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R’’(t)= -GM/(r(t))^2 =>r(t)=?

  1. Aug 8, 2010 #1
    My question is the title of the thread.

    Sorry, and this one too:
    f’’(t)=GM/((Α-f(t))^2
     
    Last edited: Aug 8, 2010
  2. jcsd
  3. Aug 8, 2010 #2

    Mark44

    Staff: Mentor

    Integrate twice with respect to t to get r(t).
     
  4. Aug 8, 2010 #3
    I want the answer, not the method to solve it. Thanks though.
     
  5. Aug 8, 2010 #4
    It's a differential equation, you can't just "integrate twice"! (The right-hand side depends on r).

    The solution r(t)=(x(t), y(t), z(t)) is not expressible in terms of elementary functions. What you can do is find the trajectory y=y(x), z=0. (by a suitable choice of the orthonormal basis).
    The result is a parabola, an ellipse or an hyperbole, depending on initial conditions. Try googling:

    "Motion in central gravitational field"
    "Motion of planets"
    "Euler's trajectory of a planet"

    Etc.
     
  6. Aug 8, 2010 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Why? Knowing an answer without knowing how to get it is useless.
     
  7. Aug 8, 2010 #6
    I thought you have calculators. You know any free online one except wolframalfa?
     
  8. Aug 8, 2010 #7

    Mark44

    Staff: Mentor

    That's not how this forum works. Instead, you show that you have made an effort and we help you with it.
     
  9. Aug 8, 2010 #8
    I think the solution can be expressed implicitly in terms of elementary functions. It's just the case of independent variable missing so for [itex]y''=-k/y^2[/itex], let [itex]p=y'[/itex] and arrive at:

    [tex]p\frac{dp}{dy}=-\frac{k}{y^2}[/tex]

    Now the variables can be separated and now you can integrate twice. Hope that's ok to help him this way.
     
    Last edited: Aug 8, 2010
  10. Aug 8, 2010 #9
    I hope it is not you contacting them and me mentioning it who is responsible that the wolframalfa is not giving the solution now whereas a couple of days ago it did. There are researchers out there that calculating it by hand is a sheer waste of time and energy.
     
  11. Aug 8, 2010 #10

    Mark44

    Staff: Mentor

    No, I have never contacted those folks, and I doubt that it would give you two different answers to the same problem.
     
  12. Aug 8, 2010 #11
    Check the attached file in the present post. It has the wolframalfa solutions it gave me a couple of days ago.

    At the case which of f’’(t)=GM/((A-f(t))^2, I entered f’’(t)=2/((5-f(t))^2, The initial conditions are f(0)=0 and f'(0)=0, therefore c1=c2=0. But that solution gives...? (-4)/(-4)=t^2 and no f(t)=...?
    Now it gives me the answer again! here it is:
    (sqrt(c_1) sqrt(f(t)-5) (c_1 (f(t)-5)-4)+4 sqrt(c_1 (f(t)-5)-4) log(2 (c_1 sqrt(f(t)-5)+sqrt(c_1) sqrt(c_1 (f(t)-5)-4))))^2/(c_1^3 (c_1 (f(t)-5)-4)) = (c_2+t)^2

    At the case of r’’(t)=-GM/((r(t))^2, if I remember well I entered GM=2 again. Here
    r(0)=A and r'(0)=0, thus c1=c2=A/2. Hm...this seems to have a solution r(t)=... so the first equation is ... impossible?
    It still does not give me the answer for this one now.
     

    Attached Files:

    Last edited: Aug 8, 2010
  13. Aug 9, 2010 #12
    Wolfram Alpha is only a sub-set of Mathematica. Ver 7, gives:

    DSolve[y''[x]==-k/y[x]^2,y,x]

    as:

    [tex]
    \begin{aligned}&
    \text{Solve}\biggr[\left(-\frac{1}{C[1]^{3/2}}k \text{Log}\left[2 k+2 C[1] y[x]+2 \sqrt{C[1]} \sqrt{C[1]+\frac{2 k}{y[x]}} y[x]\right]+\frac{\sqrt{C[1]+\frac{2 k}{y[x]}} y[x]}{C[1]}\right)^2\right.== \\
    &\hspace{200pt}\left.(x+C[2])^2,y[x]\biggr]
    \end{aligned}
    [/tex]

    That is, Mathematica is giving the solution as an implicit inverse via the Solve function.
     
