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Math Help - Differential equation solving related to 2nd Newton's law

  1. #1
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    Question Differential equation solving related to 2nd Newton's law

    Without taking into account friction and resistance to the air, but while taking into account the gravitation force and the altitude. The weight W is decreasing in terms of x (altitude):
    P(x)=\dfrac{mgR^2}{(R+x)^2}
    m: Mass of the object
    g: Gravitational force (at sea level)
    R: Radius of Earth
    x: Altitude

    What is the speed of the object in terms of its altitude (we're looking for v(x))?

    I'm trying hard to resolve this equation but just can't find how to relate it to the altitude instead of the time.

    Thanks to anyone who takes the time to answer this question. Good night,
    Last edited by bazingasmile; February 17th 2011 at 10:22 PM. Reason: Thanks topsquark for finding the typo and resolving my problem!
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  2. #2
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    Quote Originally Posted by bazingasmile View Post
    P(x)=\dfrac{mgR^2}{R+x^2}
    A slight change in the forumla (I presume this is a typo.)
    \displaystyle P(x)=\dfrac{mgR^2}{(R+x)^2}

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bazingasmile View Post
    Without taking into account friction and resistance to the air, but while taking into account the gravitation force and the altitude. The weight W is decreasing in terms of x (altitude):
    P(x)=\dfrac{mgR^2}{(R+x)^2}
    m: Mass of the object
    g: Gravitational force (at sea level)
    R: Radius of Earth
    x: Altitude

    What is the speed of the object in terms of its altitude (we're looking for v(x))?

    I'm trying hard to resolve this equation but just can't find how to relate it to the altitude instead of the time.

    Thanks to anyone who takes the time to answer this question. Good night,
    \displaystyle F = ma \implies a = \frac{gR^2}{(R + x)^2}

    Now we know that a = dv/dt, so
    \displaystyle \frac{dv}{dt} = \frac{gR^2}{(R + x)^2}

    And using the chain rule:
    \displaystyle \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{dv}{dx} \cdot v(x)

    So we have the differential equation:
    \displaystyle v(x) \cdot \frac{dv}{dx} - \frac{gR^2}{(R + x)^2} = 0

    Can you take it from here?

    -Dan
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    Talking Thank you

    Thank you very much.
    I'm not stuck anymore!
    Thanks for also pointing out my typo.

    G.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bazingasmile View Post
    Thank you very much.
    I'm not stuck anymore!
    Thanks for also pointing out my typo.

    G.
    Always happy to help out.

    -Dan
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