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Free falling particle near Earth's surface (Diff. Eq.)

  1. Feb 26, 2016 #1
    1. The problem statement, all variables and given/known data
    The free fall acceleration of a mass ##m## above the earth's surface in one dimension can be represented by ##m\ddot{y}=-\frac{mMG}{y^2}## where ##M## is the mass of the earth and ##G## is the gravitational constant. With ##\dot{y}(t=0)=0## and ##y(t=0)=y_0##.. (1) Find an equation for ##y(t)##, then ##y(t)## can be represented by the equation ##y(t)=R+\alpha(t)## where ##R## is the radius of the earth. (2) Solve the differential equation in (1) using this, that ##\alpha_0## and ##\alpha(t)## are ##<<## than ##R##, ##y(t=0)=R+\alpha_0## and the taylor expansion ##f(t)=f_0+tf'(t)##

    2. Relevant equations
    3. The attempt at a solution


    The first part of this question is quite trivial, multiplying the first equation by ##\dot{y}## and cancelling common factors then integrating we obtain $$\int_{0}^{\dot{y}}\dot{y}dy=-MG\int_{y_0}^{y}\frac{1}{y^2}$$
    which after solving for ##y(t)## results in $$y(t)=\frac{2MGy_0}{\dot{y}^2y_0+2MG}$$ Continuing on to the second part of the question now we have ##R+\alpha(t)=\frac{2MGy_0}{\dot{y}^2y_0+2MG}##, note that since ##y(t)=R+\alpha(t)## applying the initial conditions we obtain ##\alpha(t=0)=\alpha_0## and taking the derivative we find that ##\dot{y}=\dot{\alpha}(t)\Rightarrow \dot{y}^2=\dot{\alpha}(t)^2## which gives us and then using the relation ##y_0=R+\alpha_0## $$R+\alpha(t)=\frac{2MGy_0}{\dot{\alpha}^2y_0+2MG}=\frac{2MG(R+\alpha_0)}{\dot{\alpha}^2(R+\alpha_0)+2MG}$$

    Now I'm pretty sure I can get rid of the ##\dot{\alpha}^2\alpha_0## term on the denominator since it should be negligible compared to the other terms. I'm quite confused as to what to do with the taylor expansion though since I don't understand why we are using the Maclaurin series expansion (my professor told us to use this) when we haven't defined that ##t\approx 0##, regardless, if I plug in the expansion and remove the ##\dot{\alpha}^2\alpha_0## term I obtain $$R+\alpha_0+t\dot{\alpha}=\frac{2MG(R+\alpha_0)}{\dot{\alpha}^2R+2MG}$$

    After some tedious simplification I arrived at $$\dot{\alpha}^2R^2t+\dot{\alpha}(R^2+\alpha_0R^2+2MGt)=2M\alpha_0(1-G)$$

    Which just seems way too complicated (I have no clue how I would go about solving this) so I'm assuming I have made a mistake somewhere, if I had to guess I would say it has to do with the taylor expansion.
     
  2. jcsd
  3. Feb 26, 2016 #2

    ehild

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    you missed a dot in the left side and a dy from the right side.
    Divide both the numerator and the denominator by 2MG, and also divide the whole equation by R, so as you have factors in form 1+small quantity. You can consider both ##\alpha/(2MG) ## and ## \dot \alpha^2 R/(2MG) ## small with respect to 1. Do the Taylor expansion, ignore higher order terms.
    I can not follow you... try the expansion, and integrate for ##\alpha##.
     
  4. Feb 26, 2016 #3
    Okay multiplying the numerator and denominator of the RHS by ##\frac{1}{MG}## and dividing by ##R## I get $$1+\frac{\alpha}{R}=\frac{1+\frac{\alpha_0}{R}}{\frac{\dot{\alpha^2}(R+\alpha_0)}{2MG}+1}$$

    For the taylor expansion of ##\alpha## I have ##\alpha(t)\approx\alpha(0)+\alpha^{'}t##, ##\alpha(0)=\alpha_0## and since ##y(t)=R+\alpha(t)\Rightarrow \dot{y(t)}=\dot{\alpha(t)}\Rightarrow \dot{y}(0)=\dot{\alpha}(0)=0\Rightarrow \alpha(t)\approx\alpha_0##

    I feel like this taylor expansion isn't correct and that the entire ##\frac{\dot{\alpha^2}(R+\alpha_0)}{2MG}## term should go to zero.
     
