I googled for the circumference of an ellipse and found in a couple of places that the approximate value (apparently there's no exact analytical formula if it isn't a circle) is
2*pi*sqrt[ (a^2 + b^2)/2 ] where a is the semimajor axis and b is the semiminor axis.
The "average distance" usually means (a + b)/2, which is not the same as sqrt[ (a^2 + b^2)/2 ]. The square root of the average of the squares is called the root mean square.
As a numerical example, suppose a = 2b. Then (a + b)/2 = 3b/2 = 1.5b. But sqrt [ (a^2+b^2)/2 ] = sqrt[ (4b^2 + b^2)/2] = b * sqrt(5/2) = 1.58 b.
Therefore, I would assume that the average distance is not found by dividing the orbit length (circumference) by 2*pi
However, if you are working with bodies that have orbit that are almost circular (e very close to 0), then dividing the orbit length by 2*pi would give you a very good approximation.
It sort of depends upon what we're talking about here when you ask about the average.
As the Earth goes around the Sun, the 'average' distance is 93 million miles / 149 million kilometers. However, at any given instance, it will be less or more.
When at perihelion during the winter season in the Northern Hemisphere, we are closest to the Sun at 91 million miles / 147 million km.
When at aphelion during the summer season in the Northern Hemisphere, we are farthest from the Sun at 94.5 million miles / 152 million km.
Over the course of a year, the average of this will always be the same.
If the orbit were more or less elliptical, again the average will always be the same, even as the figures get larger and smaller.
Also, remember with an ellipse, there is not one center, but two focal points. The Sun is at one of the two focal points around which Earth's elliptical orbit travels.
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Interesting question. The answer is no.
I googled for the circumference of an ellipse and found in a couple of places that the approximate value (apparently there's no exact analytical formula if it isn't a circle) is
2*pi*sqrt[ (a^2 + b^2)/2 ] where a is the semimajor axis and b is the semiminor axis.
The "average distance" usually means (a + b)/2, which is not the same as sqrt[ (a^2 + b^2)/2 ]. The square root of the average of the squares is called the root mean square.
As a numerical example, suppose a = 2b. Then (a + b)/2 = 3b/2 = 1.5b. But sqrt [ (a^2+b^2)/2 ] = sqrt[ (4b^2 + b^2)/2] = b * sqrt(5/2) = 1.58 b.
For an elliptical orbit, the "mean" distance is the semi-major axis (average of aphelion distance and perihelion distance).
The circumference of an ellipse can be found with a complicated calculation, including a double integral.
http://en.wikipedia.org/wiki/Ellipse#Circumference
Therefore, I would assume that the average distance is not found by dividing the orbit length (circumference) by 2*pi
However, if you are working with bodies that have orbit that are almost circular (e very close to 0), then dividing the orbit length by 2*pi would give you a very good approximation.
It sort of depends upon what we're talking about here when you ask about the average.
As the Earth goes around the Sun, the 'average' distance is 93 million miles / 149 million kilometers. However, at any given instance, it will be less or more.
When at perihelion during the winter season in the Northern Hemisphere, we are closest to the Sun at 91 million miles / 147 million km.
When at aphelion during the summer season in the Northern Hemisphere, we are farthest from the Sun at 94.5 million miles / 152 million km.
Over the course of a year, the average of this will always be the same.
If the orbit were more or less elliptical, again the average will always be the same, even as the figures get larger and smaller.
Also, remember with an ellipse, there is not one center, but two focal points. The Sun is at one of the two focal points around which Earth's elliptical orbit travels.