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Universal Gravitation

  1. Aug 30, 2005 #1
    At what height above the Earth's surface would an object's weight be one half the value at the surface?

    W = m(2)g = Gm(1)(m2) / r^2 is the formula I am using.

    I found 1/2 weight to be 2.92825 x 10^25

    Then I found m(1) to be 1/2gxW (not sure if I should've used 1/2g here or not) and then I did the rearranging and other such things.

    I got an answer of 4.4 x 10^7. It seems a bit high to me, but I don't have a great understanding of the concept, lol.
  2. jcsd
  3. Aug 30, 2005 #2


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    The force of weight is,
    w = mg,
    where m is the objects mass and g is the gravitational acceleration, which can be calculated as,
    g = (G*M) / R^2

    If you raise something up high into the air, well above the Earth's surface, the mass will not change, but the acceleration due to gravity will.
    So to find the distance above the surface that will make the object's weight 1/2 of what it is at the surface, you must find the distance away from the center of Earth to make g one half its "normal" value.

    Normally g = 9.81 m/s^2, so we are looking for a distance R that will make it 4.906 m/s^2.
    4.905 = (G*M) / R^2, where M is the mass of the Earth, 5.97 E24 kg.
    just solve for R and you have the distance from the center of Earth.
    You then need to subtract out the distance from the Center to the surface of Earth, to get the altilitde above Earth's surface.
  4. Aug 30, 2005 #3
    Your answer is close, and I probably just used differant constants to you but a few thousand Km's up is a reasonable answer; The radius of the earth is very large and we have to raise something less than half that distance to half (just over a quarter) the force due to gravity. The inverse relationship means that the Force on each mass will rapidly decrease as the distance R gets larger.
  5. Aug 30, 2005 #4
    2638.82 km above the earth is what I got. I checked some sites and it seems reasonable. Thanks for the help.
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