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We (homeschool students and I) are trying to find the answer to this question. Does a star also create Solar Analemma figure 8 when photographed daily for a year?

Has anyone published an experiment where the same star is photographed every night at the same time?

We tried answering this ourselves by googling it, but only come up with websites mentioning the sun. Is this because it is only possible with the sun?

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The short answer: No

The long answer:

The Sun only has an analemma (the figure-8 shape traced on the sky as a result of imaging the Sun at the same time (good luck avoiding clouds) each day) because of two factors.

  • The Earth is tilted on its axis and that axis remains pointed in the same direction in space (toward the direction of Polaris).

(A more precise explanation is that the axis does precess (or "wobble" like a spinning top) ... but in a very long cycle that takes nearly 26,000 years to complete. So on the scale of single human lifespan ... it doesn't appear to move by any noticeable amount and requires very precise measurements to detect.)

  • The Earth's orbit is an ellipse ... not a circle.

While nearly circular ... it isn't perfect. The point along an orbit when an object is nearest to the barycenter of the orbit (in our case... to the Sun) is called the periapsis. But since this periapsis is for an object orbiting the Sun, it gets a special named called the "perihelion". The point of the orbit located farthest from the barycenter is called the apoapsis. Again... it gets a special name because it's an orbit around the Sun so it is called the "aphelion".

But the consequences of the Earth getting nearer to the Sun during half of its orbit ... then getting farther from the Sun during the other half of its orbit ... means Earth's velocity through space changes. As we get nearer to the Sun it is as if we are "falling" so we speed-up (our velocity through space is faster) and as we get farther from the Sun we slow-down (our velocity through space is slower).

Earth needs roughly 365 days (365.24) to complete an orbit. There are 360° in a circle. This means that each ordinary day on Earth (formally known as a "solar day") is 24 hours. In that 24 hours, we've moved forward in our orbit around the Sun by just slightly less than 1°.

It turns out that the amount of time needed for the Earth to spin 360° on it's axis (one "sidereal day") is just slightly less than a solar day. It's roughly 23 hours and 56 minutes (roughly 4 minutes short of a solar day ... that's a roughly rounded value).

So why the difference between a "solar" day vs. a "sidereal" day?

Side note: sidereal translates to mean "of the stars". This comes from the notion that ancients noticed that the debris left by a "falling star" (meteorites) that landed on Earth was mostly made of iron. They surmised that stars must be made out of iron (not realizeing that meteors are not actual stars that fall from the sky). Their word for iron is "sidero" (or "sider" pronounced like "cider" ... the juice you squeeze from apples and possibly ferment into a beverage. It is pronounced the same. Sidereal is not pronounced like "side-real". It is pronounced like "cider-eal". Alas, I digress). Sidereal day means the "day of the stars". In other words, if you note the time when a star passes through the meridian (the imaginary north/south line that separates the "west" side of the sky from the "east" side of the sky) ... and then wait to see when that same star passes through the meridian on the following day, the amount of time that has passed will be roughly 23 hours and 56 minutes.... not 24 hours.

Once the Earth has completed one 360° revolution on its axis (one sidereal day) it will also have moved forward in its orbit by nearly 1°. This means that while distant stars would appear in the same point in the sky, the Sun will not... it will be slightly behind the meridian. We will need to let the Earth spin for another 4 minutes until the Sun returns to the same position. But alas... even this has exceptions...

Recall that the Earth speeds up and slows down as it travels through space between it's perihelion and aphelion points. This means the Sun wont precisely be at the same position in the sky... it'll be fractionally ahead or behind depending on if the Earth is speeding up ... or slowing down. This accounts for the left/right variation in the Sun's position in the analemma. The Earth's axial tilt accounts for the up/down variation in the Sun's position. And when you trace out a whole year, you get a shape that resembles a figure-8.

Stars, on the other hand, are much farther away. If you image a star once every 24 hours (assuming no clouds block your view) then each night, the star will shift forward by roughly 1° ... after enough days the Star will be below the horizon and you will no longer be able to image the star. This means the shape you'll get... is that that the star will trace out an long arc ... but after a few months it will no longer be visible at the same time (you would have to observe it at a different time of the day ... and/or it may be lost in the daytime sky depending on the location of Earth in our annual orbit around the Sun.

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    $\begingroup$ Thank you Tim Campbell. Based on your answer, can we assume that if a camera were placed on every planet and a timed photograph of the sun was taken based on that planet's rotation, that a Solar Analemma would result? $\endgroup$ – Bookaholic Oct 22 at 5:52
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    $\begingroup$ Yes but the shape will be different. This page shows the analemma for Mars: en.wikipedia.org/wiki/Timekeeping_on_Mars You can see it isn't a figure 8 ... but it's rather egg-shaped. The shape is determined by the relationship between the direction of the axial tilt (the solstices) and the dates of perihelion and aphelion. $\endgroup$ – Tim Campbell Oct 22 at 18:42
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An analemma shows the position of the Sun in the sky from a fixed location on Earth at the same mean solar time. It gets its north-south shape because of the tilt of Earth's axis makes the declination of the sun change across the year, and the sideways shape due to the orbit nonuniformity.

The star analemma would get a point at the same time every sidereal day (slightly different from the solar day). To a first approximation that fixes the star's right ascension: it will not move along the celestial equator. The celestrial equator of course rotates one turn per year as seen from the ground, so in one sense the analemma is just a circle around the pole of the sky. But if you look at it in the coordinate system of the celestial sphere the star is fixed.

