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Simple intergration I think

  1. Mar 25, 2005 #1
    Here is a stupid question
    How would you solve for [tex] \frac{d^2r}{dt^2}=G\frac{M}{r^2}[/tex]
    I'm new at this stuff, can anyone tell me??
    Assume G and M are constants
  2. jcsd
  3. Mar 25, 2005 #2


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    Yes, they are constants. G is your universal gravitational constant. M is the mass of the body your object is being attracted to.

    The distance away from the center of the body (r) is the only thing that changes.
  4. Mar 25, 2005 #3


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    It's not that simple to integrate that ODE.Technically,it's nonlinear second order.Since that equation appears in the Kepler problem in classical mechanics,i would advise you to look for a book on celestial mechanics which would deal nicely with this equation.

  5. Mar 25, 2005 #4
    [tex]\frac{d^{2}r}{dt^{2}}=\alpha r^{-2}[/tex]

    Let [tex]w=dr/dt[/tex].

    Then [tex]dw/dt=d^{2}r/dt^{2}[/tex].

    Chain Rule:

    [tex]\frac{dw}{dr}w=\alpha r^{-2}.[/tex]

    [tex]\int wdw=\int \alpha r^{-2}dr[/tex]

    [tex]w^{2}=-2\alpha r^{-1}+C_{1}[/tex]

    At this point, solve for the constant using the initial velocity of the body.

    [tex]dr/dt=\sqrt{-2\alpha r^{-1}+C_{1}}[/tex]

    [tex]\int \frac{dr}{\sqrt{-2\alpha r^{-1}+C_{1}}}=\int dt[/tex]

    If you type in
    (-2 a x^(-1) + c)^(-1/2)
    to integrals.wolfram.com , you get something. Realize that what you get is t on one side and r (which is x in the website) is in the mess on the left hand side. I doubt that you can solve for r as a function of time. You can still get a graph though.

    It's kinda cool to see a graph whose *acceleration* is changing. Or maybe that's just me...
  6. Mar 25, 2005 #5
    Thanks guys, it helps a lot
  7. Mar 25, 2005 #6
    I ran into another problem, I was doing a problem similar to this
    [tex]\int \frac{dr}{\sqrt{\alpha r^{2}+C}}=\int dt[/tex]

    [tex]\frac{\log{[2\sqrt{\alpha}r+2\sqrt{C+\alpha r^2]}}}{\sqrt{\alpha}}+C_{1}=t[/tex]
    and my lst constant is 0

    and my r and t are zero
    but I can't solve for [tex]C_{1}[/tex] because log 0 is undefined, even if I let r approach to 0, it's value will still be negative infinity.
    How would I solve this problem because I need a numerical value for t when r=6.37*10^6
  8. Mar 25, 2005 #7
    I found out another thing
    If I plug in 0 for my constant fisrt, then the intergral will be different
    [tex]\int \frac{dr}{\sqrt{\alpha r^{2}}}=\int dt[/tex]

    the intergral is
    [tex]\frac{r\log{r}}{\sqrt{\alpha r^2}}+C_{1}=t[/tex]
    why is that
  9. Mar 26, 2005 #8


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    It can't be.

    [tex] \int \frac{dr}{\sqrt{\alpha r^{2}}} =\frac{1}{\sqrt{\alpha}}\int \frac{dr}{|r|} =\frac{1}{\sqrt{\alpha}}\ln |r| +C [/tex]

    ,which is something else.

  10. Mar 26, 2005 #9
    but still, ln 0 still is undefined
    Is there anyway to solve C?
  11. Mar 26, 2005 #10


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    Depends on the initial conditions.Give them and u can find all integration constants (there must be 2).Notice that [itex]r\neq 0[/itex] which can be seen from the ODE.

  12. Mar 26, 2005 #11
    What if the only known value is t=0 and r=0, because since something is going to the center of the earth, then when it's 0 second, the distance it traveled must be 0
  13. Mar 26, 2005 #12
    r=0 doesn't work. Let's say I have a perfect point particle of arbitrary nonzero mass. What is the gravitational force on another point particle of arbitrary mass (again nonzero) at the exact same location as the first particle?
    Last edited: Mar 26, 2005
  14. Mar 26, 2005 #13


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    Gravitational potential (Coulomb type) is not finite in the origin (= the point where u have the pointlike source) (it "blows up",just like the electrostatic one).Unfortunately it's not renormalizable either...:yuck:

  15. Mar 26, 2005 #14
    you guys are right, I thought r=x but that does't mean that x=r, but still, when r approches to 0, the gravity should also gets smaller. Instead I got negative numbers when my r is less than 1
    Last edited: Mar 26, 2005
  16. Mar 26, 2005 #15


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    Earth is not a pointlike gravity source.Coulomb potential will not apply.

  17. Mar 26, 2005 #16
    Well...I don't know, but I think it was proven outside of the earthbulk, the graviation field is equivalent to the whole mass concentrated at the center of mass...(of course with the Newtonian potential only, for the relativistic equivalent one I don't know)

    For the differential equation, I think you can use first a guess :

    r(t)=B*t^A...giving in particular A=2/3.

    Then u can use the variation of the constant to find the second solution...but that's a bit of work.
  18. Mar 26, 2005 #17
    Where did your differential equation came from?
    and actually, I'm assuming that the mass of the earth is the same all throughout
  19. Mar 26, 2005 #18
    When you're outside of a large body, such as a planet, you can essentially treat it as a point mass located at its center. When you're inside that planet, the force you experience is proportional to your distance from the center. Both facts directly follow from Gauss' law (assuming Newtonian gravity, of course).
  20. Mar 27, 2005 #19
    dcppc :

    I'm just trying a solution by chance or maybe by experience (you can also call it an Ansatz)...it's just a guess, it turns out it works...but it could also not....(monomial ansatz are quite useful, also for functional equations, aso...)
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