Planck's constant in planetary orbits

[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.

 

[Bystander]

How are you defining Δp and Δx?

 

[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 ?

 

[Bystander]

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

Relevant equations
ΔxΔpx ≥ħ/2

 

[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.

 

[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.