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Math Help - Is this equation correct? d/dt(-μr/|r|) = (μ/|r|)(rṙ)r - μṙ/|r|

  1. #1
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    Is this equation correct? d/dt(-μr/|r|) = (μ/|r|)(rṙ)r - μṙ/|r|

    Is this equation correct?

    d/dt(-μr/|r|) = (μ/|r|)(rṙ)r - μ/|r|,

    where μ is an arbitrary constant, and r = r(t) is a vector.

    Apparently is it, but I don't see how, please explain...
    Thanks.
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  2. #2
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    Quote Originally Posted by feyomi View Post
    Is this equation correct?

    d/dt(-μr/|r|) = (μ/|r|)(rṙ)r - μ/|r|,

    where μ is an arbitrary constant, and r = r(t) is a vector.

    Apparently is it, but I don't see how, please explain...
    Thanks.
    Pull out the - \mu and consider \frac{d}{dt} \frac{<x,y,z>}{\sqrt{x^2+y^2+z^2}} = \frac{<\dot{x},\dot{y},\dot{z}>}{\sqrt{x^2+y^2+z^2  }}  - \frac{<x,y,z> (x \dot{x} + y \dot{y} + z \dot{z})}{(x^2+y^2+z^2)^{3/2}}.

    You see it now?
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by feyomi View Post
    Is this equation correct?

    d/dt(-μr/|r|) = (μ/|r|)(rṙ)r - μ/|r|,

    where μ is an arbitrary constant, and r = r(t) is a vector.

    Apparently is it, but I don't see how, please explain...
    Thanks.
    That equation is correct.

    WLOG, suppose \mathbf{r}(t)=\left<x(t),y(t),z(t)\right>

    Note that \frac{d}{dt}|\mathbf{r}|=\frac{1}{2|\mathbf{r}|}\c  dot (2x\dot{x}+2y\dot{y}+2z\dot{z})=\frac{\left<x,y,z\  right>\cdot\left<\dot{x},\dot{y},\dot{z}\right>}{|  \mathbf{r}|}=\frac{\mathbf{r}\cdot\dot{\mathbf{r}}  }{|\mathbf{r}|}

    So apply quotient rule to \frac{d}{dt}\left[-\frac{\mu\mathbf{r}}{|\mathbf{r}|}\right] keeping in mind what I showed you above. You should then be able to justify this equation.

    Can you take it from here?
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