Fermi estimate: eccentricity of earth's orbit around the sun

I was reading about Tesla's rocket when I noticed some numbers.

Someone asked about the orbit of the car launched by the rocket Falcon Heavy in February 2018, and it said that

After orbiting the Earth for 6 hours, a third-stage burn-to-depletion was completed at approximately 02:30 UTC Feb 7, placing the dummy payload in a heliocentric orbit having a perihelion of 0.99 au and aphelion ~1.7 au.

and I thought, wait a pony picking minute, 0.99 au? Isn't earth "exactly" 1 au away from sun? Surely, the car got launched by accelerating from LEO, so when it enters the heliocentric orbit, its perihelion must be almost exactly the same as earth-sun distance.

Well, it's not precisely 1 au, and earth's orbit around the sun has eccentricity... Fermi estimate time!!!

1. Earth's temperature is in equilibrium with the sun's radiation. Earth's temperature is around 300K, with variations of about 2K sufficient to cause significant climate change. Source: climate change scientists say that 2 degrees will cause great change, enough to cause an ice age or an ice-less age, something the Paris Agreement is trying to prevent.
2. $\frac{2}{300} = 0.7\%$, so that's the variation in temperature. Consider then the black-body radiation law, which is $\P \prop T^4$, so the variation in power is $0.7\% / 3 = 0.2\%$.
3. The power input from the sun to earth is $P \prop R^{-2}$ where $R$ is the distance from the sun to earth.
4. So the variation in $R$ is $0.2\% / 2 = 0.1\%$. That means the eccentricity is about $0.1\%=0.001$

What's the right answer?

The present eccentricity is 0.017 and decreasing.

Well, that was a fail...