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Math Help - Gravitational Law

  1. #1
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    Thumbs down Gravitational Law

    An object is fired vertically upwards from the surface of a planetary body; it moves under the action of Newton’s Gravitational Law, without
    resistance, so the equation is z'' = -gR^2 / (z + R)^2 . Find the relation between v = z' and z and use this model, and the relation that you have
    found, to obtain a numerical estimate for the escape speed on the surface of the Earth.

    Its the wording of the question I don't get? I assume I integrate z'' twice to find the velocity and distance respectively, but the escape speed to escape the surface of the earth? is this like when g (9.81) begins to decrease as it moves from the planetry body? I havn't got a clue at all. Any help appreciated
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  2. #2
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    The escape velocity is the initial velocity necessary to "escape" to infinity. But you don't need to integrate twice. Just integrate once to get the formula for the velocity. The escape velocity is the smallest initial velocity such that the velocity is never 0 for any finite t.
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  3. #3
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    How can I integrate this though? I have z'' on the LHS and z on the RHS, but I can't integrate z because I don't know it? I know I can write z'' as vz' using the chain rule. Anymore help appreciated, thanks
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  4. #4
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    Quote Originally Posted by mitch_nufc View Post
    How can I integrate this though? I have z'' on the LHS and z on the RHS, but I can't integrate z because I don't know it? I know I can write z'' as vz' using the chain rule. Anymore help appreciated, thanks
    Substitute \frac{d^2 z}{dt^2} = v \frac{dv}{dz} where v = \frac{dz}{dt}.
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    Quote Originally Posted by mr fantastic View Post
    Substitute \frac{d^2 z}{dt^2} = v \frac{dv}{dz} where v = \frac{dz}{dt}.
    So you have v \frac{dv}{dz} = f(z) \Rightarrow \frac{1}{2} v^2 = \int \! f(z) \, dz + C.

    Solve for v as a function of z: v = v(z).

    Now solve v(z) = \frac{dz}{dt} \Rightarrow \int \, dt = \int \! \frac{1}{v(z)} \, dz.
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  6. #6
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    I sort of get what you're saying

    so I'd have 1/2v^2 = RHS? Or not... Hmm I used to be good at applied questions too
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  7. #7
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    Quote Originally Posted by mitch_nufc View Post
    I sort of get what you're saying

    so I'd have 1/2v^2 = RHS? Or not... Hmm I used to be good at applied questions too
    f(z) = -gR^2 / (z + R)^2. This is what you first have to integrate.
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  8. #8
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    so I get (gR^2)/(z+R) = (v^2)/2

    (2gR^2)/(z+R) = v^2

    Not sure what this is telling me though
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  9. #9
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    The missing element

    Your mathematics is correct however what 's missing from the discussion

    is that vdv is integrated from v0 to 0 the initial velocity to a velocity of 0

    at z = infinity

    The Rhs is integrated from 0 to infinity i.e. we want to integrate z over a range where it completely escapes the gravitational pull of the earth which is at infinity.

    We obtain v0^2 = 2gR

    or v0 = (2gR)^(1/2)

    Make sure to convert g to km/s^2 by dividing 9.8/1000 and use R =6400km

    You obtain 11.2km/sec which is the accepted value
    Last edited by Calculus26; May 6th 2009 at 04:35 AM.
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