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View Full Version : Stars within 200 light years of earth could see earth transiting the sun.

moozoo
2012-Jan-12, 04:36 AM
I was curious as to which stars within 200 light years of earth could see earth transiting the sun.
Logic being that any intelligent life would probably have their own exoplanet search program and hence focus their seti efforts on planets they might find with the habitable zone (liquid water)

These stars must be plus and minus some angle from the ecliptic.

My calculation is that the angle is given by
Tan(theta) = (Rs-Re)/SunToEarthDistance.
i.e. +-15.83 arc min [EDIT I originally wrote +-31.67 which is twice the value]

200ly = a parallax of 16.308

This is the SIMBAD Criteria query that returns them
plx > 16.308 & region(box ecl,0.0 +0.0,179d 31m)

You might change the simbad output options to include the parallax and ecliptic co-ordinates.
The two closest stars are Van Maanen's star, a white dwarf at 14 lys and HD 4628 (LHS 121) at 24ly.

redshifter
2012-Jan-12, 06:30 PM
I was curious as to which stars within 200 light years of earth could see earth transiting the sun.
Logic being that any intelligent life would probably have their own exoplanet search program and hence focus their seti efforts on planets they might find with the habitable zone (liquid water)

Interesting exercise, but I'm not sure it's accurate to say 'any intelligent life' will probably have their own exoplanet search, for any number of reasons. Such as, not technologically advanced enough (say a pre-industrial society), or at all (pre-historic), or advanced enough but not overly curious about exoplanets for whatever reason. Or they did an exoplanet search 50M years ago, discovered Earth, found no intelligent life, and moved on.

IsaacKuo
2012-Jan-12, 09:37 PM
I was curious as to which stars within 200 light years of earth could see earth transiting the sun.
Logic being that any intelligent life would probably have their own exoplanet search program and hence focus their seti efforts on planets they might find with the habitable zone (liquid water)
Nice exercise!

My calculation is that the angle is given by
Tan(theta) = (Rs-Re)/SunToEarthDistance.
i.e. +-15.83 arc min
Not that it makes a significant difference, but the formula should actually be:

Sin(theta) = (Rs-Re)/SunToEarthDistance

Or, if you don't require a full transit, it should actually be:

Sin(theta) = (Rs+Re)/SunToEarthDistance

But who cares? For all three formulas, the results are practically the same.

200ly = a parallax of 16.308

This is the SIMBAD Criteria query that returns them
plx > 16.308 & region(box ecl,0.0 +0.0,179d 31m)

You might change the simbad output options to include the parallax and ecliptic co-ordinates.
The two closest stars are Van Maanen's star, a white dwarf at 14 lys and HD 4628 (LHS 121) at 24ly.

Nice. I was unaware of this query tool.

moozoo
2012-Jan-13, 12:46 AM
Interesting exercise, but I'm not sure it's accurate to say 'any intelligent life' will probably have their own exoplanet search, for any number of reasons. Such as, not technologically advanced enough (say a pre-industrial society), or at all (pre-historic), or advanced enough but not overly curious about exoplanets for whatever reason. Or they did an exoplanet search 50M years ago, discovered Earth, found no intelligent life, and moved on.

Given cosmic time scales of billions of years, if they exist (and that's the big if) then chances of catching them in the 3 million year odd window before they get to where we are is pretty slim.
If they exist (again a big if) and its at all possible to detect life on earth remotely via its atmospheres spectrum, then they would already know we are here. They would also have their ultimate version of the Square Kilometre Array.

LHS 121 is pretty interesting (http://en.wikipedia.org/wiki/HD_4628). Only 24ly away, the star is about 80% the size of ours and about 5.4 billion years old (i.e. enough time) AND its thought to have a Jupiter mass planet at 4.29 AU out (far enough to allow the development of inner rocky planets and to act as an asteroid shield).

eburacum45
2012-Jan-13, 10:08 AM
What fraction of all stars fall within this transit window? I seem to recall it was around 2%, but that may be wide of the mark.

