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TIME of FREE FALL, variable g

  1. Jul 9, 2011 #1
    1. The problem statement, all variables and given/known data

    when g = constant , time = v / g
    if g varies how do I get t ?

    One electron is at rest at r0, 3600 km from surface of earth (r), 10.000 km from center.
    g(r) = 9.8 m/s^2 , g(r0) = 4, Δ PE = 2.25 x 10^7

    1) If it fell in a vacuum, after how many seconds would touchdown?

    2. Relevant equations
    (a) v = sqrt( 2x PE), (b) g= PE/ h

    3. The attempt at a solution
    from (a) => v = sqrt ((2 x2.25) 4.5 x 10^7) = 6,708 m/s
    from (b) => average g = 2.25 x 10^7 / 3,600,000 = 6.25 m/s^2, t= 6708/6.25= 1273 s
    1. The problem statement, all variables and given/known data

    2) ceteris paribus, if we change gravity with elecrostatic attraction, then drop of Potential becomes 0.0009305 eV,and final v seems to change from 6,7 to 18,1 km/ s.WHY?




    2. Relevant equations



    3. The attempt at a solution
    no idea
     
  2. jcsd
  3. Jul 9, 2011 #2

    gneill

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    Can you write an expression that would give the speed of the falling particle for any given radial distance? (HINT: conservation of energy)

    The speed of the particle is -dr/dt at radius r. (dr/dt is negative because the radius is decreasing).

    Looks like an integration is required...
     
  4. Jul 9, 2011 #3
    saying g varies, you imply that acceleration varies... look at newton's second law,

    F=ma, where a is now function of R, such as a=G/R^2




    ****hint, setting up the DE****

    dV/dt=-ma(R)

    and a change of variables:

    (dV/dt)=(dV/dR)(dR/dt) = V(dV/dR)

    gives V(dV/dR)=-mG/R^2
     
  5. Jul 10, 2011 #4
    Hello simpatico, welcome to Physics Forums.

    You have not said what your maths level is , but I am guessing this is a high school question?


    Have you covered the equations of motion for constant acceleration?

    When the acceleration is not constant each of the quantities v (edit =instantaneous velocity); s (=distance); a (=acceleration) will vary with t (=time). A graph may be drawn connecting any two of these four quantities. If this can be done the other quantities can be deduced from this graph.

    Alternatively the result may be calculated by integration as already indicated, but this approach is not taught in the UK at high school level.

    Averaging will not work as it did to obtain the constant acceleration formulae.

    Over to you.
     
    Last edited: Jul 10, 2011
  6. Jul 10, 2011 #5
    Thanks Studiot,
    The problem is high-school,
    but a Prof in Mechanics in UK, in another Forum ,solved the problem using average g finding 1021.5 sec.
    I was not convinced and gave it a try, and found 1273, WHICH is right or better?

    Just what I need, if you haven't the time to work out the exact result, is to know

    Using the procedure of average g., the way I calculated,you say : it won't work, ok, but....

    can I obtain a good approximation for TIME? ( say: 0,0 1)
    Thanks

    P.S. Can I legitimately expect that someone will give me the right answer?
     
    Last edited: Jul 10, 2011
  7. Jul 10, 2011 #6
    I have already suggested drawing a graph, if you want to avoid integration.
    Note I made a slight correction to my earlier post in the velocity term.

    The area under an acceleration - distance graph equals yields the velocity.

    You can draw this since you have been given the acceleration as a function of distance in post#3

    The proof uses integration, but you do not need to do so

    [tex]\int {ads = \int {v\frac{{dv}}{{ds}}} } ds = \int {vdv = \frac{1}{2}} {v^2}[/tex]

    So the area under the accel-dist graph from zero to any distance s' gives [tex]\frac{1}{2}{v^2}[/tex] where v is the velocity at s'.
     

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  8. Jul 10, 2011 #7
    Can someone PLEASE answer this simple question:

    If (instead of integrating variable g)

    I use average g (ΔPE= KE/ h) and the formulas I normally use when g = constant

    what approximation do I get for TIME (say 0.03)?

    am I asking too much?
     
    Last edited: Jul 10, 2011
  9. Jul 10, 2011 #8
    This was answered in post#4
     
  10. Jul 10, 2011 #9
    I got so far, my question ( in post #5) is very simple too, HOW BAD doesn't work ?
    If I need a rough solution, is it always OK?
     
