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Real free falling time

  1. Mar 29, 2007 #1
    hello fellow physitians,
    i am having a problem answering the following question, which perhaps have been answered before but which solution i have not found in either this forum or the net.
    i have asked this to my teacher but she only solved the case for a linear variation of acceleration, which does not solve the problem, but, anyway: i pretend to know what is the real time of free falling of a particle in a gravitational field. the problem lies in finding acceleration as a function of time, because, as you know, it is only given in function of distance. maybe this is a simple problem and i didn't give enough thought to it but i though it would be interesting to post this here, since i might find interesting answers.
     
  2. jcsd
  3. Mar 29, 2007 #2

    berkeman

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    Thread moved to Homework Help forums. DaNiEl, please be sure to post your homework and coursework questions in the Homework Help forums, and not in the general PF forums.

    What equations of motion determine how an object falls (ignoring air resistance I think in your question)? If an object falls from a height of a few meters to the ground, then the acceleration is constant, and written as "g". Do you know the value of g?

    But if the object falls from orbit, then the acceleration due to gravity is not constant. Do you know the equation that gives the force on an object due to gravitational attraction of another object?
     
  4. Mar 29, 2007 #3
    you missed the point of the question. it is not a home-work (at this time i have only studied the falling of an object under the influence of constant acceleration), it is rather a question of curiosity, so to speak, since is has barely any application in real life, that is, if we are talking about planet earth.
    insted, i tend to know the real falling time of a particle, in which aceleration is given by Gm/r^2. could this be one of those cases of double integrals which i have heard about (with 2 variables)? i am sorry if what i said is wrong, but i have no clue with what to do with this.
     
  5. Mar 29, 2007 #4

    berkeman

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    DaNiEl!, when you said "I have asked my teacher", that's a pretty good indicator that it's coursework of some kind. In any case, we try hard here at the PF to keep the general forums clear of schoolwork (except for advanced college projects and a few other special cases). The Homework Help forums get good help and work well (just click on "Staff" at the top of the page to see how many Homework Helpers there are!).

    And your question absolutely has application in real life! The kinematic equations of motion and their application to gravity-based problems show up in many places.

    I think at the root of your question is how do you go from an acceleration to a velocity and a position, especially when the acceleration is varying with distance as you show in your correct equation for gravitational acceleration. Yes, you integrate the acceleration to get the velocity, and integrate the velocity to get the position. Have you taken calculus yet? You should be able to show your question to the calculus teacher at your school, and have them walk you through the integrations. Hope that helps.
     
  6. Mar 29, 2007 #5
    i am having trouble finding the correct integrals (namely acceleration as a function of time). i have something of the sort:
    v = S ( GM/(R_inicial - v.t)^2 ).dt
    am i on the good path?
     
  7. Mar 29, 2007 #6

    berkeman

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    It seems like the last time I tried to do this math, I got in trouble. But I'll try to set it up the way I'd approach it, and see if others poke holes in it.

    The problem is that you have an equation that shows how acceleration varies with position, but you want to find how the objects position varies with time, as it falls in that varying gravitational field. So you have to be careful to keep track of what is varying with time, and what is varying with position. When quantities interact like that, you often end up with a differential equation that you have to solve to find the final answer.

    So I'd start something like this (assuming the object is released from rest at some distance r0, and the attracting body does not move -- like dropping a rock from orbit):

    [tex] a(r) = \frac{Gm}{r^2} [/tex]

    [tex] v(t) = 0 + \int \frac{Gm}{r(t)^2} dt [/tex]

    [tex] v(t) = \int \frac{Gm}{r(t)^2} dt [/tex]

    [tex] r(t) = r_0 - \int v(t) dt [/tex]

    [tex] r(t) = r_0 - \int \int \frac{Gm}{r(t)^2} dt dt [/tex]

    This final equation is all in terms of r(t), which is what you want to find. To solve it, you differentiate both sides twice, and solve the resulting 2nd order differential equation.

    I need to bail for today (no, really!), but I'll check back tomorrow to see if you've got it figured out. And like I said, I could be making mistakes here, or making it more complicated than it needs to be. I can also ping a couple PF hotshots tomorrow if we're still struggling.
     
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