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Using the gravitational formula

  1. Jul 14, 2013 #1

    462chevelle

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    1. The problem statement, all variables and given/known data
    if I was wanting to calculate the force of gravity at certain points relative to the earths surface. what number would I plug in the the constant of G in the formula. so far I have some terrible numbers and I want to make sure my G is correct. which I don't think it is. this isn't for homework. just for fun but I figured this was the best place to put it.


    2. Relevant equations
    Fg=G(m2m1/r^2)



    3. The attempt at a solution
    here is what I am plugging in
    G = 1.2 x 10^-4 http://en.wikipedia.org/wiki/Gravitational_constant
    that's where I got that. but I have the feeling ill need to use the other numbers. but I don't understand what that (80) is in there for
    r=6371 km
    m2=5.972 x 10^24
    m1=90 kg
    the answer I got was 1589016271417625
    the number I was hoping for was. appx 9.8 m/s^2
    then once I know im doing it right I want to check and the relationship between being deeper and deeper into the earths surface and gravity.
    thanks,
    Lonnie
     
  2. jcsd
  3. Jul 14, 2013 #2

    462chevelle

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    if I just plug in 6.67384 x 10^-11 I still get crap numbers.
     
  4. Jul 14, 2013 #3
    [itex][/itex][itex]1.2 \times 10^{-4}[/itex] is, as stated on the Wiki page, the standard uncertainty of the gravitational constant, the value of the constant itself is on the line above it, [itex]G\approx 6.67\times 10^{-11} \;\mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}[/itex]. Remember to check your units, [itex]r=6371 \;\mathrm{km} =6371\times 10^3 \;\mathrm{m}= 6371000 \;\mathrm{m}[/itex].

    You won't be getting the value you were hoping for, as you're calculating the force, not the acceleration due to gravity (which has the value of approx. 9.8 ms^-2 near the surface of the earth). Try to use F=ma to get the acceleration :)
     
    Last edited: Jul 14, 2013
  5. Jul 14, 2013 #4

    462chevelle

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    oh ya. forgot about converting the units.
    excuse my math skills but how is that formula computed? do you mind helping explain the exponents on the units.
     
  6. Jul 14, 2013 #5
    Hm? Can't you just plug the values you now have into a calculator? Or are you looking for a derivation of the Newton's law of gravtitation, ie [itex]F=G\frac{m_1 m_2}{r^2}[/itex]?

    Are you asking what they mean or why G has the units it does? If the latter: F, force, has the units [itex][F]=\mathrm{N}=\mathrm{kg} \frac{\mathrm{m}}{\mathrm{s}^{2}}[/itex], mass has the units [itex][m]=\mathrm{kg}[/itex] and distance has the units [itex][r]=\mathrm{m}[/itex].

    [tex]F_G=G\frac{m_1 m_2}{r^2}\implies [G]=\frac{[F] [r]^2}{[m_1] [m_2]}=\frac{\mathrm{kgms}^{-2}\mathrm{m}^2}{\mathrm{kg}^2}=\mathrm{kg}^{1-2}\mathrm{m}^{2+1}\mathrm{s}^{-2}=\mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}[/tex]

    If the former, well, I don't really know what to say, that's just the way units work. When we, for example, multiply two physical quantities together, the unit of the resulting quantity is the product of the units of the quantities we multiplied.

    Was that helpful at all?
     
    Last edited: Jul 14, 2013
  7. Jul 14, 2013 #6

    462chevelle

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    do I put 6.67x10^-11 in the calc or what do I do something with those units
     
  8. Jul 14, 2013 #7
    You have to be consistent; when you put 6.67×10^-11 in your calculator, you have to have your other values as meters and kilograms, in order to get Newtons as a result but as long as you do that, everything should work.
     
  9. Jul 14, 2013 #8

    462chevelle

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    just plugging it in I get 883
     
  10. Jul 14, 2013 #9

    462chevelle

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    awesome that is perfect. thank you!!
     
  11. Jul 14, 2013 #10
    Yes, that's about the magnitude of the gravitational force that is exerted on an object with a mass of 90 kilograms. If you want to get the acceleration due to gravity, you divide by 90 kg as per F=ma. The result should be close to the value you mentioned earlier.

    EDIT: Glad I could help :]
     
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