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Value of g on Earth with a smaller radius

  1. Nov 1, 2009 #1
    What would be the value of g if the radius of the earth was decreased by half while maintaining the same mass.

    This is what I did

    g = MG / R^2

    g = MG / (0.5R)^2

    g = MG / 0.25R^2

    0.25g = MG / R^2

    I know this is not correct and the actual answer is 4g because as you get closer to the center of the Earth the gravitational force will increase but I cannot figure out why this attempt is wrong.
     
  2. jcsd
  3. Nov 1, 2009 #2

    tiny-tim

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    Hi Jimmy25! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Why did you put that 0.25 on the LHS? :confused:
     
  4. Nov 1, 2009 #3

    Pengwuino

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    Gold Member

    What you have initially is the gravitational acceleration due to the Earth while you're at the surface, or at a distance R. You then manipulated the equation to ask not what the value of g would be at the surface if the radius was half as large, but what would the value of the acceleration be at that same distance R if the radius was half as large.

    What I would do is setup a ratio, g' and g, where g' is at the surface of some planet with radius R' with the same mass, but then take R' = 0.5R and divide the two equations.
     
  5. Nov 1, 2009 #4
    you confuse the values of g before and after the shrinking of the earth

    you have g_old = MG/R^2

    now g_new is MG/(0.5*R)^2

    and (0.25)*g_new = MG/R^2 = g_old so

    g_new = 4*g_old, as expected
     
  6. Nov 1, 2009 #5
    Ok I see. So gold was equal to one quarter gnew

    Thanks, was racking my brain over this one. Sometimes the simplest question can be the most challenging.
     
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