Astronomy/Math question: Where would 1 pc = 1 ly

Factual Questions
1

The length of a parsec, ~3.26 LY, is based upon the distance from the Earth to the Sun. It works out to 3606060*π AUs (I think). The length of a light year is derived from a different terrestrial source: it’s the distance light travels in the time it takes Earth to travel once around the Sun. If we were on another planet, with a different distance to the Sun, and a different length of year, we’d derive different values for our local parsec and light year.

At what distance from the Sun would 1 parsec = 1 light year? Is it possible within our solar system, or can that only happen in a different system where the star has a different mass? Please show your work, I want to understand how to set up the problem.

No, this is not a homework question. School is long behind me. :slight_smile:

*** Ponder

2

You’ll have to straighten out your terms first. One Astronomical Unit is the earth-sun distance. One parsec is one “paralax of one arc-second”. It’s the length of the long side of a triangle where the short angle is 1 arcsec and the short side is one AU.

3

I love this question, as it’s something I’ve never thought about before, but it’s the kind of thing I would think about.

The relationship between distance from the Sun, a, and orbital period, P, is not linear. So there is a chance that at some distance the two properties are changed in proportion to each other in a way that arrives at the coincidence you ask about. If the relationship were linear, then the relationship between a and P, (and thus a “local parsec” and a “local light-year”) would always be the same.

If we stick to worrying about our own Sun, we can take advantage of the fact that the square of the orbital period squared is proportional to the cube of the distance from the Sun. In other words, P[sup]2[/sup] is proportional to a[sup]3[/sup], and so P is proportional to a[sup]1.5[/sup]. For example, Venus is 0.723 times as far from the sun as the earth, and the orbital period is 0.723[sup]1.5[/sup] * 365.24 days = 224.7 days.

To find out the distance necessary to get the “local light-year” to match the “local parsec”, you need to find the distance where the ratio of the two has diminished to 1/3.262 of what it is for Earth. (Because here it’s 3.262, and we want it to be 1). Since the “local light-year” is proportional to P, it is therefore proportional to a[sup]1.5[/sup]. The “local parsec” is simply proportional to a. So the equation of interest reduces to simply:

parsecs/light-year = a/P = a/a[sup]1.5[/sup] = 1/3.262

This reduces to a[sup]1/2[/sup] = 3.262, or a = 10.64 AU.

So a planet 10.64 times as far from the Sun as the Earth would have the coincidence you describe. There are no planets there, but Saturn is closest at 9.54 AU. Let’s run a quick check on Saturn’s orbital parameters and see if that ratio is indeed close to 1:

Saturn is 9.54 times as far from the sun as Earth, so a “Saturn parsec” would also be 9.54 times as long. which is 31.1 “Earth” light-years). Saturn’s orbital period is 29.46 Earth years, and so a “Saturn light-year” is 29.46 “Earth light-years”. Those are, as expected, pretty close to 1. A “Saturn parsec” is 1.06 “Saturn light-years”, compared to the 3.26 ratio here on earth.

So, if we orbited our Sun at 10.64 AU (just outside Saturn), we would have defined the parsec and the light-year to be the same distance. For a different star, you can’t use the Earth’s (or any other planter in our solar system) orbital period and distance as a basis to simple ratios, and would have to calculate them directly based on the mass of the star and the equations of motion.

4

Couple of points:

  1. If I’m not mistaken, parallax is measured by apparent movement of the astronomical object in question relative to a much more distant (and hence ‘fixed’ for purposes of this measurement) six months apart. So the short side would be 2 A.U. – the diameter of Earth’s orbit, not its radius.

  2. Granted that a light year, as a measure of distance, has to do with the distance light travels in a vacuum in 365.25 standard Earth days (of 24 3600-standard-second hours). It is therefore tied to Earth in particular. A planet with an orbital period (;year’) of slightly over 60 days might makes six complete orbits of its star in the time it takes light to reach it from a red dwarf a light year distant.

  3. A parsec is of course the inverse of the parallax value in seconds of arc. Something at two parsecs distance subtends 1/2 arc-second, something at 50 parsecs subtends only 1/50 arc-second. Alpha Centauri subtends about 3 arc seconds parallax. The value for a parsec is of course based on Earth’s orbit (and, interestingly, the diameter of that orbit is almost precisely 1000 light seconds). But, as the OP notes, different orbit diameters would result in different values for what constitutes a parsec based on their

Given all that, I hear the OP as asking, "Suppose we define a ‘light-annum’ as the distance travelled by light in a vacuum during a given planet’s orbital period. Earth’s light-annum would be one light year; Jupiter’s, 12 LY. Given that, is there a value for orbital mechanics where one light-annum for a given planet would be equivalent to the distance representing one arc second parallax with the baseline the diameter of that planet’s orbit.

I have a hunch it would not, owing to orbital speed – I don’t think there is a value where the distance travelled by light in one local ‘year’ would be equal to a distance of an object subtending one arc second parallax in half that period. But it’ll take someone with a much better grasp of celestial mechanics than I to demostrate that.

5

According to Kepler’s law the square of the orbital period is equal to the cube of the semi-major axis of the orbit. So assuming a circular orbit the length of a planet’s light year is equal to it’s orbital radius (in AUs) raised to the 3/2 power. The length of a planet’s parsec will be very close to proportional to orbital radius since the tan x is approximately equal to x when x is very small.

That means the ratio of the light year to the parsec increases like the square root of the orbital radius. For earth the parsec is 3.26 times as long so we need to be 3.26^2 = 10.6 times as far. This is just about where Saturn is.

6

Thanks, aerodave and OldGuy! That’s exactly what I was looking for. Your explanations were very easy to follow.

Did I goof when I reversed the formula for a parsec and came up with: 3606060*π AUs?

I wasn’t sure either. I was trying to work through it, using pretty much the same steps you outlined, but I was getting bogged down while trying to use the above formula for a parsec. I find it quite neat how relatively simple the final equation was: just square Earth’s pc/ly ratio to get the final answer. All that mucking about I did to find the direct relation between an AU and a parsec turned out to be irrelevant.

I had an early intuition that the answer would turn out to be somewhere closer in to the Sun, rather than waaaay out there, so I was a bit surprised at how the actual math turned out. And then I realized the equation we’re dealing with is quadratic. And quadratics have 2 solutions. Where’s the other solution? At AU = 0. So my intuition was right! :smiley:

I was thinking about how a race that happened to have evolved at that perfect distance from its sun might consider the coincidence of a 1/1 ratio to be somehow divine. Of course, they would have also needed Babylonians somewhere in their history to invent the 360-degree circle, divided up into arcminutes and arcseconds. Our circular measurement system is just as much happenstance as the actual values of our AU and LY.

*** Ponder

7

An earlier thread of mine:
http://boards.straightdope.com/sdmb/showthread.php?t=121172&highlight=parsec