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Free fall trajectory in non constant gravitational field

  1. Jul 29, 2013 #1
    Rusty on the math...
    I am working on a problem and need the derivation of a free fall trajectory for an object at a distance above earth where the change in acceleration is not negligible. How do I integrate the distance/ acceleration formula taking into account the change in g to get the total distance traveled.
    The method and final formula...?
    I'm looking for total distance after time t, trying to see how it scales as what power of t.
    thanks.
     
  2. jcsd
  3. Jul 29, 2013 #2

    QuantumPion

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    You need to know what the function of g with respect to position is, and then plug it into your integral instead of assuming it is a constant.
     
  4. Jul 29, 2013 #3
    Let me get this right. The object is tracing a trajectory where 'g' changes according to the distance from the Earth?
     
  5. Jul 29, 2013 #4

    QuantumPion

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    Over long enough range, g is not constant as the Earth is not of uniform density. Although I suspect what the OP was interested in was the decreasing effect of gravity with altitude. In which case you would substitute newton's law of gravitation in for the constant g in your integral g(x)=G M1 M2 / x^2.
     
  6. Jul 29, 2013 #5
    You possibly can't determine the density at every point. Has to be g(x). Like QuatumPion said, put it in the equation in terms of x and integrate, In this case, it'll be in terms of 'y' as 'x' would be the horizontal distance traveled by the projectile.
     
  7. Jul 29, 2013 #6

    QuantumPion

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    Probably, but even for the first case if you have a gravity map and know the trajectory/inclination than you could make a function of g with respect to position, although that would be fairly complicated (depends on orbital parameters, rotation of earth, etc).
     
  8. Jul 29, 2013 #7

    HallsofIvy

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    Sidhartha23 and QuantumPion have argued back and forth but we haven't heard back from Creator so I am going to look at the specific case Creator asked about: An object falling from such a great height the we must use the general law for gravity: [itex]F= -GmM/r^2[/itex], rather than treating F as a constant, still satisfies "Force= mass times acceleration". Here "m" is the mass of the object, M is the mass of the earth so we have [itex]ma= -GmM/r^2[/itex] or
    [tex]a= \frac{dv}{dt}= -\frac{GM}{r^2}[/tex]

    We would like to integrate that but we need to integrate with respect to t and have "r" on the right. To fix that, we can use a method called "quadrature". By the chain rule, dv/dt= (dv/dr)(dr/dt)= v(dv/dr) becaues v, the speed, is dr/dt. Now we have
    [tex]v\frac{dv}{dr}= -\frac{GM}{r^2}[/tex]
    which we can write, in "differential form" as
    [tex]vdv= -\frac{GM}{r^2}dr= -GMr^{-2}dr[/tex]

    Integrating both sides,
    [tex]\frac{1}{2}v^2= GMr^{-1}+ C[/tex]
    assuming that the object falls from a standstill at height R, v= 0, so [itex]0= GMR^{-1}+ C[/itex], [tex]C= -GMR^{-1}[/tex].

    From that,
    [tex]v= \frac{dr}{dt}= \sqrt{2GMr^{-1}- 2GMR^{-1}}= \sqrt{2GM}\sqrt{r^{-1}- R^{-1}[/tex]

    We can write that as
    [tex]\frac{dr}{\sqrt}2GM}\sqrt{r^{-1}+ R^{-1}= dt[/tex]

    Integrating both sides of that will give t as a function of r which can then solved for r as a function of t. However, the left side cannot be integrated in terms of "elementary functions". It is an "elliptic integral" (the name, unsurprisingly, connected with the fact that planets move in ellipses around the sun). There used to be multi-volumes of tables of elliptic integrals.
     
  9. Jul 29, 2013 #8
    Thanks for all the quick responses....
    You are correct Siddarth23 and Quantum Pion....about my question.
    Creator
     
  10. Jul 29, 2013 #9

    Thank you Halls ...that is exactly what I was looking for...however, I only followed you up to this eqn. (4th one)...
    \frac{1}{2}v^2= GMr^{-1}+ C

    And your next step (5) I don't understand since I thought the v here was final velocity...

    Your last two eqns. remained in ITEX form and did not print correctly....could you give the last two again. Maybe its my browser...mot too famailiar with mathjax.
    Thanks;
    Creator
     
    Last edited: Jul 29, 2013
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