Calculate Earth's distance travelled in a year This is probably more of a math question than astronomy, but I'd like to know the distance the Earth travels in one revolution around the sun (i.e. a year).

According to the data in Wikipedia, I think there may be several ways for me to calculate this but I'm not sure whether they are correct or how to do them:


Calculate by velocity and time: the average velocity is given as $107200\ \mathrm{km/h}$ (a suspiciously round number; imprecise?) and one revolution takes $365.256363004$ days. That gives $\text{distance} = 107\,200\ \mathrm{km/h} \times (365.256363004 \times 24\ \mathrm h) = 939\,731\,570.736691\ \mathrm{km} \approx 939.731\ \mathrm{Gm}$
Calculate by semi-major axis and eccentricity: my math is too weak for that. I searched the web and didn't find a solution for that (calculating the circumference of an ellipse exactly seems to be a problem?)
Any other way using the orbital data given in the Wikipedia article?


In essence, this can be made into a pretty generic question: How do I calculate the distance an object travels in one revolution using the usual orbital parameters?
You need to find the perimeter of an ellipse. This might be accurate enough for you (recalling that an ellipse is only an accurate description of an orbit for a two-body system.

An ellipse has no simple formula to find the perimeter. There appear to be several commonly used approximations based on infinite series (see for example "Maths is Fun").

e.g. Define
$$ h = \frac{(a-b)^2}{(a+b)^2},$$
where $a$ and $b$ are the semi-major and semi-minor axes and $b = a\sqrt{1-e^2}$, where $e$ is the eccentricity.

Then the result you want is
$$ p = \pi(a+b)\left[1 + \frac{h}{4} + \frac{h^2}{64} + \frac{h^3}{256} ...\right]$$

All this is general, but might be overkill for your problem involving the Earth. If my calculator skills are working, then for the Earth's orbit around the Sun, $a = 1.00000011$ au, $e = 0.01671022$, $b = 0.99986048$ au and $h = 7\times 10^{-5}$.

Using this and just the first 2 terms in the series, I get a perimeter of $1.999895\pi$ au. So hardly a correction from $2\pi a$.

However, if you were serious about accuracy then you would have to add to this about 12.5 orbits around the barycentre of the Earth-Moon system and a further correction for the motion of the Sun around the solar system barycentre.