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If the semimajor axis is a and the eccentricity e, then each focus is off the centre either side by an amount equal to ae (but still on the major axis).

So, the distance between the foci is 2ae , therefore the ratio you want is 2ae/2a, which is e or the eccentricity.

Taking the Earth's orbit (approximately), a = 149.6 million km, e = 1/60, so each focus is 2.49 million km either side of the centre; the Sun is at one focus and there is nothing at the other focus.

The semiminor axis b is given by b = a sqrt (1 - e2). For the Earth, the semiminor axis is 149.58 million km.

So the orbit is very nearly a circle but the main effect of the eccentricity is to make the Earth's distance from the Sun vary by ±2.49 million km, closest in January and furthest away in July.

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Q: What is the ratio of the distance between the foci and the length of the major axis which defines a planet's elliptical orbit?
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