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Gravitational force with 1/r^3

  1. Jul 24, 2012 #1

    nearc

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    Gold Member

    apologies if this has been covered elsewhere in the forums, but i was unable to find it. i've seen the usual gravitational force equation [i.e. F=gMm/r^2] with essentially r^3 in the bottom. an example of this can be seen in http://www.mathworks.com/moler/exm/chapters/orbits.pdf [Broken] on pages 8 and 9.

    i'm looking for any info [paper/webpages] that might explain how to use this approach, thanks
     
    Last edited by a moderator: May 6, 2017
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  3. Jul 24, 2012 #2

    rbj

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    it's a vector equation and the author (who happens to be the guy who created the first MATLAB) is not being conventional with his math symbols.

    Newtons gravitational law can first be stated:

    [tex] F = G \frac{m_1 m_2}{r^2} [/tex]

    where [itex]F[/itex] is the magnitude of the force vector that runs along the line that connects [itex]m_1[/itex] and [itex]m_2[/itex]. and [itex]r[/itex] is the Euclidian distance between them. this is a scaler equation and is the simple inverse-square gravitational equation you see most everywhere.

    now suppose that the body with mass [itex]m_1[/itex] is located at point [itex]\vec{p_1}[/itex] and the body with mass [itex]m_2[/itex] is located at point [itex]\vec{p_2}[/itex]. the coordinates of the points are

    [tex] \vec{p_1} = ( \ x_1, \ y_1, \ z_1 \ ) [/tex]

    [tex] \vec{p_2} = ( \ x_2, \ y_2, \ z_2 \ ) [/tex]

    the distance between the two points [itex]\vec{p_2}[/itex] and [itex]\vec{p_1}[/itex] is

    [tex] r = | \vec{p_1} - \vec{p_2} | = \sqrt{(x_1-x_2)^2 +(y_1-y_2)^2 +(z_1-z_2)^2 } [/tex]

    now, the difference vector that connects [itex]\vec{p_2}[/itex] to [itex]\vec{p_1}[/itex] is [itex]\vec{p_1}-\vec{p_2}[/itex], so [itex]r[/itex] is just the magnitude of that difference vector.

    lastly to get a force vector [itex]\vec{F}[/itex] that has magnitude [itex]F[/itex], we need to create a unit vector that runs along the line that connects [itex]\vec{p_2}[/itex] to [itex]\vec{p_1}[/itex]. that unit vector is the difference vector [itex]\vec{p_1}-\vec{p_2}[/itex] normalized by dividing by its length [itex]r=|\vec{p_1}-\vec{p_2}|[/itex]. this is where the extra power in [itex]r^3[/itex] comes from.

    so the force acting on the body with mass [itex]m_1[/itex] located at point [itex]\vec{p_1}[/itex] by the body with mass [itex]m_2[/itex] located at point [itex]\vec{p_2}[/itex] is

    [tex] \vec{F} = -G \frac{m_1 m_2}{r^2} \frac{\vec{p_1}-\vec{p_2}}{|\vec{p_1}-\vec{p_2}|} [/tex]

    or

    [tex] \vec{F} = -G \frac{m_1 m_2}{|\vec{p_1}-\vec{p_2}|^3} (\vec{p_1}-\vec{p_2}) [/tex]

    that is a vector equation, but it is still the inverse-square law. the extra power of [itex]r[/itex] in the denominator is needed to normalize the vector connecting point [itex]\vec{p_2}[/itex] to point [itex]\vec{p_1}[/itex]
     
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