Planck's constant in planetary orbits

1. Sep 28, 2015

QuantumBlink

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
p=mv

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

The attempt at a solution

Δx=1.496E11
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. Sep 28, 2015

Bystander

How are you defining Δp and Δx?

3. Sep 28, 2015

QuantumBlink

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 ?

4. Sep 28, 2015

Bystander

... and, how are those quantities defined and used in the expression describing the uncertainty principle?

5. Oct 1, 2015

artfullounger

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:

Δp=p1-p2
Hence

Δp=m1v1-m2v2

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

p1-p2=m(v1-v2)​

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

Similar treatment with Δx leads to:

Δx=Δd
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?

edit:

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
6. Oct 1, 2015

artfullounger

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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