Here's the calculation:

https://upload.wikimedia.org/wikipedia/commons/2/21/Geometri...

R ≈ 6.4E6

h = 0.3

d = √(2Rh + h²) ≈ 2.0E3

γ ≈ sin γ = d/(R + h) ≈ 3.1E-4

extra rope length = 2 × (d − γR) = 2R × (tan γ − γ) ≈ 2R × γ³/3 ≈ 0.13E-3

Follow-up question - after extending the rope by 0.12mm and pulling it up to allow the cat to pass under it, how far away are each of the two points where the rope lifts off the ground?

As the article explains, these points are about sqrt(2Rh) = sqrt(2 x 6400km x 0.3m) = sqrt(3840000m^2) ~ 1.96km away from the cat.

Did a back of the envelope trigonometry calculation as an approximation, seems to check out :)

Haven't done any calculations but I would guess half of the Earth circumference? because if you want to minimize the length of the rope then you have to maximize distance between those two points.

That would be the same as the 'how far away is the horizon' calculation at the top of this thread, right? (assuming weightless rope.)