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Accelaration of gravity g(t)=?

  1. Aug 5, 2010 #1
    H. Young (page 319) says g=GM/R^2. This means g(R)=GM/R^2. So, g(t)=?
    Last edited: Aug 5, 2010
  2. jcsd
  3. Aug 5, 2010 #2


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    Yep. If an object's height changes, it will be subject to a changing gravitational acceleration.
  4. Aug 5, 2010 #3
  5. Aug 5, 2010 #4
    g(t) = t^2 when the object is falling toward the earth.
  6. Aug 5, 2010 #5
    No, g(t)=(considered constant)=g and then Δx(t)=(1/2)gt^2.

    But this "constant" is a simplification according to the GMm/R^2. The g(t) is an unfamiliar solution of an differential equation. What that equation and its solution is exactly, was my question.
  7. Aug 6, 2010 #6


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    g(t) depends on the problem. Assuming a particle that its trajectory is described by position vector r(t), then [tex]g(t)=\frac{GM}{\abs{r(t)}^2}[/tex]
    Last edited: Aug 6, 2010
  8. Aug 6, 2010 #7


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    Couldn't have said it better :approve:
  9. Aug 6, 2010 #8


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    Yes well hm, he probably is interested in the case of free fall where the differential equation is [tex]\frac{d^2r}{dt^2}=\frac{GM}{r^2}[/tex]. Hmm my math abandons me here, what is the solution to that ODE? Boundary conditions [tex]r(0)=c, r^{'}(0)=0 [/tex]
  10. Aug 6, 2010 #9
    R(t) is not known either. There is a problem with which symbolisms are the less wrong...anyway.

    where A is the initial R where there was immobility.

    You can correspond the problem with a weird spring that pulls a mass and it is supposed that g(R)=GM/R^2. Here you may discover the possibility that the logic of H.Young is wrong, but let’s find the solution according to that supposition first.

    When I put e.g. f’’(t)=2/((5-f(t))^2 at
    I get a complication which is indefinite even when replacing c1=c2=0
  11. Aug 6, 2010 #10


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    The usual method to solve this type of ODE is to define [itex]v=r'(t)[/itex], and the use the chain-rule to transform this 2nd order ODE into a 1st order separable ODE for [itex]v(r)[/itex]:

    [tex]\frac{d^2 r}{dt^2}=v\frac{dv}{dr}=\frac{1}{2}\frac{d}{dr}(v^2)=\frac{GM}{r^2}[/tex]

    Then the fundamental theorem of calculus gives you exactly what you might expect from conservation of energy:


    where the constant can be interpreted as the particle's mechanical energy per unit mass. You can then solve this algebraic equation for [itex]v[/itex], giving you a somewhat ugly, but separable 1st order ODE for [itex]r(t)[/itex].

    You need to be a little careful with setting the constants to any old value you feel like as they have physical interpretations, and hence only certain values are physically allowed. The same goes for [itex]f(t)[/itex]: what happens as it approaches 5?:wink:
  12. Aug 6, 2010 #11
    PLEASE do not mix "potential and kinetic energy" with the problem. It is definitelly not needed. Newton alone is enough here, and potential energy was invented afterwards.

    It does not approach R=5. R=5 is where the free fall starts or correspondingly the distance A from the balance point of that weird string. It aprooaches R=0. Δx=f(t) and u(t)=f'(t), and
    f(t=0)=0 and u(t=0)=0. So my answer to my problem should be ready just by substituting c1 and c2 at the calculator of wolfarmalfa.com. But in this case c1=c2=0 and the substitution gives what?
    Last edited: Aug 6, 2010
  13. Aug 6, 2010 #12
    I don’t think that gabba was relying on the concept of potential or kinetic energy to derive his equation [ (1/2)(v^2) = -GM/r + constant ] in the last post. As you say, the problem can be analyzed solely by reference to the idea of acceleration (here including variable acceleration) and the Law of Universal Gravitation.

    It seems to me that his equation is right, and it makes sense that it agrees with what was later found out about energy conservation. But the equation was derived by integrating the original statement of r’’ = GM/(r^2), rather than by invoking energy conservation.

    By the way, it seems to me that (1/2)(v^2) should increase as (r) decreases, because (v) should increase as (r) decreases.
    So I think you have to modify the sign in the original integral so as to make it:

    r’’ = -GM/(r^2) , and accordingly: (1/2)(d/dr)(v^2) = -GM/(r^2) ,

    which will then be integrated to: (1/2)(v^2) = - (-GM/r) = GM/r.

    The modified statement of the integral is equivalent to saying that the force is attractive. The statement without the negative sign is not “wrong”, but it’s just the magnitude of the acceleration at a given distance and doesn’t specify its direction.

