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    Interestingly this now gives the result we want, i.e. v ~ r^-4, so if it a mistake its a pretty good one
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    (Original post by spread_logic_not_hate)
    Ok now I get

     \frac{d^2r}{dt^2} = \frac{c^2}{r^3} - \frac{2c^2a^2}{r^5}

    Do you get the same?
    Ah sorry ignore that comment I made, if you do this you are differentiating w.r.t r not t, we want it w.r.t t.
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    Ah sorry ignore that comment I made, if you do this you are differentiating w.r.t r not t, we want it w.r.t t.
    No you're completely right - use implicit differentiation and product rule, i.e.

     \frac{d^2r}{dt^2} = \frac{-2c}{r^3} \frac{dr}{dt} \sqrt{a^2-r^2} + \frac{c}{r^2}\times \frac{1}{2\sqrt{a^2-r^2}} \times -2r\frac{dr}{dt}

    and then put in our expression for dr/dt.
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    (Original post by jmz34)
    Sorry, in my attempt to keep the question short I've forgotten to mention that. V is a function of r and r only.
    If V(r) is a solution, so is V(r)+C for any constant C.

    So I don't see how you can deduce that V(r) varies as 1/r^3.
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    (Original post by spread_logic_not_hate)
    No you're completely right - use implicit differentiation and product rule, i.e.

     \frac{d^2r}{dt^2} = \frac{-2c}{r^3} \frac{dr}{dt} \sqrt{a^2-r^2} + \frac{c}{r^2}\times \frac{1}{2\sqrt{a^2-r^2}} \times -2r\frac{dr}{dt}

    and then put in our expression for dr/dt.
    Ah yeah I see it now. So eventually subbing everything back in and integrating I get:

    V=k/r^4 + m/r^2 where k and m are constants. Just to clarify can you then just say V is proportional to r^-4 even though there is an r^-2 term?
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    Ah yeah I see it now. So eventually subbing everything back in and integrating I get:

    V=k/r^4 + m/r^2 where k and m are constants. Just to clarify can you then just say V is proportional to r^-4 even though there is an r^-2 term?
    We have this:

     \frac{d^2r}{dt^2} = \frac{c^2}{r^3} - \frac{2c^2a^2}{r^5}

    Sub this into original expression, using that

     p' = \frac{c}{r^2}

     \frac{c^2}{r^3} - \frac{2c^2a^2}{r^5} - r\frac{c^2}{r^4} = -\frac{dv}{dr}

    implies


     \frac{c^2}{r^3} - \frac{2c^2a^2}{r^5} - \frac{c^2}{r^3} = -\frac{dv}{dr}


    Notice that the first and last term cancel out, leaving:

      - \frac{2c^2a^2}{r^5} = -\frac{dv}{dr}

    i.e.

      v = \int \frac{2c^2a^2}{r^5} dr = \frac{-c^2a^2}{2r^4}

    so we have that v ~ r^-4.

    Bit of an effort to get there though hey?!
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    Again, many thanks for your help
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    Again, many thanks for your help
    No probs, sorry about all the mistakes - made a bit of a meal of it tbh!
 
 
 
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