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Gravity Calculation Question

  1. Jul 19, 2005 #1
    I am trying to figure out why F=mg corresponds to question number two below.

    First lets assume a object with mass 70kg and then calcutate the force of gravity in two different scenarios:

    Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is standing at sea level, a distance of 6.37 x 106 m from earth's center.

    Crunching the numbers yields 688.1 N


    Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is in an airplane at 40000 feet above earth's surface. This would place the student a distance of 6.38 x 106 m from earth's center.

    This yields: 686 Newtons at a distance of 40,000 feet.

    Now use the 2nd law

    F=mg (70kg)(9.8m/s^2) equals 686 N

    Why is the 2nd Law calculation the same as the calculation at 40,000 feet?

    I realize that this is a very small difference (less than 1%) between the two

    Last edited: Jul 19, 2005
  2. jcsd
  3. Jul 19, 2005 #2
    The 2nd law is the same in all inertial coordinate frames.
    You can try using F=G*m*M/(r^2) (F=mg is approx. formula near the earth's surface). Is there a noticable difference?
  4. Jul 19, 2005 #3
    g changes with distance
  5. Jul 19, 2005 #4
    I know but what I am saying is that the force of gravity in #2 above (40,000) feet above the earth is closer to f=mg than example one at sea level.

    It would seem that the sea level calculation would be closer to f=mg.
  6. Jul 19, 2005 #5


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    The acceleration of gravity is not an absolute constant. It would be constant if the earth's density was spherically symmetric and if the earth were accurately a sphere and if there was not centrifugal component (earth not spinning). None of these 3 are true. In fact the variation in g is 1/2% from pole to equator at sea level. So though your 1/3% discrepancy can be accounted for by an elevation change, it can also be accounted for by a latitude change.
  7. Jul 20, 2005 #6
    g is the acceleration experienced at the surface of the earth. That's how it's defined.
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