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Calculating the gravitational accelleration on the surface of a planet

  1. Jun 19, 2006 #1
    Why doesn't this force give the gravitational acceleration on the surface of the earth?

    [tex]a = G \cdot \frac{m}{r^2}[/tex]

    I would think that it could be derived from the law of universal gravitation:

    [tex]F = G \cdot \frac{m_1 \cdot m_2}{r^2}[/tex]

    Since [tex]F = m_2 a[/tex]
     
  2. jcsd
  3. Jun 19, 2006 #2
    When you put in the right numbers, the accelaration should come to around 9.8ms^-2.
     
  4. Jun 19, 2006 #3

    George Jones

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    It does. I get g = 9.79 m/s^2 using your formula and accepted data for the mass and radius of the Earth.

    For more accuracy, the fact the density of the Earth is not spherically symmetric has to be taken into account.
     
  5. Jun 19, 2006 #4
    It seems I screwed my units up. I had 9.77e6 m/s^2. I should have had 9.77 m/s^2.

    New question:

    The earth's radius is larger at the equator that from pole to pole. Does the acceleration due to gravity vary from, say, 9.77 m/s^2 to 9.83 m/s^2?
     
  6. Jun 19, 2006 #5

    tony873004

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    Yes, and it also varies because the velocity from Earth's rotation is higher at the equator, and 0 at the poles. This adds to the effect.

    This is also probably the reason that you don't see g expressed as more digits. If instead of calling it 9.8 you called it 9.81756423, then you'd have to specify where on Earth you were talking about.
     
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