    Last edited: Aug 6, 2010
  14. Aug 7, 2010 #13
    My solution was very probably wrong.

    As for the minus sign, it seems that the need for it is not something to do with the Energies (I said PLEASE, now I say PLEASE PLEASE do not mix or mention the Energies, it is me who asked the question of the thread), but that r is reduced with the passing of time. Because at the spring it seems that they (H.Young p.354) say x’’(t)= -(k/m)x(t) because the distance from the balance point is reduced. Hm...but then it is increased. Anyway, so it seems that (according to H.Young's g=GM/r^2) for the same reason the solution for gravity should be
    g(t)=r’’(t)= -(k/m)/(r(t))^2=-GM/(r(t))^2

    Not sure though.
    Last edited: Aug 7, 2010
  15. Aug 7, 2010 #14
    Yes, the acceleration is such that it tends to make the distance between the objects smaller. The analogy to a Hookean spring is correct in that regard.

    Though of course a Hookean spring isn’t a full analogue to gravity (unlike the Coulomb force for example, which is a total analogue) because there the restoring force is proportional to (r), where r = Δx. Whereas it’s proportional to 1/(r^2) for gravity.

    As to the problem of the numerical integration not working out:

    The equation |r’’(t)|=2/(5-r(t))^2 doesn’t seem to make sense to me as a representation of gravitational acceleration at distance r = r(t). I assume that you’ve used (5-r(t)) because r(0)=5.
    But by changing the expression for r’’(t) in that way, you’re actually altering the acceleration as a function of distance, rather than just defining a reference “zero-point distance” in a coordinate system.

    If that equation were accurate then:

    the acceleration would grow without bound (become “infinitely large”) as r→5;

    the acceleration would be larger than it actually is (i.e. proportional to 1/(r^2) at distances greater than 5 units;

    at distances less than 5 units, the acceleration would start off being “infinitely large” and would continually decrease and approach a value proportional to 1/(5^2=25) as r→0, rather than growing without bound as r→0.

    Obviously none of the above are seen to be the case when it comes to gravity.

    I think that |r’’(t)|=2/(r(t)-5)^2 (i.e, inverting the signs of (-r(t)) and (5) in that part of the expression) will hold if you wish to define the location of mass (M) as being 5 units from the zero point of your coordinate system.
    But for the purposes of this problem that seems to be a superfluous step, and |r’’(t)|=2/(r(t))^2 should serve just as well.

    The absolute value bars are in place in all the above per the directional consideration I mentioned in my first comment. Like I said, I think the equation (including the attractive directional info) should be: r’’(t) = -(constant)/(r(t))^2.

    However, when I use the solver at the site to which you linked (wolframalpha.com), it doesn’t seem to work out.

    Granted, I entered the above literally as a second-order DE rather than finding and entering the first-order DE as gabba suggested. I haven’t found and entered the 1st-order DE because I haven’t been able to determine what it should be so far.

    Although I have taken a DiffEq class including the transformation of 2nd-order into 1st-order DEs like that, I must admit it doesn’t yet intuitively make sense to me. I’m trying to grok it and to work out an equivalent 1st-order statement that can be integrated numerically (e.g. via your website link). I’ll post again when I have it. Though someone who fully understands the concept already will probably beat me to it.
  16. Aug 7, 2010 #15


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    Look, this is a silly request. The problem is a differential equation, and the energies are constants of motion which make the equation easier to solve. If you don't allow the use of energies then I don't think that you can get a closed-form solution and the best you can do is a numerical differentiation. If that is all you want then just plug in the equaqtion Delta² gave into your favorite math package, otherwise energy is going to be involved.
  17. Aug 7, 2010 #16
    Actually, the silly is that Energies are needed. Otherwise is like you are saying Newton would need the Energies to support his universal attraction. And that he could not solve it with elementary calculus. Potential Energy was invented afterwards, and I think kinetic energy too. You can start another thread with Energies, otherwise I will stop reading your posts if you mention Energies again.
  18. Aug 7, 2010 #17
    The only way I could solve the problem was to do exactly the same procedure with the spring, i.e. to change the frame of reference:

    The solution is r''(t)=-GM/(r(t))^2 where we pull the gravity spring at the right streching it at radious A, where we are holding the mass m of a stone in our hands before we drop it to free fall at t=0. The centre of the Earth is the balance point of the spring. This perhaps is half-wrong, but necessary for the moment (also because perhaps this is what the equation g=GM/R^2 claims). r(t=0)=A and r'(t=0)=0, thus the reference frame says that r=0 is at the centre of the Earth. So the solution is the exact respective with what the equation x''(t)= -(k/m)x(t) of the spring says. And thus this means that

    But then we need the acceleration in regard to a reference frame that the position zero is at where we are holding the stone. So, since we found r(t), we can now change the frame of reference because f(t)=Α-r(t), where f(t) is the position function in relation to time in the other reference frame.

