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Gravitational Force

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data

    How high does a rocket have to go above Earth's surfae before its weight is half what it would be on earth

    2. Relevant equations

    F=GMeMr/r^2, however there are too many variables here to use

    3. The attempt at a solution

    F/2=GMeMr/((sq root 2)r))^2, but I don;t know if this is right or how to use it without numbers
  2. jcsd
  3. Nov 18, 2007 #2


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    You can deal with this without numbers by using a comparison ratio. The weight of the rocket on the Earth's surface is

    W = GMeMr/(Re^2) ,

    where Re is the Earth's radius (that's how far the Earth's surface is from its center, which is the r we need in order to apply Newton's gravitation law here.

    At the position we're interested in,

    (1/2)W = GMeMr/(r^2) ,

    where r is the distance from Earth's center that we need to solve for.

    What happens when you divide the second equation by the first one? Can you solve the result for r? How do you apply this result to the original question?
  4. Nov 18, 2007 #3
    So I will get 1/Re^2 = 2/r^2 and then r = Re - x where x is the distance to the rocket? But then I get lost on applying it to the question.
  5. Nov 18, 2007 #4


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    So this tells you that r^2 = 2 · (Re^2) or r = (Re) · sqrt(2). So the rocket must be about 1.414 times the Earth's radius from the Earth's center. How far does that put it above the Earth's surface? That is the altitude the problem asks for.
  6. Nov 18, 2007 #5
    so r = Re(sqrt2) so r = 6.38E6*sqrt2. Which then equals 9022682.528, but when I punch that number into WebAssign it is wrong, so is it not in km, or what
  7. Nov 18, 2007 #6


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    First off, this is the distance from the center of Earth; the problem asks for the distance above Earth's surface, so you have to subtract Re. This will give you

    x = Re · [ sqrt(2) - 1 ] .

    Secondly, you are using the radius of Earth in meters. You didn't mention it in your original posting, but what units does the problem ask for?
  8. Nov 18, 2007 #7
    Oh yeah, my bad. It is in km and so my radius would be 6.38E3 and then the answer would be 2642.68 ..... which is right. Thanks for your help.
  9. Nov 18, 2007 #8


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    Great! Be sure to read these kinds of questions carefully. There is often a homework or exam problem which asks for altitude or gives information in terms of the altitude, rather than the distance to the center of the Earth.
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