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I have Sun data (lon+lat+distance) in geocentric coordinates:

this from app Ephemeris Meeus algorithm

but I want to visualize in heliocentric way. How to calculate that?

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Earth's position as seen from the Sun is directly opposite the Sun's position as seen from Earth, at the same distance. In ecliptic coordinates, $$ \begin{align} l_\oplus &= \lambda_\odot \pm 180^\circ \\ b_\oplus &= -\beta_\odot \\ r_\oplus &= \mathit{\Delta}_\odot \end{align} $$

The heliocentric position of the Sun is always at the origin ($r=0$) by definition.

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  • $\begingroup$ so, when i calculate geosentric, just skip L+180 and get heliocentric. $\endgroup$ – faizin F6 Oct 23 '20 at 23:47
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For some reason, I just saw this question only now… Maybe I’m too late, but anyway: Here goes…

To convert from heliocentric to geocentric and vice versa is more than just adding 180° to the longitude, because you’re also changing point of view. You have to first convert your spherical heliocentric (or geocentric) positions into rectangular positions. This is done with the following formulas:

$ X = R\ \text{cos}\ B\ \text{cos}\ L\\ Y = R\ \text{cos}\ B\ \text{sin}\ L\\ Z = R\ \text{sin}\ B $

Where R is the distance to the object (which is lacking from your table; hopefully, you have the data somewhere else), B is its latitude, and L is its longitude.

Then you need to somehow find the heliocentric position of the Earth or the geocentric position of the Sun. Let’s call those $X_0$, $Y_0$, and $Z_0$.

Then the geocentric (or heliocentric) position is found by $X_h = X + X_0$, $Y_h = Y + Y_0$, $Z_h = Z + Z_0$, which you can convert to longitude, latitude, and distance once again by:

$ r = \sqrt{X_h^2 + Y_h^2 + Z_h^2} \\ l = \displaystyle \text{atan2}\ \frac{Y_h}{X_h} \\ b = \displaystyle \text{asin}\ \frac{Z_h}{r} = \text{atan2}\ \frac{Z_h}{\sqrt{X_h^2 + Y_h^2}} $

Where atan2 is the second arctangent function that’s available in most programming language, and which gives you the proper quadrant.

Hope this helps.

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