  5. Feb 26, 2016 #4

    haruspex

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    It says "find an equation for y(t)". That suggests to me y as a function of t, not as a function of ##\dot y##. Admittedly, that is quite a challenge, unless you are allowed to leave it in integral form.
     
  6. Feb 26, 2016 #5
    Edit: misread

    My professor wanted us to first find the differential equation in terms of ##\dot{y}## and then use the simplifying details along with the taylor expansion to solve it. The question was from a midterm a few days ago so the wording isn't exact.
     
  7. Feb 26, 2016 #6

    haruspex

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    Ok, then it should have said find an equation for ##y(\dot y)##.
    Be wary of dropping small terms too soon. There will be a lot of cancellation of the largest terms.
    For the Taylor expansion, you might need to include the ##\ddot \alpha## term, which you can approximate as MG/R2.
     
  8. Feb 26, 2016 #7

    ehild

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    Multiply the equation with the denominator and divide it by 1+α/R. α/R is small compared to 1 so you can approximate 1/(1+α/R) by 1-α/R and ignore the product (α/R)α(0)/R. Take the square root odf the equation. You get a separable linear differential equation for alpha.
    Nobody said that you have to replace alpha by its first order Taylor expansion. Do ##\dot \alpha=p+qt##. Integrate to get α(t). Substitute into the equation and compare the powers of t.
    At the end you get the usual equation for the height of a falling body.
    Edit:
    As ##\dot y(0)=0## was said in the OP, ##\dot \alpha=qt##.
     
    Last edited: Feb 27, 2016
  9. Feb 27, 2016 #8
    Okay so
    $$
    1+\frac{\alpha}{R}=\frac{1+\frac{\alpha_0}{R}}{\frac{\dot{\alpha^2}(R+\alpha_0}{2MG}+1}\\
    1+\frac{\dot{\alpha^2}(R+\alpha_0)}{2MG}=\frac{1+\frac{\alpha_0}{R}}{1+\frac{\alpha}{R}}=(1+\frac{\alpha_0}{R})(1+\frac{\alpha}{R})^{-1}\approxeq (1+\frac{\alpha_0}{R})(1-\frac{\alpha}{R})\\
    1+\frac{\dot{\alpha^2}(R+\alpha_0)}{2MG}=1-\frac{\alpha}{R}+\frac{\alpha_0}{R}-\frac{\alpha\alpha_0}{R}\approxeq=1-\frac{\alpha}{R}+\frac{\alpha_0}{R}\\
    \dot{\alpha^2}\frac{R+\alpha_0}{2MG}=\frac{1}{R}(\alpha_0-\alpha)\Longrightarrow \dot{\alpha^2}=\frac{\frac{2MG}{R}(\alpha_0-\alpha)}{R+\alpha_0}\\
    \dot{\alpha}=p+qt\Longrightarrow \alpha=pt+\frac{qt^2}{2}+C
    $$

    I'm still not sure what to do with my expression for ##\dot{\alpha^2}## since I have ##\alpha## in there, I tried factoring an ##R## out of the denominator so I could sub in ##g=\frac{MG}{R^2}## and then use the binomial approximation again like so...

    $$
    \dot{\alpha^2}=\frac{2\frac{MG}{R^2}(\alpha_0-\alpha)}{1+\frac{\alpha}{R}}=2g(\alpha_0-\alpha)(1+\frac{\alpha_0}{R})^{-1}\approxeq 2g(\alpha_0-\alpha)(1-\frac{\alpha_0}{R})
    $$
    but this doesn't seem to be going anywhere useful.
     
  10. Feb 27, 2016 #9

    ehild

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    Substitute p+qt for ##\dot \alpha## and α(t) = α(0) + pt +qt2/2. Compare the terms with equal power of t.
    By the way, it was said that ##\dot y(0) = 0 ## in the OP, so you can take ##\dot \alpha = qt##.
    Edit: you can also ignore α(0) with respect to R in the denominator.
     
