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Dimensional Analysis Question

  1. Aug 29, 2005 #1
    First off, I'm glad I found a forum like this! Thank jeebus!

    I'm a new college physics student thats been out of the science/math arena for about 2.5 years. I'm a Software Engineer by day but I"m going back to get my mechanical engineering degree. Unfortunately I build business systems all day and physics nor calculus is used much in these applications that I build, hence why I'm kind of off the rocker on this stuff now.

    I've been pouring over these problems ALL DAY LONG and I've beat my self up (mentally) because I cannot figure them out.

    Just so you know, I'm not looking for the answer, I'm looking for the direction to go down to get hte answer. I want to learn this stuff as best as I can, but I need to know how it works!

    With the introduction to the side, here I go...

    In my physics book, I have a quesion that says:

    "Newtons law of universal graviatation is represented by

    F = GMm/r^2

    Here F is the magnitude of the gravitational force exrted by one small object on another. M a m are the masses of the objects and r is a distance. Force has the SI units kg * m/s^2.

    What are the SI units of the propotionally constant G? "

    Here's what I've tried..

    kg * m/s^2 = GMm/s^2

    => kg * m/L^2 = GMm/s^2

    This is where my math skills are a little rusty. I know the dividing by a fraction is a little weird, but I think i remember how to do it. (I took Calc I and Calc II a couple years back, but like i said, I'm just rusty. As soon as I figure it out, I remember everything about it).

    I'm not sure if it goes like this (which is how i've done it)

    => kg * s^2 = GMmm/r^2 => Gm^3/r^2

    => kg * s^2 = Gm^3/r^2

    => (kg * s^2)/G = m^3/r^2

    => 1/G = m^3 * (kg * s^2) / r^2 * 1

    But this answer is COMPLETELY WRONG. I MEAN, WAY OFF... The answer is in the back of the book as:

    m^3 / kg * s^2

    I've tried moving around The G, replacing r^2 with L (length) and setting s^2 = L (lenth) and I've moved around fractions, numerators, demonminators, etc etc and I CANNOT get close to this answer at all.

    What might I be doing wrong? TO tell you the truth, I"m not really sure what this quesiton is asking. Is it asking me to solve for G or what is it asking me to do? The question is under the section of dimensional analysis so i'm assuming that I need to make sure both sides are dimensionally correct. My answers never seem to add up!

    I've tried thinking of all possibilities, solving for other variables, and I cannot wrap my head around this. The entire chapter never touches on anything like this at all.

    Any help would be appreciated!

  2. jcsd
  3. Aug 29, 2005 #2


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    Homework Helper

    Coincidence or not, but the same question was posted here yesterday.

    As I said in the other topic, I'd work out for G first like this:

    [tex]F = G\frac{{m_1 \cdot m_2 }}{{r^2 }} \Leftrightarrow G = \frac{{F \cdot r^2 }}{{m_1 \cdot m_2 }}[/tex]

    Now just replace everything on the RHS by their dimensions and simplify a bit.
  4. Aug 29, 2005 #3
    We have, [tex]F = G\frac{Mm}{r^2}[/tex], and we want to solve for G.

    [tex]G = \frac{Fr^2}{Mm}[/tex]

    The SI unit for mass is kg (kilogram) and for length it's m (meter). We know that:
    [tex] {F = \frac{kg*m}{s^2}[/tex]

    With all this in mind we have that:

    [tex]G = \frac{\frac{kg*m}{s^2}*m^2}{kg*kg}[/tex]

    Now all you have to do is to simplify.
  5. Aug 29, 2005 #4

    Thats weird, but assuming that this is the first semester of this year, its understanable that someone would have the same book and possibly the same class. But in all honesty, I dont know who this is.

    I'm going to Solve for G here. These images/equations you posted pointed out a very big math flaw that I had, which makes A TON of difference in this case.
  6. Aug 29, 2005 #5

    AHHHHHHHHHHH !!! it makes sense! I wasnt putting G back into the eqaution. I was solving for G and then stopping. I was only doing half of the problem! Oh man, I feel like an idiot now!

    Thanks for the help guys, I appreciate it.

    Also, where do you find the codes to post these images?

  7. Aug 29, 2005 #6


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    I keep track of which "m" stands for the variable, "mass" ,
    and which "m" stands for the unit "meter" (which you goofed up)
    by always putting units inside brackets:
    [m] means meter, which can never cancel a mass m .
  8. Aug 29, 2005 #7


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    Have a look here.
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