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Freefall in a newtonian gravitational field

  1. Jun 17, 2013 #1
    1) The problem statement :

    For this problem, We use Newtonian Mechanics. We are placed in a geocentric frame of reference. An object (of which the mass is irrelevant for this problem) is released into the earth's gravitational field at an altitude p(0) with no velocity whatsoever.
    The question is how do you describe the motion of this object ? In other words, how to find the function p(t) which associates position p to any instant t of the free fall considering that the acceleration is given as a function of position a(p) = (G*M)/(p^2) where G is the gravitational constant and M the mass of the earth.

    2) Relevant equations :

    a(p) = (G*M)/(p^2)

    3) Attempted solution :

    I realized i was confronted with a differential equation and that its solution would be the function I'm looking for. However, I'm new to differential equations and I could only go this far :

    a(p) = (G*M)/(p^2) means that p''(t) = (G*M)/(p(t))^2

    So I would have to solve this :

    y'' = k/(y)^2 ( with k = G*M )

    This where I need your help.

    Many thanks for considering my request.
  2. jcsd
  3. Jun 17, 2013 #2
    a = dv/dt.

    Hint: What happens when i multiply the RHS numerator and denominator by dx.
  4. Jul 30, 2013 #3
    Integration difficulties

    Hi !

    Thanks to your hint, I was able to make some good progress towards the solution of my differential equation.

    If we assume an initial height "h" (y(0) = h) , and a initial speed of zero (v = dy/dt (0) = 0), we can start solving it :

    [itex]\frac{d^{2}y}{dt^{2}}[/itex] = [itex]\frac{dv}{dt}[/itex]

    We then apply the chain rule as you suggested in order to get :

    [itex]\frac{d^{2}y}{dt^{2}}[/itex] = [itex]\frac{dv}{dy}[/itex] * v

    [itex]\frac{dv}{dy}[/itex] * v = [itex]\frac{- k}{y^{2}}[/itex]

    We integrate both sides with respect to dy, and get :

    ∫ [itex]\frac{dv}{dy}[/itex] * v dy = ∫ [itex]\frac{- k}{y^{2}}[/itex] dy

    So :

    [itex]\frac{v^{2}}{2}[/itex] = [itex]\frac{k}{y}[/itex] + C (C is a constant).

    Which means that :

    v = [itex]\sqrt{\frac{2k}{y} + 2C}[/itex]

    We know that v(0) = 0 and y(0) = h, so we can write :

    0 = [itex]\sqrt{\frac{2k}{h} + 2C}[/itex]

    Which gives us C = - [itex]\frac{k}{h}[/itex]

    Thus, we get :

    v = [itex]\sqrt{2k}[/itex] * [itex]\sqrt{\frac{1}{y} - \frac{k}{h}}[/itex]

    [itex]\frac{dy}{dt}[/itex] = [itex]\sqrt{2k}[/itex] * [itex]\sqrt{\frac{1}{y} - \frac{k}{h}}[/itex]

    We rearrange the above equation to get :

    [itex]\frac{dy}{\sqrt{\frac{1}{y} - \frac{k}{h}}}[/itex] = [itex]\sqrt{2k}[/itex] dt

    We integrate both sides :

    ∫ [itex]\frac{dy}{\sqrt{\frac{1}{y} - \frac{k}{h}}}[/itex] = [itex]\sqrt{2k}[/itex] * ∫ dt

    ∫ [itex]\frac{dy}{\sqrt{\frac{1}{y} - \frac{k}{h}}}[/itex] = [itex]\sqrt{2k}[/itex] * ( t + C' )

    C' is a constant.

    So solving the problem comes down to being able to integrate the LHS and then writing y as a function of t. I am not able to do that. I used Wolfram and got a function that gives a complex output, I don't know how to interpret that or use it in my equation.
    The link to the Wolfram integration I did : http://integrals.wolfram.com/index.jsp?expr=1%2FSqrt[1%2Fx+-+k%2Fh]&random=false

    Thank you.
  5. Jul 30, 2013 #4

    D H

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    By taking the square root, you are implicitly assuming velocity is positive.

    The object is released at time t=0 with zero velocity. It falls downward. Since positive y is upward, the velocity will be negative up to the point where the object impacts the Earth.

    You should be using the additive inverse, [itex]v=-\sqrt{2k/y + 2C}[/itex].
  6. Jul 30, 2013 #5
    Oops ! My bad. Thanks for spotting that !

    So that gives us :

    ∫ [itex]\frac{dy}{\sqrt{\frac{1}{y} - \frac{1}{h}}}[/itex] = - [itex]\sqrt{2k}[/itex] * (t + C')

    And I have the LHS integration problem again.
  7. Jul 30, 2013 #6
    Try ## \cos u = \sqrt {y/h} ##.
  8. Jul 30, 2013 #7

    D H

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    And try solving for t as a function of y rather than y as a function of t.
  9. Aug 2, 2013 #8

    So I finally came up with this, pfew :smile: :

    [itex] t = \frac{\sqrt{h^{3}}(arccos\sqrt{\frac{y}{h}} + \sqrt{\frac{y}{h} - \frac{y^{2}}{h^{2}}} ) }{\sqrt{2k}}[/itex]

    This is good but isn't the whole point of studying motion to be able to write the position as a function of time ?
  10. Aug 2, 2013 #9
    Being able to find an explicit formula for position as a function of time is an exception rather than the rule. In most cases, you cannot even solve the differential equation in any way.

    In this case, the position is an implicit function of time, which still much better than the differential equation you started from. You can plot the function.

    You could simplify your equation somewhat by defining ## x = ct, \ z = y/h ##, and the plot of x(z) will work for any such problem, it just needs scaling.
  11. Aug 2, 2013 #10

    D H

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    That's a transcendental equation. You aren't going to be able to find the inverse of that equation, one that gives position as a function of time, in the elementary functions.
  12. Aug 2, 2013 #11
    Many thanks !

    Okay !

    Thank you very much for all the help ! :biggrin:

    I hope I'm not getting too off topic here but, since in most cases you can't even solve the equation (this is news for me, i'm shocked), what would an engineer do to predict complicated motion described by complicated DEs ? (Computational methods, which are only approximations of reality ?)

    Again, thank you so much, I've been stuck with this problem for longer than I can confess. :wink:
  13. Aug 3, 2013 #12
    This is the reality of engineering. This is why we invented computers to begin with.
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