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Thread: Differential Equation involving mechanics

  1. #1
    Junior Member
    Mar 2009

    Differential Equation involving mechanics


    I need help with the following question. I have attached my question. I need help with part b)

    this is what i have managed to do so far, but i don't think i am doing it correctly.

    $\displaystyle m\ddot{r}=\frac{mMG}{r^{2}}$

    dividing both sides my $\displaystyle m$

    $\displaystyle \ddot{r}=\frac{MG}{r^{2}}$

    $\displaystyle \int\ddot{r}dt=\int\frac{MG}{r^{2}}dt$

    $\displaystyle \dot{r}=\frac{MG}{r^{2}}t+C

    any help is appreichated


    Attached Thumbnails Attached Thumbnails Differential Equation involving mechanics-capture.gif  
    Last edited by cooltowns; May 12th 2010 at 10:27 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
    Mar 2009
    near Piacenza (Italy)
    This problem can be solved as in...

    Setting for semplicity sake $\displaystyle M G = \mu$ the DE becomes...

    $\displaystyle r^{''} = -\frac{\mu}{r^{2}}$ (1)

    Now because is...

    $\displaystyle r^{''} = \frac{d r^{'}}{dt} = \frac{d r^{'}}{dr} \frac{d r}{dt} = r^{'} \frac{d r^{'}}{dr}$ (2)

    ... we obtain from (1)...

    $\displaystyle r^{'} dr^{'} = -\frac{\mu}{r^{2}} dr$ (3)

    ... i.e. a separable variables DE the solution of which is...

    $\displaystyle r^{' 2} = \frac{2 \mu}{r} + c_{1}$ (4)

    ... just like in Your textbook. The further step could be finding $\displaystyle r(t)$... a little less confortable step ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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