Wolf-S
2012-Jan-14, 12:51 AM
Should the proper motion of the stars be taken into account over long time scales? At each point in time the amount of such systems which could see the Earth could be vastly different...

moozoo
2012-Jan-14, 04:08 AM
@eburacum45
The full sky has ~41253 square degrees.
The area in which the earth's transits are visible is ~ 360*(31.87/60) or ~190 square degrees.
So I work it out as ~190/41253 = .45% of the sky i.e. 1/217th
Note: I'm approximating the transit area on a thin cylinder, Its actually a strip around a sphere which means it will be slightly less than this figure. If someone wants to volunteer the correct formula I'd be grateful.

@Wolf-S
Yep I had the same thought and its an important factor re seti.
The sun takes ~250 million years to go around the galaxy and each star is travelling at a different speeds. So I doubt the transits are visible from any particular star for very long on this time scale.
For any particular star, you could work it out via its proper motion.

@IsaacKuo
I don't have a lot of time in my life at present to triple check everything before I post.
So I'd welcome anyone correcting my maths.
That said.. I still think its tan(theta), but as you said, the difference in the result is very small.
I was only considering a full transit. Considering partial transits does increase the angle by a small amount.

moozoo
2012-Jan-14, 11:25 AM
This is the SIMBAD Criteria query that returns them
plx > 16.308 & region(box ecl,0.0 +0.0,179d 31m)

This was not getting all of them. It should have been region(box ecl,0.0 +0.0,360d 31m) But the region box must not be handling large r.a. dimensions. [Edit I had 259d, should have been 360d]
I did the query for plx > 16.308 and then handled the region check myself.
There are 54 stars.

The query below gets all of them bar HR 7327
plx > 16.308 & (
region(box ecl,0.0 +0.0,45d 31.67m) |
region(box ecl,45.0 +0.0,45d 31.67m) |
region(box ecl,90.0 +0.0,45d 31.67m) |
region(box ecl,135.0 +0.0,45d 31.67m) |
region(box ecl,180.0 +0.0,45d 31.67m) |
region(box ecl,225.0 +0.0,45d 31.67m) |
region(box ecl,270.0 +0.0,45d 31.67m) |
region(box ecl,315.0 +0.0,45d 31.67m) |
region(box ecl,360.0 +0.0,45d 31.67m))

eburacum45
2012-Jan-14, 12:41 PM
Yes, for nearby stars proper motion would affect the transit window significantly, especially for very nearby stars like Van Maanen's Star. but stars thousands of light years away would appear to move much more slowly.

eburacum45
2012-Jan-14, 01:05 PM
Incidentally, I found an answer to my query above;
http://en.wikipedia.org/wiki/Methods_of_detecting_extrasolar_planets#Transit_me thod

For a planet orbiting a sun-sized star at 1 AU, the probability of a random alignment producing a transit is 0.47%. About a quarter of what I recalled, which just goes to show.

eburacum45
2012-Jan-14, 01:15 PM
@eburacum45
The full sky has ~41253 square degrees.
The area in which the earth's transits are visible is ~ 360*(31.87/60) or ~190 square degrees.
So I work it out as ~190/41253 = .45% of the sky i.e. 1/217th almost exactly the figure given on wiki. Thanks!

moozoo
2012-Jan-16, 12:23 AM
Yes, for nearby stars proper motion would affect the transit window significantly, especially for very nearby stars like Van Maanen's Star. but stars thousands of light years away would appear to move much more slowly.

My rough calculations for this are:
31.67 arc minutes = 1900.2 arc sec = 1900200 mas.
The stars 54 stars have total proper motions between 4700 mas/year and 11 mas/year.
If they moved through the transit area as fast as possible in a straight line from top to bottom, the time taken is between 400 years and 168,000 years. The average being ~18500 years.
Since they travel though the transit area on a slope the actual times will be longer than this.
For the closer ones most are more than 2000 years.

The .47% of the sky figure probably includes partial transits. With that you have to decide how much of the earth's disc has to overlap the sun before you can detect it.
To keep it simple I stuck with a full transit. The other thing with a partial transits is that they don't last as long.

