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Uncertainty priciple help

  1. Nov 23, 2008 #1
    At any given instant, the position and momentum of the Earth are somewhat uncertain becasuse of quantum effects. This implies that a year (defined as the time it takes the Earth to complete one orbit) is somewhat uncertain. Give a rough estimate, as a percentage, for this uncertainty.

    I am assuming that the measured time is t=365.25days and you need to find (delta)t and divide to find teh percentage of uncertainty but I am unsure as to how to find this.

    Thank you in advance for any help.

    -D
     
  2. jcsd
  3. Nov 23, 2008 #2

    duo

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    hi.

    well there are two ways i can think of doing it. one is to get the angular velocity of the earth, which shouldn't be too hard since you know it revolves through 2 pi in 365 days. get the radial velocity from that and mutliply it by the earths mass to find its momentum. then use the uncertainty principle to get the uncertainty in the position. roughly the uncertainty in a year will be the ratio of this uncertainty to the circumference of the earth's orbit around the sun.

    i guess you could also find the uncertainty in energy and relate it directly to time if you use the radial velocity to get the earth's kinetic energy. i don't know if that's the right energy to use.

    but even if that is helpful dont take it for granted that i am right, ive probably made some gross error in my reasoning and someone much smarter on this forum will set you right soon, if that is the case. i hope!
     
  4. Nov 24, 2008 #3
    Heisenberg taught us that simultaneous measurements of position and momentum interfere with one another because they both cause some effect which alters the conditions of the other measurement. In the case of the Earth, it is so massive that the two measurements will have an almost immeasureably small effect on each other. The Uncertainty Principle certainly operates but the percentage uncertainty must be vanishing small.
     
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