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How to find the jerk of the acceleration of gravity?

  1. Oct 14, 2009 #1
    Good evening,

    The definition of jerk is change in acceleration with respect to time. So, we can define it as a derivative:
    [tex]\vec{j} = \frac{d\vec{a}}{dt}[/tex]
    In absolute value:
    [tex]j = \frac{da}{dt}[/tex]
    For most physics problems, the acceleration of gravity, g, is considered constant, since the height variations involved are relatively small. However, since g varies with the inverse of the square of the distance to the Earth's center (g = GM/r²), there is a negligible change in it while a body is falling in the ground. Thus, there is a jerk.
    We are just curious: we were wondering how we could calculate the jerk of the acceleration of gravity with respect to time. What should its value be?
    The variation of g with respect to the radius interval travelled would be given by dg/dr, but this is not what we are looking for.
    Any ideas?

    Thank you in advance.
    (NOTE: we know some calculus from self-teaching - basic notion of derivative and integral and how to calculate them)
     
    Last edited: Oct 14, 2009
  2. jcsd
  3. Oct 14, 2009 #2
    well you could treat it as a single body problem (ie, it is just "you" that is moving, where the earth is still) and then calculate the force due to gravity with the inverse square law and set it equal to mass*accelleration via newton's law. Then you have an equation of motion which will essentially tell you all you want to know about the system given some initial conditions.

    The jerk that you are referring to will be proportional to velocity (so will depend on initial conditions)
     
  4. Oct 17, 2009 #3
    Thank you for the answer.

    [tex]mg = \frac{GMm}{r^2}[/tex]

    But, if I want to take the differential of g with respect to time, wouldn't I have to write the equation for the force with respect to time somehow in order to find the jerk?

    Thank you in advance.
     
  5. Oct 17, 2009 #4
    Precisely.

    The equation which allows you to do that is Newton's Second Law, ie: F=ma.

    You will want to treat it as a one dimensional problem (set r=x, a=x'').

    then you can find an expression (in terms of x,x') for jerk.
     
  6. Oct 17, 2009 #5
    Is this what you mean?
    [tex]\frac{\mathrm{d}^2x}{\mathrm{d}t^2} = -\frac{GM}{x^2}[/tex]

    [tex]\frac{\mathrm{d}a}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\left(-\frac{GM}{x^2}\right)[/tex]

    Inside my basic calculus knowledge there are no differential equations. So, I searched the internet and I found some techniques, and, based on them, I get this (tell me if I'm wrong):
    [tex]\frac{\mathrm{d}v}{\mathrm{d}t} = -\frac{GM}{x^2}[/tex]

    [tex]\frac{\mathrm{d}v}{\mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}t} = -\frac{GM}{x^2}[/tex]

    [tex]v \mathrm{d}v = -\frac{GM}{x^2} \mathrm{d}x[/tex]
    Integrating both sides:
    [tex]\frac{v^2}{2} = \frac{GM}{x} + constant[/tex]
    But I have no idea on how to proceed from this to jerk.
    Can you give any more help?
    By the way: can I just integrate both sides of the first equation with respect to t two times for obtaining the expression for x?
    Thank you in advance.
     
  7. Oct 17, 2009 #6
    So you are very close, and the math in your previous post looks correct, but you are making it a bit more complicated than it needs to be.

    Indeed, in order to solve the differential equation

    [tex]
    \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = -\frac{GM}{x^2}
    [/tex]

    you will need a bit of help with differential equations in order to get an expression for x(t).

    If, however, we are just seeking for an expression for the jerk (da/dt) we can start with this expression:

    [tex]
    \frac{\mathrm{d}a}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\left(-\frac{GM}{x^2}\right)
    [/tex]

    which you wrote. On the left we have jerk, and on the right we have an explicit time derivative. Since x=x(t), we find:

    [tex]
    \frac{\mathrm{d}a}{\mathrm{d}t} = \left(-GM\right) \left(\frac{-2}{x^{3}}\right)\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)
    [/tex]

    and thus,

    [tex]
    \frac{\mathrm{d}a}{\mathrm{d}t} = \frac{2GM x'}{x^{3}}
    [/tex]

    which you can plug in for a given position and velocity. remember that here we are assuming that the mass of the earth is a point mass, in which case if we want our equations to hold true, x must be greater than the radius of the earth.

    is this what you were after? the differential equation

    [tex]
    \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = -\frac{GM}{x^2}
    [/tex]

    is second order and nonlinear, so there may be a couple of ways to solve it. If you wish to proceed from the point where you were, ie:

    [tex]
    \frac{v^2}{2} = \frac{GM}{x} + constant
    [/tex]

    would be to set v=dx/dt, take the square of both sides and get:

    [tex]
    \frac{\pm 1}{\sqrt{\frac{2GM}{x}+C}} \mathrm{d}x = \mathrm{d}t
    [/tex]

    which you can then integrate to find x(t) and you will have two constants of integration to solve for base on initial conditions.

    this is not a very pretty way to do it, and (i think) there is a prettier way to find the solution using lagrangian mechanics
     
  8. Oct 17, 2009 #7
    yes, I'm exactly after an expression for the jerk. But I don't understand how you go from this expression:

    [tex]
    \frac{\mathrm{d}a}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\left(-\frac{GM}{x^2}\right)
    [/tex]

    to this expression:

    [tex]
    \frac{\mathrm{d}a}{\mathrm{d}t} = \left(-GM\right) \left(\frac{-2}{x^{3}}\right)\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)
    [/tex]

    Where did the -2 come from?
    Thank you in advance.
     
    Last edited: Oct 17, 2009
  9. Oct 17, 2009 #8
    ok so that step was done using the chain rule.

    basacally: d/dt f(x) = f'(x)*(dx/dt)

    I am on my phone on a train right now so I hope that explanation will be sufficient. in our case, f(x)=GM/x^2
     
  10. Oct 17, 2009 #9
    Thank you for the response. It was really helpful.
    I get it now.
    I may ask some more questions later, if you don't mind.
     
  11. Oct 18, 2009 #10
    I understand how you arrived to the expression of jerk, but I'm not very sure on how this expression would be applied.
    Should I find x' at a given x by using the expression below (that I found by isolating v in that expression I wrote)?
    [tex]v = \pm\sqrt{\frac{2GM}{x} + C}[/tex]
    (note: I put "C" instead of "2C" because it is just a constant)
    in this case, I'm not very sure on how I would discover the value of C. It is simple for constant acceleration, for example, v = v0 + at (C = v0 and the term at can be +at or -at depending on the orientation of acceleration with respect to the velocity), but I'm not sure on how to do this in this case (does the plus/minus of taking the square root account for the possibility of the acceleration to be positive or negative?).
    OK, I made many questions in only one paragraph...
    Thank you for your patience.
     
    Last edited: Oct 18, 2009
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