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Change in GPE of the moon as the radius increases. Getting extreme answers

  1. May 17, 2012 #1
    Okay, first of all, sorry for not posting here, I unknowingly posted this q in the wrong section and got an infraction. Sorry again


    1. The problem statement, all variables and given/known data

    Question : Calculate the change in gravitational potential energy of the moon, when the radius of it's orbit increases by 4cm.

    Radius : 3.85*10^8 m
    Mass of the Earth : 6*10^24 kg
    Mass of the moon : 7.36*10^22kg

    We were first told calculate the force between the moon and the earth using
    F = Gmm/r^2. Comes to around 2*10^20 N
    We were then told to find the change in G.P.E when the radius increases using the force which we calculated.

    2. Relevant equations

    F = GMm/r^2

    GPE = GMm/r

    3. The attempt at a solution

    Method I : Work done = Force * Distance. 2*10^20 * 0.04 = 8*10^18J

    Now this is what my friend did, and why I think this is incorrect? Because he assumes that the force is constant as the radius changes, which clearly isn't. Why do I think he is correct? Because he does what was asked, he used the force to find the change in gpe.

    Method II (This is what I did) : Potential energy of a mass m at a distance r is = -GMm/r
    (Asked to calculate the change, hence i'll use the modulus of that and ignore the minus)

    GMm(1/r2 - 1/r1) [Where r2 is 3.85*10^8 + 0.04 and r1 is 3.85*10^8)

    When you go ahead and solve this using your calculator, you realize that the change in the radius is soooo less, even the calculator won't show up the change. Hence the value you get is 0J.

    Why I think i'm correct? Because the change is the GPE shouldn't be much because the radius didn't change a lot. And because I used a perfect formula. (?)
    Why I think i'm wrong? Because I didn't use the force to calculate.

    My dilemma is clear, these are 2 extreme answers, 8*10^18 J and 0J.

    Someone please help me out here!
    Thanks a lot :D
     
  2. jcsd
  3. May 17, 2012 #2

    gneill

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    Staff: Mentor

    Since the force changes with distance, why not calculate the energy symbolically first by integrating the force over the distance? Suppose the Moon starts at radius ro and moves to radius ro + x, where x will be your 4cm. You might find an opportunity to spot a useful simplification/approximation in the result given the significant figures available.
     
  4. May 17, 2012 #3

    collinsmark

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    Homework Helper
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    Yes, you are technically correct. :approve:

    Keep in mind though, the difference in the force in this case is negligible, moving such a short distance of 4 cm. Realize that the approximation your friend made (and was told to do, I presume), is the same approximation that is made every time for modeling g = 9.81 m/s2 on the surface of the earth, and that gravitational potential energy = mgh. Every time you model gravitational potential energy using mgh, you're making the same sort of approximation.

    But you are correct, you could make the argument that a distance h above the surface of the earth, compared to the surface of the earth, that [itex] \mathrm{Change \ in}\ P.E. = -GMm\left( \frac{1}{r_e + h} - \frac{1}{r_e} \right) [/itex] (where [itex] r_e [/itex] is the radius of the Earth) is more accurate. And it is. At least it works for very large and small h alike, even where h is comparable to, or greater than the radius of the Earth. But for small h (where h << [itex] r_e [/itex]), mgh is a very good approximation.
    Try this: Find a common denominator for (1/r2 - 1/r1). It will work much better with your calculator that way. :wink:
    Find a common denominator for r1 and r2 in your formula. When finished, and the formula is slightly re-written, it shouldn't cause any problems with your calculator.
     
    Last edited: May 17, 2012
  5. May 17, 2012 #4

    Janus

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    Staff Emeritus
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    Gold Member

    The problem is that your calculator doesn't carry out enough digits.

    Try this: add 3.85e8 + 0.04 on your calculator, you'll likely get an answer of 385000000. Here's why: The calculator rounds to 10 significant digits, and 385000000.04 has eleven. Since the last digit is smaller than 5, it rounds it down to zero and you essentially lose the 0.04.

    If instead, you were to add 3.85 + 0.05, you would get an answer of 385000000.1, because now it will round up the last digit.
     
  6. May 17, 2012 #5
    Thank you very much for the replies.

    @Janus : I predicted so, thanks for the info though :)

    @collinsmark : I thought my method was more accurate. Although, is getting an answer very close to 0 J practical? And the method my friend used is clearly incorrect isn't it? I mean, does using this formula work better Work = ΔF * Δd ? What he did was W = F*Δd. Big difference clearly. A change of 10^18J when the radius changes my 0.04m, does that even make sense to you guyz?

    And thanks for the common denominator method, will try that now.

    @gneill : Will try that, thanks.
     
  7. May 17, 2012 #6
    Okay, i get it, it does make sense actually. I used wolfram alpha to calculate the tiny change and when multiplied by GMm, it comes to 10^18J. Whoa, that is something huge, didn't expect that.
    Fd gives the same answer. Bleh, losing those 2 marks :(
     
  8. May 17, 2012 #7

    collinsmark

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    In case you didn't understand what I meant about the common denominator, I'll give you a prod.

    Start with

    [tex] \left ( \frac{1}{r_2} - \frac{1}{r_1} \right)[/tex]

    The common denominator is the multiplication of both denominators, [itex] r_1r_2 [/itex]

    [tex] \left( \frac{1}{r_2} - \frac{1}{r_1} \right)= \left( \frac{r_1}{r_1r_2} - \frac{r_2}{r_1 r_2} \right) = \left( \frac{r_1 - r_2}{r_1r_2} \right)[/tex]

    Noting that [itex] r_1 - r_2 = -0.04 \ \mathrm{m} [/itex], you should be able to plug that into your calculator without causing truncation problems.

    Yes, your method is more accurate. I'm just saying that the approximation works very well for small Δd.
    No. That's a truncation problem caused by your calculator's inability to store enough significant figures (as Janus and gneill have mentioned).
    I wouldn't call it incorrect, no. Rather I'd call it a very good approximation, albeit still an approximation.
    I'm not sure if you are in a calculus based class, but if you are, the *real* work formula is

    [tex] dW = \vec F \cdot \vec{ds} [/tex]

    (I substituted s for d, because dd just looks confusing). So in this situation,

    [tex] W = \int_{r_1}^{r_1+d} G\frac{Mm}{r^2}dr = -GMm\left( \frac{1}{r_1 + d} - \frac{1}{r_1} \right) [/tex]

    which is essentially the same thing that you are doing.
    [Edit: The moon is pretty massive. So an answer of order of magnitude of 1018 J sounds pretty reasonable to me.]

    And a separate exercise, try doing the common denominator thing on the more accurate "surface of the Earth" equivalent to this problem.

    [tex] \mathrm{Change \ in}\ P.E. = -GMm\left( \frac{1}{r_e + h} - \frac{1}{r_e} \right) [/tex]

    See if you can understand why it simplifies to mgh for h << re. :wink: If you do you'll realize why mgh is a good approximation.
     
    Last edited: May 17, 2012
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