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Homework Help: Dimentional Analysis problem

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data

    2. An object, initially still at a distance from the sun of [tex]1,5\times 10^{8}[/tex] km.
    Suppose the body is only influenced by the sun's gravitational field:
    2.2 Recurring to dimensional analysis, make a prediction on the time it takes for the body to fall in the sun (in days).

    - The answer expected is aproximately 58 days

    2. Relevant equations

    I'm guessing :
    [tex]F = m a[/tex]
    [tex]F = G\frac{m1.m2}{R^{2}}[/tex]

    and possibly:

    [tex] x = x_{0} + v_{0} + \frac{a \times t^{2}}{2} [/tex]

    3. The attempt at a solution

    I'm pretty much at a loss here:

    The statement clearly asks for dimentional analisys, but i cannot relate time with a length only, since I have to make a prediction:
    I'm guessing the prediction means to try to find an equation that's dimentionally correct, without concern for the adimentional constant.

    So, what I tried so far has pretty much nothing to do with dimentional analisys, because i really can't see what to do.

    I have, indeed tried some things but don't get anywhere near the supposed solution (aproximately 58 days).

    What I tried :
    (note, this is not what the problem asks, since it isnt dimentional analisys)

    [tex]x = \frac{a \times t^{2}}{2}[/tex]

    and using 1) and 2) :

    [tex]a = G\frac{m2}{R^{2}}[/tex]

    x = R

    [tex]R = G\frac{m2 \times t^{2}}{2R^{2}}[/tex]
    [tex]t = \sqrt{\frac{2R^{3}}{G.m2}}[/tex]

    and, substituting R for the value given on the statement, G for the gravitational constant, and m2 for the mass of the sun, the result has nothing to do with the expected (58 days)
  2. jcsd
  3. Jan 17, 2010 #2


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    Homework Helper

    If the question asks you to use dimensional analysis, that means you don't need to use any equations relating to kinematics or forces. Just get the units right.

    Intuitively, you can conclude that the time, t, depends on G, the mass of the Sun, and the initial distance. It shouldn't depend on the mass of the falling object. G has units of m^3/kgs^2, mass kg, distance m. If you play with G, m, and d, you'll get

    t = \sqrt{\frac{R^{3}}{G.m2}}

    which is 58 days. Note that [tex]
    t = \sqrt{\frac{2R^{3}}{G.m2}}
    [/tex] is equal to 82 days, which is very close to 58 as far as approximations are concerned.
  4. Jan 17, 2010 #3
    This question seems strange to me, you are basically just making the most extreme approximation possible by setting everything equal to one? Why does this work? I would never think to use dimensional analysis to obtain an approximation like that.
    Last edited: Jan 17, 2010
  5. Jan 17, 2010 #4
    well, i'm not sure myself. I think it's somewhat introductory to dimentional analisys, so probably it's not concerned about results, and more about process.

    And thanks!
  6. Jan 17, 2010 #5
    Dimensional analysis is simply the act of analyzing the units of your equation. This is not any way I have ever used dimensional analysis, and I think its almost strange to call it that.
  7. Jan 17, 2010 #6


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    Homework Helper

    Yeah, pretty much. And it doesn't always work. But for simple problems you can usually get away with making a rough approximation this way, for a few reasons:

    - Most simple physics formulas are purely multiplicative, no addition or subtraction (or if there is, usually one of the terms is by far the largest so the others can be neglected)
    - If you know your physics, it's easy to make an intuitive guess as to what any given quantity depends on
    - Often there's only one way to put the various quantities together to get the right units in the end
    - Most numerical constants are close to 1

    Analyzing the units of your equations to make sure they're correct is only the most obvious application of unit analysis.
  8. Jan 18, 2010 #7

    now that I checked the math, are you sure that it's 58 days?

    I mean, if we substitute for the given data:

    t = \sqrt{\frac{(1.5 \times 10^{8})^{3}}{(6.67 \times 10^{-11}) \times {1.98 \times 10^{30}}}}

    we get :

    [tex]t = \sqrt{25555.40412} [/tex]

    [tex]t = 159.86 [/tex]

    which, in seconds is the equivalent to little more than 2 minutes, and it's a quite absurd result, i think
    did i get the math wrong?
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