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Reading this Wikipedia article and following the derivations (https://en.wikipedia.org/wiki/Solar_irradiance#Milankovitch_cycles), I have found an equation I can't see where it comes from. Perhaps it's simple but my positional astronomy is on its basics.

It says that if $\theta$ the conventional polar angle describing a planetary orbit ($\theta = 0$ at the vernal equinox), $\epsilon$ is the obliquity, the declination $\delta$ as function of the orbital position is $\delta = \epsilon \sin\theta$.

Could you please explain or point to some resource? Thx.

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Briefly, the declination varies smoothly from $-\epsilon$ to $\epsilon$ & back as $\theta$ ranges from 0 to $2\pi$. The function $\delta=\epsilon\sin \theta$ is the simplest way to do that. However, this formula is just a reasonable approximation. Using a spherical right triangle, we get $$\sin\epsilon = \frac{\sin\delta}{\sin\theta}$$ so $$\sin\delta=\sin\epsilon\sin\theta$$

Wikipedia's Position of the Sun has some good info, as well as more refined calculations. Also see Celestial coordinate system for the formulas to convert from ecliptic latitude & longitude to right ascension & declination. The ecliptic is the Sun's apparent path, so its ecliptic latitude is zero (if we ignore the precession of the equinoxes), which simplifies the calculations.

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  • $\begingroup$ Appreciate your comment, but do you know of some book or reference about it? My spherical trigonometry is ok $\endgroup$ – David Feb 19 '19 at 12:28
  • $\begingroup$ Let me a comment (I cannot insert a draw) in order to see if I am understanding: $\theta$ is the hypotenuse of that right triangle, measured over the ecliptic, $\delta$ is the cathetus from the Equator to the point where the Sun is and $\epsilon$ is the angle opposite to $\delta$. The fact that we measure $\theta$ along the ecliptic instead of along the Equator is slightly weird to me... $\endgroup$ – David Feb 21 '19 at 16:51
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    $\begingroup$ @David Pretty much. The vertex of the triangle with the angle $\epsilon$ is where the ecliptic crosses the (celestial) equator. If it were a flat triangle, we'd have $\sin\epsilon=\delta / \theta$. We measure $\theta$ along the ecliptic because the Sun "travels" along the ecliptic. $\endgroup$ – PM 2Ring Feb 21 '19 at 17:06

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