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The protagonist in my SF story is on an asteroid gazing towards the Sun and inner planets. For purposes of the narrative he needs to locate the position of the earth at different times in its orbit.

He is standing on the asteroid, about 2.5 Astronomical Units out, in the asteroid belt, viewing towards the Sun

enter image description here

Since he’s basically outside looking in with regard to the Earth, wouldn’t all the inner planets line up in a straight line passing through the sun?

Wouldn’t the Earth (and Mercury, Venus, and Mars) move back and forth along that imaginary line (red arrows), their location along that line depending on what point they are in the orbit? For instance, Venus could appear to be inside Mercury, as in my drawing, correct?

Edit: the asteroid has little or no inclination.

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    $\begingroup$ Congratulations! You have discovered the Zodiac from purely theoretical considerations. (The Zodiac is a belt and not a line because while it's true that the planets are close to being in the same plane there are actually differences of a few degrees between the planes of their orbits. For the same reason the transits of Venus are rather rare.) $\endgroup$ – AlexP Mar 20 '17 at 19:59
  • $\begingroup$ Right. Either that or the Ecliptic plane. $\endgroup$ – catsteevens Mar 20 '17 at 20:03
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    $\begingroup$ There was a highly related question on SpaceEx.SE earlier today: space.stackexchange.com/q/20719. Note in that question how the "side view" shows the inner planets are not in a line. $\endgroup$ – cobaltduck Mar 20 '17 at 20:09
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    $\begingroup$ @catsteevens if it was tidally locked with the sun you would have rotation that matches the orbital period, so from the perspective of the observer no rotation. $\endgroup$ – Joe Kissling Mar 20 '17 at 20:31
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    $\begingroup$ You are saying that your observer's location has no inclination. That doesn't matter, as the three non-earth planets in view do have some inclination. Therefore your observer will see their paths as tight ellipses, not as lines. $\endgroup$ – cobaltduck Mar 20 '17 at 20:31
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Yes, that's approximately how it would look, with some caveats:

  • The asteroid is orbiting the sun, too, so your protagonist isn't a completely stationary observer (although it will orbit slower, since it is further out).
  • The sun is bright, so you may need to have something to block it to observe carefully.
  • Any planets closer to the sun than you will have phases, like we observe on Venus. The observer would be able to determine, then, whether the planet is in front or the back part of their orbit around the sun.
  • All the planets are very slightly tipped, so they likely won't actually transit (pass directly across) the sun on every orbit. Compare the frequency of transits of Venus and transits of Mercury.
  • Unless other asteriods come close, Earth's moon is probably the only other object large enough to distinguish, at least toward the sun.
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  • $\begingroup$ I calculate that the Sun would be roughly magnitude minus 22, at 2.5 a.u.'s distance. Is that anywhere near correct? If so, that is bright. $\endgroup$ – catsteevens Mar 21 '17 at 2:44
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    $\begingroup$ @catsteevens Wikipedia puts the Sun at -26.74 apparent magnitude on Earth. Here is the formula for difference in apparent magnitude. Doubling the distance (1 AU to 2 AU) reduces the amount of light seen by four times, and going from 1 AU to 2.5 AU reduces it to on the order of 1/(2.5^2) or about 1/6. Without having done the math myself, your apparent magnitude of -22 of the Sun viewed at 2.5 AU thus seems roughly correct. $\endgroup$ – a CVn Mar 21 '17 at 7:42
  • $\begingroup$ I made another error--I thought that intensity and distance goes as the cube, but apparently it's only the square? So 6.25 (2.5 squared) dimmer is about 1.6 mags, therefore the sun is minus 26 or so at that distance? It really would blot out nearby plot I reckon. $\endgroup$ – catsteevens Mar 21 '17 at 15:15
  • $\begingroup$ Not my day I guess... I was told the Sun is -24.8 apparent at that distance. $\endgroup$ – catsteevens Mar 21 '17 at 21:00
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No, this wouldn't occur, due to something called orbital inclination. None of the planets orbit in exactly the same plane. Their orbits are a little tilted relative to each other. Some are easier to notice than others, but Mercury has a pretty big tilt (6 degrees) that would definitely be noticeable. This handy figure shows the inclination for all the planets:

orbital inclination

What is more, as you can see from the 3D map below, not only are their inclinations different, they are bent in different directions, throwing the line even more off (which is why the angles on the Wikipedia page above don't seem to line up):

enter image description here

As you can see from the 3D map, if you include Pluto then the whole thing is way off, it has a huge orbital inclination of 17 degrees (depending on how you measure it)

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    $\begingroup$ While your final sentence is certainly true (Pluto really is the odd one out), you are going to be hard pressed to view Pluto by naked eye from 2.5 AU from the Sun looking in any direction, let alone toward the Sun. $\endgroup$ – a CVn Mar 21 '17 at 7:45
  • $\begingroup$ Besides, Pluto is not really considered to be a planet these days. $\endgroup$ – catsteevens Mar 21 '17 at 13:57
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    $\begingroup$ Nothing in your answer is factually false, per se, but I think your conclusion ("no, this wouldn't occur") is unsupported, as a practical matter, for an observer that simply "needs to locate the position of the earth at different times in its orbit". Yes, there are very minor inclination differences, but yes, the planets will appear to be nearly in a straight line. $\endgroup$ – BradC Mar 21 '17 at 15:34
  • $\begingroup$ @BradC: I don't think 6 degrees is "very minor". And all the differences would be much larger than the size of the planets themselves. $\endgroup$ – TheBlackCat Mar 21 '17 at 15:38
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    $\begingroup$ @TheBlackCat I'm not disagreeing with your facts, just disagreeing with you on how important they would be to OP's story. Even to a careful observer, the inner planets will appear to be going back and forth in (nearly) a straight line, with some measurable but small deviations. Mercury's inclination will seem lessened by being closest to the sun. For comparison, see this pic of Jupiter's moons. In that case the deviation is because of Jupiter's tilt in relation to us, but they are clearly in a (non-perfect) line. $\endgroup$ – BradC Mar 21 '17 at 15:47
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What you described would more or less occur, provided the light from the sun can be blocked out, the planets would appear to line up. Over the course of the year(s) they would appear to move back and fourth. However, the planets would be much less visible, if at all, at their closest approaches to the observer because the sunlight would not be reflecting off of them.

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  • $\begingroup$ Right, didn't think about that fact, especially if the planet lines up between the sun and the asteroid, right? It would go through phases like the moon. $\endgroup$ – catsteevens Mar 20 '17 at 20:23
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    $\begingroup$ Correct, they would not be visible if thew were between the observer and the sun. You would also get waxing and waning phases as moved closer or further from the observer. As pointed out it would not be a straight line due to slight orbital inclination variations of the inner planets, but for the purposes of the question, it would be pretty close. $\endgroup$ – Joe Kissling Mar 20 '17 at 20:27
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Not quite.

Inferior planets (ones closer to the sun than your orbit) will appear to move a set number of degrees on either side of the sun. You show Earth as moving on a short path. That path will take across the sun to the other side.

From an asteroid at 2.5 AU, the earth would move from about 22 degrees on one side of the sun to 22 degrees on the other side. Venus and Mercury will be proportionally less.

You can figure out what side of the orbit they are on by brightness. While further away when the planet is on the far side, most of the disk you see is facing the sun, on the nearside most of the face you see is dark.

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  • $\begingroup$ I suppose the drawing is confusing, I just meant the arrows to show that the planets move back and forth, roughly on the white line, depending on what point they are on their orbit. Length was arbitrary. 22 degrees on either side; good, I needed that calculation. $\endgroup$ – catsteevens Mar 21 '17 at 19:56

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