    One weird thing is that when choosing f(t) at the first place, then f(t=0)=0 and f'(t)=0, so for that or another reason it is impossible to find the differential equation that solves the problem.

    But I am not sure the logical step x''(t)= -(k/m)x(t)=>(logical step)=>r''(t)=-(GMm/m)/(r(t))^2=r''(t)=-GM/(r(t))^2 is correct, and the worlframalfa does not give the answer now. Yestarday it gave me an answer (!), but is was not of the form r(t)=(no r(t) here). You know any other free online calculator for second order dif. eq.?
    Last edited: Aug 7, 2010
  19. Aug 7, 2010 #18


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    There are posts with solutions for the time it took for two objects at rest to collide due to gravity, one for fixed sized objects, the other for point (zero size). I'll see if I can find those old threads.
  20. Aug 8, 2010 #19
    @DaleSpam: I could be wrong, but it does seem to me that an analytical solution (as opposed to simply numerical integration) is possible with this sort of DE, without having to rely on principles of energy conservation. Of course the solution will also make sense if you look at it in terms of energy conservation, but you don’t necessarily have to know that in order to solve the problem.

    [Tangent: By analogy (which isn’t really a terribly similar problem, but meh) if you’re dealing with one of those situations where a bullet undergoes an inelastic collision with a block of wood that’s strung from the ceiling like a pendulum, you don’t need to use [(m.block)*g = (1/2)(m.bullet)(v.initial.bullet)^2].
    If you have data for Δx/Δt = v.avg.x and Δy/Δt = v.y , where the differences are sufficiently small, you can get a decent numerical-integration estimate of the initial speed of the bullet, using the fact that Δv.avg.x/Δt = cos(θ)*T.avg and Δv.avg.y/Δt = sin(θ)*T.avg.
    Of course all the variables in the above paragraph should technically have index subscripts (e.g. Δx.i/Δt.i) to emphasize that there are as many of them as you have data points, but I think that would’ve made my expressions even more annoying to read.
    In practice it’s quite hard to measure a system like that with enough temporal granularity to make your estimate useful, since it’s going to be moving really fast. So it’s easier, if more recondite, to use energy conservation.]

    I'm still trying to work out the solution though. I haven’t done 2nd-order DEs in a while. I think that gabba’s suggestion of reducing the 2nd-order equation to a 1st-order equation is the best way to proceed. But I need to go back and consult my textbook in order to really grasp how that works.

    Also, I think luckis11 is specifically trying to solve the problem via a numerical integration estimate, rather than by finding the analytic solution.


    If you’re taking a course where you need to do numeric integrals like this, I suggest you get a full computational program. MATLAB (which I have) is available for a pretty reasonable price if you’re a currently-enrolled student somewhere (though the non-student price is fairly ridiculous).
    Via a quick check I found:
    http://en.wikipedia.org/wiki/Genius_(mathematics_software) ; http://www.jirka.org/genius.html

    But I’m pretty sure the problem isn’t one of the software not working properly. It’s rather that we’ve fed it some DEs that either described the situation incorrectly, or didn’t specify the boundary points you need.
    It doesn’t look like there is a way to specify the boundary points via WolframAlpha; that’s probably one of the features for which they hope you’ll buy Mathematica, etc.

    Also, if you’re on campus, there might be library or lab computers somewhere that have MATLAB or Mathematica or another numerical-integration-capable program.

    However, I realize that when I submit r’’(t) = -(constant)/(r(t)^2) to the wolframalpha program, the graphs of (r) vs. (t) and (r’) vs. (r) solutions for the case of r(0)=1 and r’(0)=0 appear plausibly consistent with gravitational acceleration from rest.

    In particular the (r) vs. (t) graph looks correct.

    However the (r’) vs. (r) graph looks like it inflects the “wrong” way; i.e. it seems that r’(t) should be reduced as (t) increases and (r) decreases. But perhaps it does work out because it’s a graph of (r’) vs. (r) , rather than (r’) vs. (t), and (r) does not follow a linear path with respect to (t).
    It’s also frustrating that there are no reference points marked on the graphs.
    Once again, the decrease I’m referring to above arises from the fact that the force is attractive.
    Last edited by a moderator: Apr 25, 2017
  21. Aug 8, 2010 #20
    how do satellites transmit their signals...??
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