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Homework Help: Help Solving Gravitational Constant Equation

  1. May 27, 2012 #1
    Hi, I was just hoping for some help in explaining the steps to solving an equation using the universal gravitational constant. I can easily find the answer online but not easily find out how to actually solve it, since I want to understand what I am doing.

    I understand the basics of equation solving and know enough that if a step by step explanation is given then I can grasp what you are saying.

    The question involves solving the gravitational force between two masses of 70 kg standing one meter apart. The equation is -

    F=6.67*10^-11*(N*m^2)/kg^2*(70 kg * 70 kg)/(1 m)^2

    I know I probably do not need to state what any of the letters mean to anyone who can explain this to me but just for clarity's sake - N (Newtons) ; m (meters) ; kg (kilograms)

    Huge thank you to anyone who takes the time to help and explain how this is done!
  2. jcsd
  3. May 27, 2012 #2


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    I don't quite understand what you are asking here. If the problem is simply to compute the force, then since you have all the numbers, and you've already plugged them into the equation, then why don't you just compute it?
  4. May 28, 2012 #3
    Sorry I guess I should have been clearer and maybe I did not post in the right place because it is more of a math problem, though it pertains to physics.

    If you were to solve this yourself, how would you do it step-by-step? I am just not sure how to resolve some of the numbers such as the N*m^2/kg^2 and 70 kg *70 kg/(m)^2.

    70 kg * 70kg / (1 m) ^2 = 4900 kg/ms/ms?

    (N*m^2) / kg^2 = ?

    Or am I just going about it wrong trying to solve it that way? Again, thank you.
  5. May 28, 2012 #4
  6. May 28, 2012 #5
    Okay, thank you for that insight it helps immensely. I will probably still have a question or two but I will study that first and give it some time and try to resolve it before I post back.
  7. May 28, 2012 #6
    After a lot of research and much studying, I have come up with two possible understandings.

    The gravitational constant can be understood more simply as 6.67*10^-11 N. The dimensional analysis lead me to believe that the other letters were just a way of expressing what was going on in relation to what the letters were representing.

    For instance -

    70 kg * 70 kg / (1 m)^2 is another way of saying 4900 kg traveling at a meter per second per second and since a Newton is basically 1 kg traveling at a meter per second per second then you could say that the 4900 kg is 4900 N.

    When I multiply the 4900 N by the 6.67 * 10^-11 N, I get 32683 * 10^-11 N. Add in the decimals to simplify and I get 3.2683 * 10^-7 N or 3.3 * 10^-7 N rounded up, which is the answer given.


    Do the letters cancel each other out in the equation?

    For instance -

    In N * m^2 / kg^2 * 70 kg * 70 kg / (1 m)^2, do the m^2's cancel each other out and kg^2's cancel each other out leaving only N * 4900? Which then multiplies by the 6.67 * 10^-11, resulting in the numbers I gave above.

    It seems to me that it has to be one or the other explanations as they give the correct answer, I am just not sure which one it is and why. And again, I want to understand what I am doing, not just come to the right conclusion.
  8. May 28, 2012 #7


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    No, the gravitational constant cannot be interpreted more simply as just having units of newtons. It has precisely the units that are listed there: (Nm2)/kg2

    Everything you've said above is completely nonsensical. The 4900 is a quantity that has dimensions of (mass2)/(length2). There is no need to try to get an "intuitive" sense of what that means. Newton's universal law of gravitation says that the gravitational force between two masses is proportional to the product of the two masses, and inversely proportional to the square of the distance between them. When you compute the product of the two masses and divide by the square of the distance between them, you get this number which has those units. (kg2/m2). That's all there is to it.

    Look, I don't know what you're trying to do here, but to work out the units of the final answer, just follow the rules of algebra. Let's start with the equation$$F = G\frac{m_1m_2}{r^2}$$ $$= \frac{(6.67 \times 10^{-11}~\textrm{Nm}^2/\textrm{kg}^2)(70~\textrm{kg})(70~\textrm{kg})}{1~\textrm{m}^2}$$ $$ = \frac{(6.67 \times 10^{-11}~\textrm{Nm}^2/\textrm{kg}^2)(4900~\textrm{kg}^2)}{1~\textrm{m}^2}$$

    Now let's collect all of the units together and simplify:$$ = (6.67 \times 10^{-11})(4.9 \times 10^3) \frac{\textrm{Nm}^2}{\textrm{kg}^2} \frac{\textrm{kg}^2}{\textrm{m}^2}$$ $$ = 32.683 \times 10^{-8}~\textrm{N} \frac{\textrm{kg}^2}{\textrm{kg}^2} \frac{\textrm{m}^2}{\textrm{m}^2}$$

    After simplifying the units by cancelling out things that appear in both numerator and denominator (remember the statement in bold above), we are left with:$$ = 3.2683 \times 10^{-7}~\textrm{N}$$

    So, as you can see, the units work out just fine! We're left with newtons, which is good, because this equation is supposed to be for computing the gravitational force. The reason why G has the units that it does is precisely in order to make the units work out. Because on the left hand side of the equation, you have something with dimensions of force, which means that the right hand side also has to have dimensions of force in order for the equation to be dimensionally consistent. But mass*mass/length*length doesn't have dimensions of force. The force is *proportional* to this, with the constant of proportionality G having the dimensions to make it work out. IF you look at the right hand side without G, it has units of "kg^2/m^2", which I'll call "blah" for short. Therefore, G has units of "N/blah" so that when you multiply by G, you just end up with units of N.
    Last edited: May 28, 2012
  9. May 29, 2012 #8
    Thank you very much, I appreciate you taking the time to help me better understand all of this. The problem I was having was with the algebra and I wish I had been more clear about that from the beginning to make it easier for you. I do also, however/obviously, have a need to understand the properties associated with the numbers so that I can understand the whole picture better, so I am glad you helped explain this as well.
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