Can someone solve this for me: DSolve[{r''[t] == -A/(r[t]

>>8666674
no

>>8666674
rewrite it using latex

>>8666695
[math]r''(t)=\frac{\lambda}{r(t)|r(t)|}, \\ r'(0)=0, \\r(0)=h,\\ \lambda > 0[/math]

>>8666878
That isn't a trivial equation. Maybe you can get some inspiration from newtonian gravity. But without other constraints, such as conservation of angular momentum, I doubt you'll find a closed form solution.

>>8666674
Yes, here you go
With[{h=1,A=1},NDSolve[{r''[t] == -A/(r[t] Abs[r[t]]), r'[0] == 0, r[0] == h}, r[t], {t,0,1}]]

>>8666931
>With[{h=1,A=1},NDSolve[{r''[t] == -A/(r[t] Abs[r[t]]), r'[0] == 0, r[0] == h}, r[t], {t,0,1}]]
but only for t<1.1, I need whole period

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I haven't checked. Since it seems like, for your initial conditions, the denominator is always positive, I removed the absolute value.

>>8666878
minus before lambda is missing

>>8667059
In that case it would be an oscillating function that would have to be solved piecewise. Use >>8667051 method to find the first piece, it would be an accelerating downward function, find v at r=0, substitute back in the original equation, find the second piece and so on.

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There seem to be some ugly inversions of logs in roots involved.

In any case, assuming r(t) > 0 as well as the existence of a series with
[math]r(t) = \sum_{n=0}^\infty r_n\, t^n[/math]

you can write a script to approximate the thing to arbitrary order.

Solve

[math]r(t)^2\, r''(t) = \lambda[/math]

From the start we know

[math]r(t) = h + \lambda (t/h)^2 / 2 + \sum_{n=3}^\infty r_n\, t^n[/math]

it goes on as

[math]r(t) = S(t)\cdot \frac {h^{12} \tau ^{12} }{\lambda ^6}+\frac{887 \tau ^{10}}{113400 h^4}-\frac{73 \tau ^8}{5040 h^3}+\frac{11 \tau^6}{360 h^2}-\frac{\tau ^4}{12 h}+h+\frac{\tau ^2}{2} [/math]

Where [math] \tau = t \sqrt{ \lambda } / h [/math] and [math] S(t) = \sum_{n=0}^\infty S_n\, t^n [/math] the series with the conefficients you want to compute.