If you divide the earth's radius by the sun's radius you get amount by which the light curve dips. This is about 1/11917. From memory the sun varies about 1/100000. So I guess you could come up for a figure as to how much of a dip you need to be confident it was a transit. Realising of course that the dips only happen once every year and you have to allow for sunspots etc.
With a partial transit, depending on the depth, I don't think you could work out the radius of the planet with any confidence.

[Edit changed to flares to sunspots and 1/11000 to 1/11917]

IsaacKuo
2012-Jan-16, 08:26 AM
That said.. I still think its tan(theta), but as you said, the difference in the result is very small.

It's an extremely common mistake, made by assuming the sun is a flat disc rather than a sphere. You can see the problem at short distances. For example, suppose the observer is at 2 radii away from the center of the Sun. In that case, the half angle should be 30 degrees (.524 radians). However, if you use atan instead of asin, you end up with 26.56 degrees (.464 radians).

Or consider an observer at 1 radii away from the center of the Sun. In other words, the observer is standing on top of the Sun. The half angle should be 90 degrees. The Sun covers an entire hemisphere of his view. But using atan gives a half angle of only 45 degrees. This is because using atan assumes the Sun is a flat disc rather than a full sphere.

moozoo
2012-Jan-16, 03:22 PM
@IsaacKuo
Thanks. I was assuming a flat disc. I've been playing with Cabri 2 and can see that it is indeed sine.

Noclevername
2012-Jan-17, 12:25 AM
I assume the OP is implying a species with vision equivalent to ours, who have the same telescope technology that we have. Small variations in either could skew the answers quite a lot.

whimsyfree
2012-Jan-17, 04:18 AM
@IsaacKuo
Thanks. I was assuming a flat disc. I've been playing with Cabri 2 and can see that it is indeed sine.

The difference rapdily approaches zero as the distance increases, so there is no need to worry about it for the purposes of this thread. Of course we don't have anything like an exhaustive list of stars within 200ly, though the ones we don't know about would be red dwarfs.

If you think beaming radio waves at aliens is a good idea, sun-like stars in this sample would be a good starting place.

IsaacKuo
2012-Jan-18, 09:06 AM
The difference rapdily approaches zero as the distance increases, so there is no need to worry about it for the purposes of this thread. Of course we don't have anything like an exhaustive list of stars within 200ly, though the ones we don't know about would be red dwarfs.
The distance to the other star is irrelevant. It's the distance between the Sun and Earth which is relevant, and this does not change regardless of the distance to the other star.

The real issue is that it's simply wrong to use atan rather than to simply use the small angle approximation. Atan actually makes the error *worse* than the small angle approximation, in all cases. It's a very common mistake to make, and it makes things even worse, so it should be corrected.

chornedsnorkack
2012-Jan-18, 11:33 AM
Note that the list of stars which see Earth transit Sun is almost exactly the list of stars which are seen occulted by Sun. Almost, because the orbit of Earth has nonzero eccentricity.

The stars which see partial transits should be about 1/50 of the stars which see full transits. The analogy here is seeing the star occulted from one side of Earth, but passing by from the other side.

moozoo
2012-Jan-18, 02:51 PM
@eburacum45
On this page http://kepler.nasa.gov/Science/about/characteristicsOfTransits/ they are saying for SNR reasons they need to average the transit dip over a certain minimum amount of time and that restricts the usable transit range to 86.6% percent of the total.

This means the figure is actually 0.47% * 0.866 = 0.407%
Note the 1% figure they mention is for detecting an earth and/or venus in a system that had both.

Its possible to work back from the 0.407% figure to the angle either side of the ecliptic.
0.407% of the full sky (41253 square degrees) is 167.9 square degrees.
This is the area of a strip of sky running 360 degree around the ecliptic.
Approximating this with a thin cylinder gives a height of ~0.466411 degrees or ~27.98 minutes (or +-13.99 minutes)

plx > 16.308 & (
region(box ecl,0.0 +0.0,45d 27.98m) |
region(box ecl,45.0 +0.0,45d 27.98m) |
region(box ecl,90.0 +0.0,45d 27.98m) |
region(box ecl,135.0 +0.0,45d 27.98m) |
region(box ecl,180.0 +0.0,45d 27.98m) |
region(box ecl,225.0 +0.0,45d 27.98m) |
region(box ecl,270.0 +0.0,45d 27.98m) |
region(box ecl,315.0 +0.0,45d 27.98m) |
region(box ecl,360.0 +0.0,45d 27.98m))

and it returns 44 stars.

