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Problem with gravitation equation

  1. Sep 19, 2008 #1
    I'm having a problem with a project I've been kicking around for some time now.

    This arose out of a discussion with a co-worker who was wondering how much less one would weigh at various intervals of increasing altitude above the surface of the earth.

    I set out to create a spread sheet to demonstrate.
    The spreadsheet would take inputs for the mass (weight) of the person & the desired altitude and show the weight at altitude.
    Although I created the spreadsheet using Newton's formula for gravitational attraction, I noticed a peculiar thing. That is, if I set the altitude to zero, i.e to remain at mean sea level, the field showing weight at altitude does not give the original starting weight.

    Despite having searched the web for the highest precision values I could find for the variables, I'm supposing that the problem can only be due to some inaccuracy in one or more of the values I have in the equation. (Or I messed up somewhere else)

    Here's what I have:

    F = G*m1*m2/ r^2

    G = 6.67259 * 10^-11

    m1 = Mass of Earth = 5.9736 * 10^24 kg

    m2 = Mass of object on surface of earth (at mean sea level) = 50 kg

    r = Radius of earth = 6.3674425 * 10^6 m

    1 Newton (N) = 9.80665002864 kg

    (6.67259 * 10^-11) *(5.9736 * 10^24) * 50 = 19.929692 * 10^15

    (19.929692 * 10^15) / (6.3674425 * 10^6) ^2 = 491.5532 N = F

    F = 491.5532 N BUT! 50Kg * 9.80665002864 = 490.3325 N For a difference of 1.227 Newtons.

    With this kind of error, I have to input an altitude of 7920 m to get the result to equal the starting weight.

    Can anyone see what I'm doing wrong here? Or am I just trying to get more accurate results than is possible with available data?

    Thanks for any insight.
  2. jcsd
  3. Sep 20, 2008 #2


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    I didn't go through all the numerics, I'm sure you did them right (more or less).
    But taking this general formula
    F = G*M*m/ r^2
    with M the mass of the earth, and m that of the object, you want it to reduce to the familiar
    F = m g
    if you put r = R (the earth radius)

    Comparing, you find that g = G*M/R.

    If r is close to R (i.e. close to the surface of the earth) then F = m g is a good approximation to G M m / r^2.
  4. Sep 20, 2008 #3
    "1 Newton (N) = 9.80665002864 kg"

    1 N = Kg m sec^-2,

    otherwise you are setting g to a constant.
  5. Sep 20, 2008 #4


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    I see two possible problems.

    R and g vary with latitiude. Were the values used consistent with each other? I.e., the value of g used corresponds to the same latitude that has the value of R used?

    Also, there was no accounting for the centrifugal contribution to g.

    To avoid using centrifugal effects, try doing the calculation at the poles (latitude 90 degrees, north or south):

    R = 6,356.750 km
    g = 9.832129 m/s^2

    FYI: http://en.wikipedia.org/wiki/Earth's_gravity
  6. Sep 21, 2008 #5
    Thanks Redbelly,

    The values you gave get me within .01%.

    I had considered centrifugal force being a factor, but if anyone asked, I was going to say that we were doing the experiment at the poles.

    Until now, I was not aware that the value of g, and therefore the method of expressing Newtons as weight in kg would vary so much with latitude.

    I was trying to use a theoretical perfect sphere (where R was the mean value between the equator & poles) for my calculations,which didn't work. But matching the R value & the g value at the poles did the trick!

    I do appreciate the help.
    Thanks again
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