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Homework Help: Planck's constant in planetary orbits

  1. Sep 28, 2015 #1
    The question is:
    If quantum mechanics were detectable in planetary orbits, how much larger than the currently accepted value would Plancks constant need to be?

    Relevant equations

    ΔxΔpx ≥ħ/2

    Earth-Sun dist 1.496E11
    Earth orbit speed 28,900ms^-1
    Earth mass 5.972E24 kg

    The attempt at a solution

    Δpx = 28,900ms^-1*5.972E24 kg

    h= 4π Δx Δpx
    h≈3.38E41 Js

    I'm not sure I've I've gone about this question correctly, it's worth a considerable amount of marks so am unsure if it's a suitable solution.
  2. jcsd
  3. Sep 28, 2015 #2


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    How are you defining Δp and Δx?
  4. Sep 28, 2015 #3
    Defining Δx as the Earth Sun distance and Δp as the momentum of the Earth (mv), I wasn't sure how to do it any other way with the information given ?
  5. Sep 28, 2015 #4


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    ... and, how are those quantities defined and used in the expression describing the uncertainty principle?
  6. Oct 1, 2015 #5
    So oddly enough I have almost the same question, except ours is a bit less precise in the values of the Earth-Sun distance etc.

    So I was taking the approach that:



    One would assume m is constant and so m1=m2, while v is an average and there are minor fluctuations. Thus we have:


    when you take the differential of that, m being a constant disappears so we get Δp=Δv

    Similar treatment with Δx leads to:

    where d=Earth-Sun Distance

    From there you would solve the inequality for ħ then use ħ=h/2π

    However I'm concerned I've been thinking about the Δ meaning wrong and hence come to the wrong conclusion, however in my problems last week with the uncertainty principle my demonstrator indicated I can't just assume Δp=p for whatever I'm doing, but I'm realizing now I'm a little fuzzy on the reasoning.

    Am I on the right track here?


    I've realized I've had brain fart (not actually had to do any calculus since first year >.> ) and the m doesn't magically disappear, so Δp DOES equal p in this case, since it would be mΔv, and here Δv=v, I think.
    Last edited: Oct 1, 2015
  7. Oct 1, 2015 #6
    Having subbed in all the values I get the same answer as QuantumBlink is greater than h, so...

    for quantum effects to be detectable (i.e. effects due to the uncertainty principle), h should be strictly greater than 3.38x10^41, I think?

    So in general Planck's constant would need to be 75 orders of magnitude larger.
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