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Homework Help: Non Uniform Acceleration as a function of distance and not time

  1. Jul 5, 2012 #1
    There are two point masses in space, each 1kg and one metre apart from each other.
    The only Force acting on the masses is the gravitational force between them.
    After what time do they collide?

    Gravitational Force F = Gm1m2/(2s)2 where m is the mass and s the distance to the center of both masses (Hence 2s the distance between them). G is the gravitational constant.
    F = G*1*1/(2*0.5)2 = G
    As F = ma and m is 1, a = G at the initial position.

    As the masses dont change I can say for all moments: a(s) = G(2s)-2
    The acceleration a is a function of s, the distance, and not time, as usual.

    From there on I was stuck.
    My guess was that a(s) = dv/ds
    And as a(t) = dv/dt = dv/ds*ds/dt and ds/dt = Velocity v
    a(t) = a(s)*v
    But I have no clue what velocity this is? Is this a totally wrong approach?

    Another thought was to take the average acceleration and use it for my calculations.
    When s = 0.5m the acceleration is 1G, and when s = 0.25m it is 4G.
    Is 4G the average acceleration?
    t = √(2s/a) = √(1/4G) = 61221.94s = 17 hours
    If this average acceleration is wrong, how do I calculate it?

    I would thank you for any help :S
  2. jcsd
  3. Jul 5, 2012 #2
    a= 6.674 EE 10-11 m/S2

    I would use the distance formula Xf=Xi+Vi*t+.5a*t2 and solve for t
  4. Jul 5, 2012 #3
    That seems to be the formula for a uniform acceleration.
    True the acceleration is 6.674 EE 10-11 m/S2, but only for the initial moment. As soon as it moves, the gravitation take more effect and the acceleration increases.

    In my first post I wrote:
    t = √(2s/a)
    which equals your equation already rearranged (inital values are 0, hence excluded)
    This question is about Non Uniform Acceleration.

    I would appreciate any help.
    Last edited: Jul 5, 2012
  5. Jul 5, 2012 #4
    Take into account only 1 particle.

    It experiences force F and it moves a distance s.

    Why do you have 2s in your equation for acceleration?

    a = G/s2

    V[itex]\frac{dv}{ds}[/itex] = [itex]\frac{G}{s^{2}}[/itex]

    V dv = [itex]\frac{G}{s^{2}}[/itex] ds

    Integrate and solve for V. What limits should you use?
  6. Jul 5, 2012 #5
    So my first attempt was correct, just needed to continue on that.
    I got 2s, as each object only moves distance s as you also said. As both object move that distance, the distance between them is 2s.

    Integrating would give me [v²/2]: = [-G/s]: (: representing the limits)
    or v²/2 + b = -G/s + c (b and c representing the constants, however it is written)
    I do not know what that v represents or would tell me if I were to solve for that.

    The limits are 1 and 0, not? I told you the question, so you know as much as me :P
    The distance starts off being 1 metre and the end is when they reach each other.
    Solving the RHS with these limits would give me -G-(-G/0) = -G+G/0, which is as s tends to 0; = infinity or just a very high number

    I dont have limits for v, do I?
    How do I get the time from all of this? :D
    A bit confused here :S
    I hope you can continue to help me, as you seem to be very competent.

    I guess I cant take 0 as the limit, as it converges to it.
    If I take a value near zero (0.01), I receive a very small number for my RHS (6.6*10-9)
    How do I solve the left hand side without limits or the constant? Or do I have the information?
    Last edited: Jul 5, 2012
  7. Jul 5, 2012 #6
    You should only consider half of the problem because it's all you need.

    If you integrate 2s you assume the 2nd particle doesn't move, so you should just leave it as s. And only focus on half the problem.

    Well V is velocity, and considering is starts from rest the limits are [itex]∫^{v}_{0}[/itex]

    Now since were dealing with only 1 particle, the s limits are [itex]∫^{\frac{1}{2}}_{0}[/itex] The reason its only a half a metre is because the distane between the 2 is 1 and they both move 0.5.

