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Naturally, the parallax is easier to measure for stars that are closer than farther away. But if all stars were the same distance from the Sun, and if there were some other reference to measure the parallax, would all stars show the same parallax?

In other words, does the position of a star (the right ascension and declination) affect the parallax measurement and calculation? For example, are stars near the ecliptic or equator or celestial pole easier to measure than other locations on the sky?

Feel free to include any equations or references if it helps.

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  • $\begingroup$ Is this limited to only earth-based observations (ie no satellites)? $\endgroup$ – cms Aug 15 at 21:34
  • $\begingroup$ @cms. You can include satellites but I think the issue is the same. The satellite is in basically in orbit around the sun, so some stars are near the orbit plane and some are near the pole. $\endgroup$ – JohnHoltz Aug 15 at 23:42
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First, let's keep things simple and consider a star with no proper motion, i.e. no motion through the Galaxy relative to Earth.

If you could observe a star continuously throughout the year (as parallax-measuring satellites like Hipparcos or Gaia do), you would find that the path of a nearby star on the sky, relative to background stars, would trace out an ellipse on the sky. For a star at exactly the ecliptic pole (line of sight from Earth is exactly perpendicular to Earth's orbital plane), that ellipse would be a circle. As you move your line of sight away from the ecliptic pole, one axis of the ellipse would shrink by the cosine of the angle you moved (or by the sine of the ecliptic latitude, the angle up from the orbital plane). When you reach a star right on the ecliptic, the ellipse would have flattened out to a straight line, i.e. the one axis would have shrunk to zero. But the length of the long axis is unaffected, so by measuring the length of that long axis of the parallax ellipse, we get the distance to the star, regardless of its position in the sky.

In practice, stars also have proper motion (or at least, any star that is close enough to have a measurable parallax will also have a measurable proper motion), so paths on the sky are those ellipses, combined with a steady linear motion, like this:

Parallax paths from Hipparcos

(from here)

So in practice, measuring the parallax involves fitting a function to the positional data that includes both the size of the parallax ellipse and the proper motion. (But with only three free parameters - two dimensions of proper motion, plus the parallax; the shape [but not the size] of the parallax ellipse is set by the known ecliptic latitude.) The parallax angle is half of the angular width of that path perpendicular to the proper motion direction.

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  • $\begingroup$ This is a great description of stellar motion but could be made a tiny bit better by making the answer to the question more explicit, yes? $\endgroup$ – cms Aug 15 at 21:47
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It's all about basic geometry.

Base for the parallax measurements is Earth's orbit around the sun, giving you a maximum of 300 Mio km. With a given base dimension, you get the best precision when the base is orthogonal to the star direction. (In the other extreme, you get no parallax at all if the base is in line with the star).

For stars close to the ecliptic you get this optimum base angle only using two specific dates, half a year apart from each other (those when the star appears 90 degrees distant from the sun).

For stars nearly perpendicular to the ecliptic, you can choose any two dates half a year apart, giving you more chances to contribute max-precision measurements.

If doing a continuous observation of the star over e.g. a year, the difference should amount to a factor of sqrt(2), if the other parameters are comparable.

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The measurement of the parallax - theoretically - does not depend on where on the position of the star in the sky.

Parallax

There is a IMHO simple geometric argument: Consider a star which is perfectly in one direction at the given distance d.

Now we want to check whether we can measure the same angle for a star at the same distance at any arbitrary point on the sphere of radius d around the sun. Do the simple thought experiment: we can reach any point on a great arc by rotating the star around the 'anchor points' for July and January. Now we can rotate the whole setup around the sun (or more accurately the normal vector of the orbital plane). And as such we have an infinite amount of great arcs, so we reach every point on the sphere while maintaining the same arc which has an angle of "2\pi".

You could visualize that with a thread, a marble glued to the middle and the two ends of the thread glued to a flying saucer (or any other disc symbolizing the Earth's orbital plane). Without rotation of the disc, the marble can do a great circle. With rotation of the disc and the marble can reach any point on a sphere.

For Earth-based telescopes you might have the practical difficulty that you need to do some observations during day or more realistically to not measure twice the parallax (thus half a year apart) but some other - yet equally well-known - angle with a smaller temporal difference, like only 3 months. Practically most of these observations are meanwhile done by spacecraft, so day and night don't play much of a role.

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  • $\begingroup$ I had to read your description a number of times, but I think this is what you are saying with the thread, marble, disc experiment. The marble represent the distant star. If at the ecliptic pole, the star sees the Earth's orbit as a circle. All points on the orbit are equal distance, so the parallax is identical all year long. If you rotate the disc so that the orbit is seen partially edge on from the star, the orbit is an ellipse. Rotate the orbit until edge on, it is a line. But the maximum points on the orbit are still the same distance, so the parallax is the same at 2 times of the year. $\endgroup$ – JohnHoltz Aug 15 at 19:16
  • $\begingroup$ yes, exactly. Near the pole, it's always the same angle for an arbitrary choice of dates 180 days apart, for one in the ecliptic plane you need to choose the right two dates (or plot positions from several measurement dates and interpolate) $\endgroup$ – planetmaker Aug 15 at 19:57

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