  14. Aug 9, 2010 #13
    Can you post the link of that Mathematica dif. eq. calculator?
     
  15. Aug 9, 2010 #14
    That output above is from Mathematica 7 (my copy) which Mathematica expresses in terms of logs I believe it does so to handle all possible cases. However below, a particular IVP is in arctan.

    Luckis, as you can see, sometimes the answer is not enough. You still need to understand all the underlying math to really handle it well. Lets look at a simple IVP just for starters:

    [tex]y''=-\frac{2}{y^2},\quad y(0)=2,y'(0)=0[/tex]

    Letting [itex]p=y'[/itex], I get:

    [tex]pdp=-\frac{2}{y^2}dy[/tex]

    Now integrate from y(0) to some y(x):

    [tex]\int_{p(0)}^{p(x)} pdp=-\int_{y(0)}^{y(x)} \frac{2}{y^2}dy[/tex]

    [tex]p^2-p(0)^2=4(1/y-1/2)[/tex]

    or:

    [tex]\left(\frac{dy}{dx}\right)^2=4(1/y-1/2)[/tex]

    taking the square root and integrating again:

    [tex]\int_{y(0)}^{y(x)}\frac{dy}{\sqrt{4(1/y-1/2)}}=\pm \int_{x_0}^x xdx[/tex]

    doing that integration, making the substitutions and simplifying (the result below is from Mathematica), I get:

    [tex]-\frac{\sqrt{-1+\frac{2}{y}} y+\text{ArcTan}\left[\frac{\sqrt{-1+\frac{2}{y}} (-1+y)}{-2+y}\right]}{\sqrt{2}}\biggr|_{2}^{y(x)}=\pm x[/tex]

    which in this case, it's in the form of arctan and note it's improper at the lower limit so I take it's limit at that value and obtain [itex]\frac{\pi}{2\sqrt{2}}[/itex]. Now, I can take either the plus or minus value. Here, I'll take the positive value because that's what agrees with the numerical calculations (below) and I get for the final answer:

    [tex]\left(\frac{\sqrt{-1+\frac{2}{y}} y+\text{ArcTan}\left[\frac{\sqrt{-1+\frac{2}{y}} (-1+y)}{-2+y}\right]}{\sqrt{2}}\right)+\frac{\pi }{2 \text{Sqrt}[2]}=x[/tex]

    and that's still not too bad because it's easy to generate a table of (x,y) values and compare to the numerical results. In the plot below, the analytical solution is depicted as dots along the numerical one:


    Code (Text):
    mysol = NDSolve[{Derivative[2][y][x] == -2/y[x]^2, y[0] == 2,
        Derivative[1][y][0] == 0}, y, {x, 0, 2}]

    lowy = N[y[1] /. mysol]

    mytable = Table[
       {(Sqrt[-1 + 2/y]*y + ArcTan[(Sqrt[-1 + 2/y]*(-1 + y))/
             (-2 + y)])/Sqrt[2] + Pi/(2*Sqrt[2]), y},
       {y, lowy[[1]], 2, 0.01}]

    lp1 = ListPlot[mytable]

    pp1 = Plot[y[x] /. mysol, {x, 0, 2}]

    Show[{pp1, lp1}]
     
    Edit: I think I may have a problem with signs above. Probably need to go over that but I'll leave a careful check of it to the interested reader.
     

    Attached Files:

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    Last edited: Aug 9, 2010
  16. Aug 9, 2010 #15
    Can you post the link of that Mathematica dif. eq. calculator? Private message will do fine.
     
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