMm/r = 1/2 mv^2 + (-GMm/Re)? but even then -GMm/Re + GMm/r is still negative because r > Re and so - 1/Re + 1/r will be negative?
     
    Last edited: Jul 23, 2016
  6. Jul 23, 2016 #5

    SammyS

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    You were correct. I had a brain cramp.

    It's the change in potential energy that you need for part e). (Be careful of signs.)
     
    Last edited: Jul 23, 2016
  7. Jul 23, 2016 #6
    If we do it as K1+U1=K2+U2 or take it as increase in kinetic energy= decrease in potential energy we still get
    1/2 mv^2 = (-GMm/Re) - (-GMm/r) which means the RHS is negative-still confused about this.
     
  8. Jul 23, 2016 #7
    Ok no, I get it now for part e. Initially the velocity was 0 so the KE at r is 0.
    So if we use K1 + U1 = K2 +U2 where 1 is at r and 2 is at Re we get
    0 - GMm/r = 1/2 mv^2 - (GMm/Re)
    So -GMm/r + GMm/Re =1/2 mv^2
    GMm ( -1/r + 1/Re) = 1/2 mv^2
    Now -1/r +1/Re is positive
    Is that correct?

    As decided for h it will be U=-GMm/(Re+h) at altitude and i) U=GMm/Re for the surface
    Then for j we can just look at the inverse proportionality between U and the distance. As you increase the distance, U increases as well (presence of negative sign). So it's greater with Re+h? To find how much, we can just subtract the two?
    k) is yes because the work done =change in energy


    Just not sure of f and g now.
     
    Last edited: Jul 23, 2016
  9. Jul 23, 2016 #8

    haruspex

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    That all looks right.
    For f, if it is thrown up with the minimum velocity to reach height h, what will its velocity be when it gets there?
     
  10. Jul 23, 2016 #9
    It will be 0 so k2 will be zero, our equation will then be
    K1+U1=K2+U2
    1/2 mv^2 + (-GMm/Re) = 0 + (-GMm/r)?
     
  11. Jul 23, 2016 #10

    haruspex

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    Right.
    What about g? Velocity is a vector, so what else can you deduce from the fact that velocity will be 0 at height h?
     
  12. Jul 23, 2016 #11
    g will decrease?

    Sorry the equation will be 1/2 mv^2 + (-GMm/Re)= 0 + (-GMm/(r-Re)
    since h=r-Re?
     
  13. Jul 23, 2016 #12

    haruspex

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    No, I meant question g.
    No, you were right in post #9, but you can cancel a common factor.
     
  14. Jul 23, 2016 #13
    But they're asking for U at h and not r?
    It's going to break up into components because it's not exactly vertical now?
     
  15. Jul 23, 2016 #14

    haruspex

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    Yes, they want the answer in terms of h, not in terms of r, but your equation in post #9 is correct and your equation in post #11 is wrong. Have another go at converting the post#9 equation from using r to using h.
    Yes, but that's not quite what I was hinting at. If you know the velocity as a vector is zero at h, what does that tell you about those components?
     
  16. Jul 24, 2016 #15
    Ohhh sorry!!! My bad, if h=r-Re then r=h+Re so the denominator of the r fraction will be h+Re

    So I got
    1/2 mv^2 + (-GMm/Re) = 0 + (-GMm/r)
    1/2 mv^2 = (-GMm/r) + (GMm/Re)
    1/2 mv^2 = GMm ( -1/r + 1/Re)
    1/2 v^2 = GM ( -Re + r / rRe) but h=r-Re so
    v^2= 2GM (h /rRe)
    and v is the square root of that.

    So vx^2 + vy^2=0?

    Components equal and opposite to one another?
     
    Last edited: Jul 24, 2016
  17. Jul 24, 2016 #16

    haruspex

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    Yes.
    Yes.
    No.
    Yes.
    No.
     
  18. Jul 24, 2016 #17
    Ok, I get it.
    I'll have to substitute for r=Re+h in (-1/r + 1/Re)

    -1/(h+Re) + 1/Re = -Re+Re+h / Re (Re+h) = h / Re(Re+h)
    So v^2 = 2GM ( h / Re(Re+h))

    Components=0
     
    Last edited: Jul 24, 2016
  19. Jul 24, 2016 #18

    haruspex

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    Yes, and yes.
     
  20. Jul 24, 2016 #19
    Yes!
    So for g it won't be able to reach this altitude then?
     
  21. Jul 24, 2016 #20

    haruspex

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    That's right, though to complete the argument you need to say why the horizontal component will not be zero at the top of the trajectory.
     
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