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Universal Law of Gravitation problem

Discussion in 'MCAT Study Question Q&A' started by kmcgrath, Jan 7, 2009.

  1. kmcgrath

    5+ Year Member

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    How do I get to this answer step by step? Any help would be greatly appreciated. Thanks!

    Q: At what height above the earth's surface is acceleration due to gravity 10% of that at sea level?

    A: 1.36x10^7 m
     
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  3. TooMuchResearch

    TooMuchResearch i'm goin' to Kathmandu...
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    I think we would just set up the normal gravitational force equation:

    ____Gm1m2
    F = -------
    _____r^2

    Next, recall that F=ma. In this case, that m probably equals the mass of an object (that isn't the earth)...lets call it m1.

    Now, m1 will cancel from each side, leaving us with:

    ____Gm2
    a = ------
    _____r^2

    Since we are given acceleration due to gravity is 10% of that at sea level (recall: g=9.8 m/s^2 or ~10 m/s^2), we know that a=g=1.0 m/s^2.

    Just solve for r and should be okay...you might have to subtract the radius of the earth from the value you determine for r since the question asks for the height above earth's surface.
     
  4. kmcgrath

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    That's funny b/c I did exactly that and got the wrong answer...I am trying it again with a fresh mind. At least I am going the right direction with this one. Probably a math error I was repeating from exhaustion. Thank you for your help.
     
  5. Vandau

    Vandau New Member
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    Don't plug radius of earth into the equation, since g is calculated at earth's surface already. 10% of g would be the distance starting from earth's surface so just solve for r, the distance above earth's surface.

    g=GM/r^2
    1=GM/r^2
    r^2=GM
    r=sq rt of GM
     
  6. Mantis Toboggin

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    Would just like to add to this that r'= sqrt (GM) is not your final answer. Because the question asks what height above the surface this occurs at, your answer will be

    h= r'-r

    To find r, use the value of a=g=10.0 m/s^2

    10= GM/(r^2)

    10r^2= GM

    (we know the square root of 10 will fall between the squares of 3 and 4, closer to 3... just guestimate it to 3)

    So r= .67 * 10^7 m

    r'= 2 * 10^7 m

    h= 1.3 * 10^7 m or 1.3 * 10^4 km
     

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