Relation of orbital speed and eccentricity Earth's eccentricity is 0. 0167 and speed at perihelion is 30.3 Km/s and at aphelion 29.3 with a difference of +/- 1. 0164 wrt average orbital speed


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*Is this a coincidence or are the variations of speed directly related to eccentricity?

*can we calculate the time elapsed from aphelion,  knowing the eccentricity of the orbit?

 A: No it's not a coincidence.
The linear eccentricity, $c$, is the distance from the centre of the ellipse to either of the foci. This diagram shows an orbit with this marked - for clarity I've made the orbit very eccentric:

The eccentricity that you quote is defined as:
$$ e = \frac{c}{a} \tag{1} $$
where $a$ is the semi-major axis.
The lower diagram shows the Earth at its closest and most distant positions. These distances are:
$$\begin{align}
 r_{\text{max}} &= a + c \\
 r_{\text{min}} &= a - c 
\end{align}$$
Conservation of angular momentum tells us that:
$$ r_{\text{max}}v_{\text{max}} = r_{\text{min}} v_{\text{min}} $$
and therefore the ratio of the velocities is:
$$ \frac{v_{\text{max}}}{v_{\text{min}}} = \frac{r_{\text{min}}}{r_{\text{max}}} = \frac{a-c}{a+c} $$
Since equation (1) tells us that $c = ae$ the above equation simplifies to:
$$ \frac{v_{\text{max}}}{v_{\text{min}}} = \frac{1 - e}{1 + e} $$
Now we use the binomial theorem to approximate $(1 + e)^{-1}$ as $1 - e$ and this gives us:
$$ \frac{v_{\text{max}}}{v_{\text{min}}} \approx (1 - e)(1 - e) \approx 1 - 2e $$
where I've dropped terms in $e^2$ and higher powers on the grounds that if $e$ is small higher powers will be much smaller and can be ignored.
So does this work? Well, if we substitute your figures for the velocities we get:
$$ \frac{29.3}{30.3} \approx 0.967 \approx 1 - 0.33 \approx 1 - 2 \times 0.165 \approx 1 - 2e $$
And that's why the ratio of the velocities is related to the eccentricity.