    Last edited: Jul 10, 2011
  11. Jul 10, 2011 #10
    Hi helpers and mentors
    I've done my homework,
    (is this formula correct?
    t = 1/ (sqrt(GM/r0^3) * (π/ 2 + sqrt (r/ (r0 * (1-(r/ r0))- arcsin(sqrt(r/ r0)) )


    Could anyone, please tell me if it is right?

    P.S.
    (In case you are reluctant to help because you think I am a teenager cheating with his homework), I'm pushing 70 !
     
    Last edited: Jul 10, 2011
  12. Jul 10, 2011 #11

    gneill

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    1258 secs is the accurate result.
     
  13. Jul 10, 2011 #12

    ideasrule

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    Using your method, you should get 1073 seconds. 6708/6.25 is 1073, not 1273.

    I can confirm that the answer should be 1258 s. (I actually got 1257 s, but the difference is negligible.)
     
  14. Jul 11, 2011 #13
     
    Last edited: Jul 11, 2011
  15. Jul 11, 2011 #14

    gneill

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    The average that you're using is a spacial one, based upon two values separated by some distance; it's not a time-averaged acceleration. So, for example, the time taken to cross the second half of the trajectory is significantly less than the time to cover the initial half. When you apply the average (but constant!) acceleration to the particle, time to cross the upper section of the trajectory is substantially reduced, and in the lower regions where the 'true' acceleration should really be higher, the particle spends much less time crossing, so its influence isn't as great. So the total time is less for the average acceleration.

    You can confirm this, in a way, by calculating the final velocity of the particle for the two cases. Use the change in potential energy to find the change in kinetic energy. In one case you use the 'true' gravitational potential change from Newton's law (GMm/r), and in the second (average) case, assuming a uniform gravitational field, ΔE = m*g*Δh.

    For your second question, I think you'll have to be more specific about the details.
     
  16. Jul 11, 2011 #15
    Thanks , you have been really GREAT help.

    The details of the second question are the same as first:
    just substitute Gravity with electricity.
    But never mind , probably someone else will take care of that!
     
  17. Jul 12, 2011 #16

    ideasrule

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    We (or at least, I) treat everybody the same way, regardless of age. I think it was reasonable to assume that this was a homework question, because you posted in the "Homework & Coursework Questions" forum, whereas general discussion of physics should go here: https://www.physicsforums.com/forumdisplay.php?f=111.

    What charge are you assuming for the Earth, and why that value? Whatever value you're using, why do you think 18.1 km/s is not a reasonable final speed?
     
  18. Jul 12, 2011 #17
    thanks ideasrule,
    (wasn't it right to use the homework section, since I needed an expert to check my calcs?)

    now...

    just suppose at center of earth you have a positve charge with attractive force equal to GM (whatever it is)

    an electron is in free fall from 3600km(.....same data...ceteris paribus means: all other data being equal) Δ PE = 0.00093 eV

    shouldn't final velocity at 6.4x 10^6 (ground) be the same namely 6.7 km/ s?
    thanks again
     
    Last edited: Jul 12, 2011
  19. Jul 12, 2011 #18

    gneill

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    The charge on the electron is not the same as its mass, so you'd need to adjust the "Earth" charge to take this into account, perhaps by setting the charge so that the force on the electron at its initial separation is the same as the force that gravity had produced (equivalently, so that the PE's are the same).

    If this is done, then since the final velocity should depend only on the change in PE, they should be the same.
     
    Last edited: Jul 12, 2011
  20. Jul 12, 2011 #19
    I can't see the point. (I chose one electron on purpose as it is
    the elementary unit of charge: the unit Coulomb is just 6,24x10^18 charge units and
    the unit of mass,: 1 electron or 1 kg receive the same acc.)


    What I thought matters is only attractive force, (that's why I left it unspecified ,Im not sure
    what is the exact charge that would produce such drop of potential
    )
    You calc it and substitute.Where do I go wrong?

    (P.S. my calcs gave 11.5 million charges = 0.0055 esu. but I think it makes no difference)
     
    Last edited: Jul 12, 2011
  21. Jul 12, 2011 #20

    gneill

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    That's fine. Suppose you substitute a proton for the electron (and change the sign of the attracting charge). The magnitude of the charge on the particle is the same as before, so the attractive force is identical to the electron case. However, the particle mass is 1836 times greater, and so the acceleration a = F/m, will be 1836 times less. This will affect the velocity profile (KE).
    Perhaps you could calculate and post a suggested value for the charge? [EDIT: Never mind, I see you added it to your previous post -- 11.5 million e+. Looks a bit small to me, I would have thought closer to 1.5 Billion e+ charges, or about 0.252 C.]
     
    Last edited: Jul 12, 2011
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