Re the figures I gave for how long stars take to pass though the transit area, they are only indicative figures of their speed vs the width of the transit area. They might not pass from one side to the other. i.e. the might dip in and then out again.

whimsyfree
2012-Jan-20, 06:17 AM
The distance to the other star is irrelevant.

If you read my post carefully you will notice that I didn't say that it was.

It's the distance between the Sun and Earth which is relevant, and this does not change regardless of the distance to the other star.

And how many solar radii is the Earth from the Sun? and were the distances you used in your example one radius and two radii?

The real issue is that it's simply wrong to use atan rather than to simply use the small angle approximation. Atan actually makes the error *worse* than the small angle approximation, in all cases. It's a very common mistake to make, and it makes things even worse, so it should be corrected.

That's a valid point but earlier you said the difference between the calculations was negligible. Now you seem to be saying it's important to use one particular calculation. This gives the impression that you work too hard trying to prove how smart you are.

IsaacKuo
2012-Jan-20, 08:18 AM
If you read my post carefully you will notice that I didn't say that it was.

And how many solar radii is the Earth from the Sun? and were the distances you used in your example one radius and two radii?
So what? I already noted that the numerical difference between atan, asin, and the small angle approximation are not significant for this example.

That's a valid point but earlier you said the difference between the calculations was negligible. Now you seem to be saying it's important to use one particular calculation. This gives the impression that you work too hard trying to prove how smart you are.
No, I said it's important to correct the mistake. I'm not trying hard to "prove how smart" I am. I'm trying hard to correct a mistake.

Not only is this a common mistake to make, it is common for the person making the mistake to insist they've got it correct. That's what happened in this case.

I first corrected the formula with a note that it's a common mistake and also noting that the difference was not significant. (In other cases, the difference is significant--such as a thread about the angular size of a planet as seen from orbit.) After that the OP mistakenly "corrected" me by insisting upon the atan formula. At that point, I would be doing a disservice to the OP by letting him continue to hold the wrong belief. Who knows? Maybe the OP is a student or a future job applicant who would use this as part of an exam/interview answer, and the wrong answer would be marked off. (The grader might accept the small angle approximation, but not a clearly wrong formula.) Or maybe some student reading this thread would see the wrong formula confirmed by a lack of rebuttal.

I don't like to leave an incorrect formula uncorrected, especially after the correct formula has been "corrected" with the incorrect one.

moozoo
2012-Jan-21, 12:07 AM
@IsaacKuo
I was only after an approximation so that I could come up with the SIMBAD query and get the topic going.
I was mostly interested in feedback on the idea that someone else out there might be seeing the earth transit the sun, know it had life and what this might mean for SETI.

I knew my initial flat disc calculation was an approximation but didn't see that the correct formula was a simple as asin instead of atan.
Once you said it was my flat disc approximation I visualised the triangle required and saw that the correct formula wasn't as hard as I thought it would be.
I'll draw a nice diagram of it and put it up on my web site.
Anyway I'm grateful for your feedback.

The diagram http://moozoo.dyndns.org/transit/transit.jpg

moozoo
2012-Jan-24, 02:03 AM
My previous queries where overlapping a number of areas to cover the desired area.

The people at SIMBAD got back to me.
The correct query should be
plx > 16.308 & region(ZONE ecl,0.0 +0.0,360.0d 27.98m)
This adds 3 more stars to make 47

moozoo
2012-Jan-24, 02:40 AM
now I just need to combine this list with

http://ssd.jpl.nasa.gov/horizons.cgi

Ephemeris Type [change] : OBSERVER
Target Body [change] : Earth [Geocenter] [399]
Observer Location [change] : Sun (body center) [500@10]
Time Span [change] : Start=2012-01-24, Stop=2013-01-24, Step=1 d
Table Settings [change] : QUANTITIES=1,9,18,20,29
Display/Output [change] : default (formatted HTML)

and I'll know when and where to point my 1TW nuclear emp powered pulse transmitter :)