...Except that there will be some stellar parallax due to Earth's different locations along its orbit during the year. The star will seem to move along a tiny ellipse when compared to distant stars in the background. I think the parallax ellipse is the closest counterpart to the sun analemma.

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  • $\begingroup$ Can you explain what you mean by orbit non-uniformity $\endgroup$ – Kai Oct 22 at 4:30
  • $\begingroup$ The orbit is an ellipse, and the speed the planet moves along it changes as per Kepler's laws. $\endgroup$ – Anders Sandberg Oct 22 at 14:29
  • $\begingroup$ Stellar parallax resulting from the Earth's orbit is really really really tiny. A major objection to the heliocentric model, historically, was that if the position of stars weren't fixed relative to the earth, it should be possible to observe stellar parallax unless stars were so absurdly far away that it should be impossible to see them unless they were absurdly big. The notion that stars were fixed relative to the earth was less of a scientific stretch than the notion that anything could exist that was large enough to be visible despite being far enough away not to exhibit parallax. $\endgroup$ – supercat Oct 22 at 21:58
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Fortunately the free planetarium software Stellarium makes it easy to simulate this experiment. If you pause its clock and repeatedly press = to step forward 24 hours at a time, the stars appear to move westward about 1° per night as the Earth moves around the Sun. Stepping by 23h56m04s sidereal days with Alt = eliminates this effect. Unfortunately you can only do this for a few months before daytime skyglow dominates the view; 3m56s per day adds up to 24 hours per year. Fortunately you can disable the atmosphere model with A and pretend to take images year round.

If you disable the ground model with G, set an equatorial frame with Ctrl M, set a 1° field of view with Ctrl Alt 9, and timestep by sidereal years with Shift Ctrl Alt ], the stars appear to move roughly eastward about 50.3" (arcseconds) per year. This is due to precession and can be eliminated by selecting a star and centering it with Space . If you step for a few centuries, you'll see some field rotation due to precession, plus some linear changes in star positions relative to each other (proper motion). Only a few stars have proper motions larger than 1" per year.

A camera on a telescope and professional-grade equatorial mount would record the effects of aberration (20.5", 1 year cycle) and nutation (17.2" × 9.2", 18.6 year cycle). Unfortunately parallax is much smaller, e.g. 0.1" for a star only 10 parsecs away. Fortunately, within a small field of view, aberration and nutation affect all stars equally and can be eliminated by aligning images on the fainter background stars. Stellarium models nutation but not aberration or stellar parallax.

Dennis di Cicco, who made a famous solar analemma photograph in 1978-79, also made analogous observations of Barnard's Star in 1994-96. The resulting wavy line, about 10,000 times smaller than the solar analemma, is plotted here. The north-south extent is due to proper motion, for which Barnard's Star holds the record. In the east-west dimension, there is a linear component due to proper motion and an oscillating component due to parallax.

In the professional arena, the Hipparcos mission did this on a large scale in the early 1990s, and the Gaia mission is currently doing it on a massive scale. The end product, a catalog of star positions with parallaxes and proper motions, comes from analysis of the wavy trails produced by each star's combined parallax and proper motion.

PS: If you set a location on Mars with F6, the = timestep changes to the Martian mean solar day of 24h39m35s, and you can see the Sun follow the Martian analemma. This has a teardrop shape instead of a figure 8 because Mars's orbit is 5.6 times as eccentric as Earth's. The other factors, axial tilt and solstice-perihelion relationship, are currently similar to Earth's.

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The stars are very very far away. So far away that only the rotation of the Earth matters for their position in the sky.

If you photograph a star every time the Earth completes a rotation, it will appear as a single point.

But that's not what happens if you photograph it "at the same time". A day is slightly longer than one rotation of the Earth, to keep up with the Sun. During 365 days, the Earth rotates 366 times.

As such, the celestial sphere appears to complete one rotation every year. Thus, your experiment will show the stars moving in circles around the celestial poles.

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  • $\begingroup$ Hello! Your answer is great, but can be a bit confusing because the word "rotation" might refer to the rotation of the Earth on itself or to the rotation of the Earth around the Sun. Would you please consider being extra explicit in your answer? $\endgroup$ – Stef Oct 21 at 15:39
  • $\begingroup$ I recommend Anders Sandberg's answer for a more robust treatment of exactly what the differences between the various rotational periods are. I'll let this answer stay as informal as possible. $\endgroup$ – SE - stop firing the good guys Oct 21 at 15:42
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    $\begingroup$ @Stef The planets revolve around the sun while rotating on their axes. $\endgroup$ – Ross Presser Oct 21 at 18:37
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As far as:

Has anyone published an experiment where the same star is photographed every night at the same time?

I don't know but if you did it would just be a dotted line across the sky. The stars cross the meridian about 4 minutes earlier at each night.

If you photographed at local midnight and you saw a star low in the East and photographed every night after that, it would move almost 1 degree along a circle each night.

360/365.25 = 0.986 degrees.

It will cross the meridian about 4 minutes earlier each night

24 * 60 / 365.25 = 3.94 minutes.

All you get is a dotted line from East to West.

And like @AndersSandberg's answer says if you photographed the same sidereal time ever day (which would be the 3.94 minutes earlier each day) then the star would stay almost exactly the same place.

The only deviations would be due to things like parallax or the Earth's rotation speeding up or slowing down, which are very, very tiny effects.

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