Heliocentric correctionΒΆ

The heliocentric correction compensates for the differences in the observer’s distance to the observed object due to movement of the Earth around the Sun. When we measure the time that elapsed between two events and the time span is in the order of days or more, we may need to compensate for the timing due to the finite speed of light. A common approach is to transform the observation time as it would be observed from the center of the Sun, which is considered to be fixed with respect to the object. The Julian date JD is transformed into Heliocentric Julian date HJD. The difference between those two dates is called the heliocentric correction HC and usually it is expressed in days.

(1)HJD = JD + HC

Figure 1 illustrates the situation. The heliocentric correction HC is the time that light needs to cover the distance \|\vec{r}\|. The vector \vec{r} can be thought of as a projection of the Earth-Sun vector \vec{s} into the direction of Earth-object unit vector \vec{n}.

Heliocentric correction

Vectors used to derive heliocentric correction. s: Earth-Sun vector; n: Earth-object unit vector; r: projection of s into direction of n

The length of the vector \vec{r} can be computed as:

(2)\|\vec{r}\| = \|\vec{s}\| \cos\theta

The cosine of angle \theta can be derived from the expression for dot product of two vectors:

(3)\vec{a}\cdot\vec{b} = \|\vec{a}\|\,\|\vec{b}\|\,\cos\theta

When we put the two equations together, we get

(4)\|\vec{r}\| = \|\vec{s}\| \frac{\vec{s}\cdot\vec{n}}{\|\vec{s}\|\,\|\vec{n}\|}

the length of vector \vec{s} cancels out and because the vector \vec{n} is a unit vector, its length is equal to 1. Then we can simplify the previous equation to

(5)\|\vec{r}\| = \vec{s}\cdot\vec{n}

Vector \vec{s} is computed using Flandern’s formulas for Earth-Sun distance R_S and Sun’s ecliptic longitude \lambda_S (see [flandern79]).

The object’s position is specified by two equatorial coordinates - right ascension \alpha and declination \delta. These coordinates must be transformed into the ecliptic coordinate system to get object’s ecliptic longitude \lambda and latitude \beta.

The easiest way to compute the dot product of two vectors in polar coordinates is to transform them into the 3-D Cartesian system first, using a trigonometric identity for cosine of the sum of two vectors we get

(6)\|\vec{r}\| = R_S\,\cos\beta\,\cos(\lambda-\lambda_S)

Using a speed of light c = 299,792,458 meters per second we can derive the formula for the Heliocentric correction. Please note, that it is necessary to convert the units to get the value in days.

(7)HC = \frac{R_S}{c}\,\cos\beta\,\cos(\lambda-\lambda_S) = -0.00577552\,R_S\,\cos(\beta)\,\cos(\lambda-\lambda_S)