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B Falling time for a free falling object through large distances

  1. May 24, 2017 #1
    Well, first of all i want to apologize about my english skills hahaha
    I have been trying to calculate a time formula for free falling object through large distances (or weak gravitational fields) and my results don´t have much sense (I test my results in a quite realistic game called Algodoo)
    What I have calculated is

    g(r) = G⋅(M/r2) and as g = dv/dt = dv/dr ⋅ dr/dv = dv/dr ⋅ v

    we got

    ∫v⋅dv = ∫G⋅M/r2 = 0.5⋅v2 = G⋅M⋅(-1/r)

    v(r) = √(2⋅G⋅M⋅(-1/r)) = dr/dt now we integrate both sides ∫dt = ∫1/[ √(2⋅G⋅M⋅(-1/r) ] ⋅dr

    and finally

    t = t0 + (2/3)⋅√(r3/-2⋅G⋅M)

    Please tell me what is wrong or if what i´m doing makes sense
  2. jcsd
  3. May 25, 2017 #2
    First, sign: a = -GM/r2: acceleration is towards the mass.
    Second, you need a definite integral: ∫-G⋅M/r2 = GM(1/r - 1/r0)
  4. May 25, 2017 #3
    Well, I have done that and now i have something that I dont know how to integrate (I´m actually not very good at calculus)


    Sorry for the photo :/
  5. May 25, 2017 #4


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    There are a few old threads about this. Note that these threads are considering the time it takes for two objects to collide (both objects accelerate towards each other), but it seems they have similar intermediate formula for velocity, M for the one object case, (m1+m2) for the two object case. For the 2 object case, if m1 >> m2, then the sum (m1+m2) ~= m1, so let M = m1. Some of these consider point objects others consider objects with non-zero radius.

    See post #11 for a Kepler's law approach and post #19 for the approach used here:

    Non constant accelleration equation(s)

    Using a different substitution here, and also allowing for non-zero radius objects:

    Rectilinear motion of two attracting masses

    See post #66, #68, and #73 for examples of both substitutions:

    Last edited: May 25, 2017
  6. May 25, 2017 #5
    Oh, thanks! :bow:

    Now I will spend a few hours trying to understand that hahaha but I´m sure that it will be worth :D
  7. May 31, 2017 #6
    Hi again
    There is something that I don't understand in the solution of the problem (the part after "using arildno's method") which integrating method did you use to solve sqrt (r/r0-r) dr?
  8. May 31, 2017 #7


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    I guess you have the integral
    $$I=\int \mathrm{d} r \sqrt{\frac{r}{r_0-r}}.$$
    Let's try the substitution
    $$r=r_0 \cos^2 u.$$
    Then you have
    $$\mathrm{d} r= -2r_0 \sin u \cos u$$
    and thus
    $$I=-2r_0 \int \mathrm{d}u \cos^2 u.$$
    Now we have
    $$\cos(2u)=\cos^2 u - \sin^2 u=2 \cos^2 u-1 \; \Rightarrow \; \cos^2 u=\frac{1+\cos(2u)}{2}$$
    and thus
    $$I=-r_0 \left [u+\frac{1}{2}\sin(2 u) \right]=-r_0 (u+\sin u \cos u)=-r_0 \left [\arccos \left (\sqrt{r/r_0} \right)+\sqrt{(r/r_0-r^2/r_0^2)} \right].$$
  9. May 31, 2017 #8
    ohhhh, okey, now i got it hahaha
    Thanks to all ^^
  10. May 31, 2017 #9


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    The prior threads showed two different substitutions. The arildno substitution is:

    $$u = \sqrt{\frac{r}{r_0-r}}$$

    That post then solves for r and then takes dr to end up with:

    $$dr = \frac{2 r_0 u \ du}{(1 + u^2)^2}$$

    This results in :

    $$\sqrt{\frac{r}{r_0-r}}\ dr = \frac{2 r_0 u^2 \ du}{(1 + u^2)^2}$$
    Last edited: May 31, 2017
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