    So now you have a value for V. Replace V with ds/dt and solve for t.
  8. Jul 5, 2012 #7
    That would mean:
    v²/2 = -G/(1/2) - (-G/0)
    v²/2 = -2G + G/0
    which is not possible.
    Lets say for the ease, or until you tell me a better way of doing it that the lower limit is 0.1 and not 0.
    v²/2 = -2G + G/0.1 = 8G
    v = √16G = 3.267*10-5
    What does that tell me?
  9. Jul 5, 2012 #8
    Yeah I ignored the 0 and negative square root.
  10. Jul 5, 2012 #9
    What? There is no negative square root when you subtract the second part. The problem is that the acceleration tends to infinity as the distance tends to zero. Thats why I used the lower limit of 0.1.
    If I continue looking at it like that:
    ds/dt = 3.267*10-5
    s = 3.267*10-5t + c
    t = (s-c)/3.267/10-5 // Wait what did I do? HELP :D
    Last edited: Jul 5, 2012
  11. Jul 5, 2012 #10


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    The next step will involve another integration, using the fact that v = ds/dt, which will get messy.
    Last edited: Jul 5, 2012
  12. Jul 5, 2012 #11
    Lets all delete our posts? :D
    v = ds/dt = 3.267*10-5 = s'(t) is what I did in my previous post.
    s(t) = 3.267*10-5t + c
    t = (s-c)/(3.267*10-5)
    Now what?

    {s as of the fourth post means the total distance between the particles.
    "v2/2 = -G(m + m)/s + c = -2Gm/s + c"
    Where did you get the (m+m) from and what does that mean?
    Isn't the integral just -G/s + c?
    And should not v²/2 also have a "+ c" on it? or "+ b" of course for a second letter.
    I still dont know what those "m"'s mean to follow on your calcs..}
    Last edited: Jul 5, 2012
  13. Jul 5, 2012 #12


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    I removed most of the stuff from my previous post since I wasn't sure what s was.

    I wasn't sure what s was. If s is the total distance, then

    force = G m m / s^2

    acceleration for the left particle is +Gm/s and for the right particle -Gm/s. The rate of acceleration of both particles towards each other would be G(m + m) / s. I wasn't sure how you wanted to approach this.
    Last edited: Jul 5, 2012
  14. Jul 5, 2012 #13
    m = 1, hence the individual acceleration at the starting position is G/s or -G/s depending on which one you're looking at.
    I am concentrating on one particle, so the force together is irrelevant.
    The problem is, that the magnitude of the force increases the more the particle moves.
  15. Jul 5, 2012 #14


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    You end up with a messy integral (C is some constant based on G and m):

    C sqrt(s / (s0 - s)) ds = dt

    where s0 = 1 meter (or .5 meter depending on how you define s).

    This can be solved by substuting:

    u = sqrt(s / (s0 - s))
    u2 = s / (s0 - s)
    s = s0 u2 - s u2
    s (1 + u2 ) = s0 u2
    s = s0 u2 / (1 + u2)
    s = (s0 + s0 u2 - s0) / (1 + u2)
    s = (s0 (1 + u2) - s0) / (1 + u2)
    s = s0 - s0 / (1 + u2)

    ds = 2 s0 u / ( 1 + u2 )2
    Last edited: Jul 5, 2012
  16. Jul 5, 2012 #15
    I dont not understand where you get that equation from.
    And what is u? What actually is c?
    And if s0 is 1metre, then what is s?
    Sorry, I dont really get what youre telling me :S

    I cant make sence of your working when I dont know where it comes from, what it means and what u is.. :S
    Last edited: Jul 5, 2012
  17. Jul 5, 2012 #16


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    It's for doing integration by substitution. I defined:

    u = sqrt(s / (s0 - s))

    and determined that

    ds = 2 s0 u / ( 1 + u2 )2

    so the integral becomes:

    C 2 s0 u2 / ( 1 + u2 )2 du = dt

    Assuming I understand how s is being used, then

    C = sqrt(s0) / (sqrt(4 G)) = 1 / (sqrt(4 G))

    The distance between the two particles. The limits of integration need to be changed for s = 0, u = 0, and for s = s0, u = ∞.
  18. Jul 5, 2012 #17
    I have never in my life done integration by substitution.
    I am still doing my AS-levels :D
    I guess thats why I dont have a clue what youre doing.