    Last edited: Feb 27, 2016
  11. Feb 27, 2016 #10
    If I take ##\dot{\alpha}=p+qt=\sqrt{\frac{2MG}{R}(\alpha_0-\alpha)}=\sqrt{\frac{2MG}{R}(-pt-\frac{qt^2}{2})}=\sqrt{-\frac{2MG}{R}(pt+\frac{qt^2}{2})}##
    I find it odd that the entire square root is negative, I could insert ##i^2=-1## but I'm not sure how I'll end up with the normal free fall equation if I do this.
     
  12. Feb 27, 2016 #11

    ehild

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    Do not take the square root. Write ##{\dot \alpha}^2=(p+qt)^2##
     
  13. Feb 27, 2016 #12
    Okay then I get $$
    \dot{\alpha^2}=q^2t^2=-\frac{2MGpt}{R}-\frac{MGqt^2}{R}\\
    q^2=-\frac{MGq}{R}\Longrightarrow q=-\frac{MG}{R}$$
    Now since ##q=\ddot{\alpha}## we get $$
    \dot{\alpha}=-\frac{MG}{R}t+\dot{\alpha_0}=-\frac{MG}{R}t \\
    \alpha=-\frac{MG}{2R}t+\alpha(0)=\alpha_0-\frac{MG}{2R}t^2
    $$
    Edit: whoops I just saw ##\dot{\alpha^2}=(p+qt)^2##, that shouldn't matter since ##p=0## right? (Or is it still useful for relationships?)
     
  14. Feb 27, 2016 #13

    ehild

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    Yes, you could have taken ##\dot{\alpha}=qt##
     
  15. Feb 27, 2016 #14
    Hmm ##\alpha=\alpha_0-\frac{MG}{2R}t^2=\alpha_0-\frac{gR}{2}t^2## seems inaccurate to me?
     
  16. Feb 27, 2016 #15

    ehild

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    You lost an R. Go back to #8.
     
  17. Feb 27, 2016 #16
    $$
    \dot{\alpha^2}=\frac{\frac{2MG}{R}(\alpha_0-\alpha)}{R+\alpha_0}\approxeq \frac{2MG}{R^2}(-pt-qt^2) \\
    q^2t^2=-gqt^2\Longrightarrow q=\ddot{\alpha}=-g \\
    \dot{\alpha}=-gt\Longrightarrow \alpha=\alpha_0-\frac{gt^2}{2}
    $$
    This was a hilarious amount of work for an equation that anyone could've guessed lol. I have a question pertaining to the final method we used to solve for ##q=\ddot{\alpha}##, is the method of using ##\dot{\alpha}=p+qt## and ##\alpha=a_0+pt+qt^2/2## then comparing coefficients a common way for solving a certain type of differential equation or was it just a simple method to solve this specific case.
     
  18. Feb 27, 2016 #17

    ehild

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    We did a lot of approximations, and one has to take care with them. Sometimes it works. It worked now.
    Do you can solve differential equations? Try to get the result by solving the equation you got for alpha in Post #3 without the taylor expansion.
     
  19. Feb 28, 2016 #18
    Hmm I can't seem to get it.

    Should I try to solve ##\dot{\alpha^2}=2g(\alpha_0-\alpha)## by finding a homogeneous solution and a particular solution or should I try to obtain a different ##DE##?
     
  20. Feb 28, 2016 #19

    ehild

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    It is not a linear equation. Take the square root, separate variables,and use u=√(α0-α) substitution in the integral.
     
  21. Feb 28, 2016 #20
    Ok so I have ##d\alpha=\sqrt{2g(\alpha_0-\alpha)}dt##, if I set ##u=\sqrt{\alpha_0-\alpha}## then ##du=-\frac{1}{2}(\alpha_0-\alpha)^{-1/2}dt## so I should rewrite ##d\alpha## as ##\frac{2g(\alpha_0-\alpha)dt}{\sqrt{\alpha_0-\alpha}}=2gu^2du## then ##\alpha=\frac{2}{3}gu^3=\frac{2}{3}g(\alpha_0-\alpha)|_{0}^{t}##, is this correct? I am going to have different powers of ##\alpha## so I am skeptical.
     
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