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Mathematical genius wanted

  1. Nov 5, 2008 #1
    http://www.science27.com/formel.jpg [Broken]
    I want to calculate the area that occurs in a curve of acceleration due to gravity, - hacted area. ( g=GxmM/r^2)
    Does somebody know a formula, or able to create one?
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 5, 2008 #2
    I’m not a mathematical genius, but it seems to me that would be a fairly easy integration of g as a function of r. You have g (r) = GmM/r^2. Integrating that gives –GmM/r. You just need to plug in your limits for r over the range you are integrating. Looking at the graph, I am not sure what units of measurement you are using for g and r and what is your reference point.
     
  4. Nov 5, 2008 #3

    Kurdt

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    If you want to know the area under a curve then you need to compute a definite integral.
     
  5. Nov 5, 2008 #4
    Sorry the the formula was g=GxM/r^2
    I don't know how to: "compute a definite integral"
     
  6. Nov 7, 2008 #5
  7. Nov 8, 2008 #6
    So fare I understand, before I can intergrate I have to calculate the hacted area
    How can I calculate this area ?
    Can I use some kind of simple formula ?
     
  8. Nov 8, 2008 #7

    cristo

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    Calculating the area beneath the curve is equivalent to computing the integral. That is, you need to compute

    [tex]\int_1^8 \frac{GmM}{r^2} dr [/tex]

    Do you know how to perform this integral? If not, may I ask where this question came from: i.e. is it homework? Either way, this is a textbook-style question, thus I am moving it to the homework forums.
     
  9. Nov 8, 2008 #8
    I don’t think this advanced way really is needed; a simple formula must could do it.
    I think the area can be calculated bases on: (max) g x r/3 [minus the difference (8x1) + (7x0,125)]
    g = acc. due to gravity.
    r = distance

    I am considering how much gravity deforms space per distance.
     
    Last edited: Nov 8, 2008
  10. Nov 8, 2008 #9

    Hootenanny

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    If one wants to accurately determine the area under a curve one must compute the integral, period. Of course one may make approximations by breaking the area down into common geometric shapes, but this will always be an approximation. There are some cases where a numerical analysis is preferable, but in this case the integral can be computed trivially and so there is no need for any approximations.
    And why would you want to do that?

    The area under your curve represents the potential of the gravitational field, rather than 'the deformation of space'.
     
    Last edited: Nov 8, 2008
  11. Nov 8, 2008 #10
    Lets say that the hatch area reach the vertical and horizontal line of the diagram
    I think the formula: “(max) g” x r/3 is 100% accurate, right?

    Now it is easy to reduce the surplus
     
  12. Nov 8, 2008 #11

    Hootenanny

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    NO! Your formula is nowhere near accurate. Have you even bothered to read any of the previous replies?
     
  13. Nov 8, 2008 #12
    Hootenanny wrote:
    I still think the simple formula I have shown can do it 100% accurate.

    I do understand you do not agree, - how much wrong have you found it to be ?
     
  14. Nov 8, 2008 #13

    cristo

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    It's not a matter of "disagreement." The simple fact is that you are wrong, and you are refusing to listen to people telling you otherwise.
     
  15. Nov 8, 2008 #14

    Hootenanny

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    Feel free to believe that - and be wrong.
    I haven't explicitly calculated any numerical values, but I do know the correct form of the solution and it is sufficiently different from your formula.
     
  16. Nov 8, 2008 #15

    Kurdt

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    I don't know how you arrived at your formula and I can say that it is not 100% accurate. If you had bothered to actually learn how to do the definite integral instead of spending time trying to come up with ever more "simpler" ways of computing the area you would have had the answer a long time ago. People spent many many years trying to compute the area under a curve. The method they came up with was to use definite integrals. If you believe you can come up with a better method while trying to solve a simple problem, then be my guest. We could all be here a long long time.

    In summary, look up how to integrate and get on with it.
     
  17. Nov 8, 2008 #16
    I was wrong and lazy, - yes the suggestion was too stupid, - now I hope I have improved .

    Well it may most of the time be necessary to use advanced mathematically equations, as well the integration suggested, - but I don’t think this time.
    Why?
    Because of acc. due to gravity it self is a very simple formula. It's not because I am unwilling "to listen"

    http://www.science27.com/images/Sqaurefinla.jpg [Broken]

    http://www.science27.com/formel.jpg [Broken]
    To keep it simple I have changed the hath area
     
    Last edited by a moderator: May 3, 2017
  18. Nov 8, 2008 #17

    Hootenanny

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    Let us be clear. There are no advanced mathematical formula involved here, and the integration required is elementary, we are not the one's attempting to over-complicate things - you are. In your efforts in oversimplifying the problem, you're making it much more complicated than it needs to be.
    No, again this is simply wrong. If you would like to discuss how you reached your formula, then please post a derivation and I'm more than happy to point out where you're going wrong.

    P.S. If you'd have bothered to read my first post, you'd have noticed that I've already said that calculating the area under the curve will not return the desired quantity.

    Edit: Please stop editing your posts after I have already responded to them.
     
    Last edited: Nov 8, 2008
  19. Nov 8, 2008 #18
    No, again this is simply wrong. If you would like to discuss how you reached your formula, then please post a derivation and I'm more than happy to point out where you're going wrong.

    OK
    I will try to understand Intewgrate stuf.,

    PS!
    My new suggestion r^2/3(max)g was only online 2 minutes, - and deleted before the "edit" stamp went on, and I think 10 minutes before your reply. - I did regret that before you had answered, - according to my opinion. So I think you may be from Texas right? (fast on the tricker)
     
  20. Nov 8, 2008 #19
    I had a feeling you were a Dane. Before you go trying to understand how to calculate definite integrals, what is the highest mathematics class that you have taken so far or are in right now?
     
  21. Nov 8, 2008 #20
    Chislam
    The problem is that it is too long time ago
    But since I like science so much I proberly have to do something about it asap.
     
  22. Nov 8, 2008 #21
    I see. Well you definitely need to get refreshed on at least your basic calculus with differentiation and integration. Once you have those basics down, you'll realize how easy your problem is, at least this one in particular.
     
  23. Nov 8, 2008 #22

    Hootenanny

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    I'm afraid that I'm not the quickest typist, I should have rechecked your posts for edits before finally posting.
    If you would like to brush up on your basic calculus skills, I would recommend getting hold of a basic undergraduate text something like Stewart or Thomas. In the meantime and for the problem in hand, check out the excellent Intro/Summary of Integration by Kurdt.

    I think that it would be worth emphasising again that determining the area under the curve, will not produce this so-called "curvature" that you're after. As I said earlier, the area under the curve represents the gravitational potential rather than any "curvature".
     
    Last edited: Nov 8, 2008
  24. Nov 8, 2008 #23
    It seems to me the most sensible thing to do is to assume G, m and M are constants - the mass won't change when the radial distance changes. Then

    [tex]
    \int_1^8 \frac{GmM}{r^2} dr
    [/tex]

    is equal to [tex]
    GmM \int_1^8 \frac{1}{r^2} dr
    [/tex]

    that is, the integral of GmM/r^2 is equal to GmM times the integral of 1/r^2.
     
  25. Nov 9, 2008 #24
    Exactly, which is very easy to integrate.
     
  26. Sep 8, 2009 #25
    NEW QUESTION

    If you accelerate 1 M. per s.
    You will after 100 second move 100 m per second
    BUT how long distance have you moved in these 100 second

    The question is: how do I tell my calculate to make a such calculation ?
     
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