    Is there a way of solving this problem without integration by substitution?
  19. Jul 5, 2012 #18


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    Not that I'm aware of. It's a substitution done to allow integration by parts. This problem has been brought up before:


    Although it matches the method used by arildno in a previous thread (where the latex is messed up), I'm not sure the answer has been confirmed by more than just one or two members here. I did check the answer using numerical integration via a spreadsheet and that result corresponded to the answer I got from the integral method.
    Last edited: Jul 5, 2012
  20. Jul 5, 2012 #19
    Thank you very much.
    I will need to learn Integration by substitution and read through both threads with this kind of question, until I understand it.
    If I in future, by any means, need more help I will contact you via this thread agains, althy I believe this to be unlikely.
  21. Jul 12, 2012 #20
    I have now looked into the maths required and ran into a different problem:
    [itex]dt = \sqrt{\frac{s_{0}}{2G(m_{1}+m_{2})}}\sqrt{\frac{s}{s_{0}-s}}ds[/itex] was the start of the problem.
    By substitution:
    [itex]dt = \sqrt{\frac{s_{0}}{2G(m_{1}+m_{2})}}\frac{2s_{0}u}{(u^2+1)^2}du[/itex]
    Integrating that with the limits of 0 to infinity as required I get:
    [itex]t = \frac{\pi s_{0}^{\frac{3}{2}}}{2\sqrt{2G(m_{1}+m_{2})}}[/itex]
    As s0 is 1 and both masses are 1, I get 96138.4 as an answer.
    The correct answer is 96136 as said many others, so I am off by two seconds. Why is that? Where is my mistake?
  22. Jul 12, 2012 #21


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    What level of precision did you use for pi etc
  23. Jul 12, 2012 #22
    I used pi as in the constant in my calculator and wolfram, and that should be quite accurate.
    And for G i used 6.674e-11 (actual value is slightly below that, but to get the right answer G needs to be 6.67433e-11)
    I can only imagine 6.67e-11, 6.673e-11 or 6.674e-11 being used in the calculation

    Master rcgldr please call in :D
    Last edited: Jul 12, 2012
  24. Jul 12, 2012 #23


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    The codata value as of today, and going back to June 5, 2011 is

    G ≈ 6.67384(80)e-11 = 6.67384e-11 ± 0.00080e-11

    Using that value I get a time ~ 96139.53 seconds.

    Link to references:



    The 96136 was based on an older (2009) codata value, G ≈ 6.67428(67)e-11

    Using the "wayback machine", click on impatient, instead of waiting for it to load all the history:


    Once at that wayback page, you can also click on the dateline above to see older and newer values.

    Some articles still use that value:

    Last edited: Jul 12, 2012
  25. Jul 12, 2012 #24
    As I said G needs to be bigger that 6.674 for the equation to work (approx. 6.67433)
    So any true value under 6.674 will be even more off the true answer.
    6.67384 will give me a time of 96139.5s

    Using the min. value at 6.67304e-11 I get 96145.3s
    Using the max value at 6.67464e-11 I get 96133.8s
    For the agreed value I need 6.67433e-11 to get the required 96136s every other thread seems to talk about. Why would they use this "random" number?

    Another Edit upon your edit:
    Using 6.67428e-11 I get an answer of 96136.4, which can be rounded to 96136 to be also correct. It seems this was the value used for the calculation "back in the time" the other topics were new.
    Last edited: Jul 12, 2012
  26. Jul 12, 2012 #25
    To sum it up the value used in the calculations in the other threads are outdated and hence a different value for G has been used.

    In 2006 the value was between 6.67361 and 6.67495
    Now the value is between 6.67304 and 6.67464
    If the value is truly between 6.67304 and 6.67361, then the prior estimating was wrong, which leads to the conclusion that this newer estimation could be "proven" wrong aswell in the future.
    The only thing I can really say is that the time needed for the two particles to be at the same position using our theoretical model is between 96130 and 96145 second and further accuracy is due to the uncertainty in G not acquireable.

    Do you confirm my statement master?
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