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I made a simple sheet that based on apparent diameter of a celestial body which shows to me which of my eyepieces I should use to view an object in my telescope. However, some of bodies like planet Mars have different apparent diameters based on current date and time. On some days Mars may be seven times bigger than on other days. Is there any way so to calculate apparent size of a planet on a particular day? Even rough approximation would make here the thing.

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    $\begingroup$ Read this question: Distance from earth to another planet in solar system. You just the distance from Earth to Mars. $\endgroup$
    – Uwe
    Apr 26, 2023 at 9:48

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Apparent diameter is inversely proportional to distance, and by trigonometry (assuming coplanar orbits), you could obtain the distance to the planet from the relative phase angle:

$$d = \sqrt{(\sin(\phi) r_{Earth})^2 + (r_{Mars} - \cos(\phi)r_{Earth})^2}$$

But since you in any case need to look up some position data, why not look up the value directly?

Asking WolframAlpha:

Mars apparent diameter from Earth April 26 2023

→ Diameter: 5.524 arcseconds
→ Distance: 1.692 au

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    $\begingroup$ Does this work for inner planets (Venus, Mercury) where a single phase angle can correspond to two very different distances? And shouldn't "circular" be added to "coplanar" in the list of approximations? How far off would this be for elliptical orbits like Mars and especially Mercury? $\endgroup$
    – uhoh
    Apr 26, 2023 at 23:21

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