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The Earth-Sun distance on a given day of a year, can be calculated using the formula (source pdf): $$\frac{d}{\rm AU} = 1 - 0.01672\cos(0.9856({\rm day}-4))$$

where $\rm day$ is the count of days from January 1.

My question is how is this formula derived?

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$$1- 0.01672*\cos(0.9856*(\text{day}-4))$$

This is an approximate expression. Term by term,

  • $1$
    The mean distance between the Earth and the Sun is about one astronomical unit.

  • $0.01672$
    This is the eccentricity of the Earth's about about the Sun.

  • $\cos$
    This is of course the cosine function, but with argument in degrees rather than radians.

  • $0.9856$
    This is $360/365.256363$, where $360$ is the number of degrees in a full rotation and $365.256363$ is the length of a sidereal year, in mean solar days.

  • day
    This is the day number of the year. Since this is an approximate expression, whether one starts with the first of January being zero or one is irrelevant.

  • 4
    The Earth currently reaches perihelion between the fourth and sixth of January, depending on the year.


So where does this approximation come from? If the Earth's orbit was a Keplerian orbit about the Sun (which it isn't), the distance between the Earth and the Sun would be given by the modern version of Kepler's first law, $$r = a\frac{1-e^2}{1+e\cos\theta}$$ where $r$ is the distance between the Earth and the Sun, $a$ is the length of semi major axis of the Earth's orbit, $e$ is the eccentricity of the orbit, $\theta$ is the angle subtended at the Sun between the semi major axis line and the current position. (In other words, the true anomaly).

We switched from using the Sun to measure time to clocks a few hundred years ago. In fact, Kepler was one of the first to say that clocks rather than the Sun is the proper measure of time. This means using days in lieu of theta is not quite right. Our days are a measure of mean anomaly rather than true anomaly. For our low eccentricity orbit, the difference between the two is less than 20 minutes. (This difference is part of the equation of time.) Using day number as a stand-in for true anomaly is a reasonable approximation of true anomaly, so long as we divide by the number of days in a sidereal year. If the cosine function takes degrees as an argument, we need to multiply by 360. Hence the 0.9856 ($360/365.256363$).

The next approximation used is that $\frac1{1+x} \approx 1-x$ for small values of x. This brings us to $$r=a(1-e^2)(1-e\cos\left(0.9856\,\text{day#}\right))$$ Next, we need the day number, with day zero marking the time of perihelion passage. That's pretty simple. The Earth currently reaches perihelion around the fourth of January or so. Finally, we need a value for $a(1-e^2)$. That's about one when distance is expressed in astronomical units. The final result is $$r=1-0.01672\,\cos\left(0.9856\,(\text{day-4})\right)$$

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    $\begingroup$ And the $\ast$'s? Just kidding. Nice answer. $\endgroup$ – 299792458 Apr 23 '15 at 12:01
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Approximations are made:

  1. Earth's orbital velocity remains the same: Angle between the Earth and the Earth's perihelion $\theta$ is increasing constantly.
  2. Eccentricity is small enough, that ellipse can be approximated to be $r=a(1-e \cos\theta)$.

Earth is at its perihelion on 4th of January, and its eccentricity is 0.0167, so the given formula can be derived, as Hammen already answered.

But if you want to calculate more accurately the distance between the Earth and the Sun at any given epoch, you need to look up for Kepler's Equation. This can be obtained using Kepler's laws of planetary motion, or just solving 2-body problem: that it would still be an approximation, but much more accurate than the one given in your question.

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