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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

    #2

    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
    numbers.

    Thanks
     
    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

    krab

    User Avatar
    Science Advisor

    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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