If you don't know the orbital period, you can't know the inclination. The orbital period and inclination both conspire to affect the duration of the transit. A short-period planet with an edge-on orbit will have the same transit duration as a longer period planet in a slightly inclined orbit (with a higher impact parameter).

This is assuming that the orbit is near circular. An eccentric orbit can have widely varying transit times, depending on which portion of the orbit is in front of the star (as seen from Earth).

@Hungry: If you have the radius of the star, you don't need the period. You can compute either from the other. Or did you find an error in my calculations from last time?

@Mongo: If you have the radius AND the period, you can at least take a stab at the eccentricity, since the velocity will be high or low compared to circular. Subtle differences in the timings of first, second, third, and fourth contact should be clues too--assuming we had something sensitive enough to measure that.

--Greg

I think that it would be a lot easier to get sufficiently precise radial velocity measurements to directly determine eccentricity than it would be to accurately determine the 1st, 2nd, 3rd and 4th contacts -- and you would also find large, close-in non-transiting planets as well, although transit timing variations might do so as well.

@ Greg, If you don't know the orbital period, you won't know the planet's orbital velocity, and thus how long it should take to go across the star. That's why a long-period planet with a high impact parameter can have an equal transit duration as a short period planet without a high impact parameter.

The transit duration of these two planets will be the same:

IIRC, under the right conditions, Kepler might be able to find planets as small as Mercury. (Granted, these conditions will be rare.) I'm not sure it will ever be possible to get radial velocity (Mongo, I'm assuming you meant RV for stellar wobbles induced by the planet - is that right?) for planets that small, and even up to Earth-mass. One of the reasons I'm hoping for an extended Kepler mission is not just finding longer-period planets, but also refining statistics for the short-period ones, which will help constrain contact timing (among a myriad of other things).

Yes. To be sure, random stellar noise from photospheric motion will dominate the RV signal for any given measurement, but averaging repeated observations over many orbits (subtracting the calculated RV effect of any other known planets for each observation) should reduce that effect by the square root of the number of observations. That should be sufficient to allow reasonably accurate mass determinations.

The big problem would be getting sufficient telescope time in the first place.

@Hungry Those two examples will have very different transit onset times, so that's how you tell them apart.

--Greg

What do you mean by transit onset times? The time inbetween first and second (or third and fourth) contacts? If so, I thought Kepler didn't have the cadence to accurately measure those.

H4I, meet short cadence. Short cadence, meet H4I.

http://arxiv.org/PS_cache/arxiv/pdf/1001/1001.0142v1.pdf

The Earth would take De/Ov seconds or 12,715 km /29.8 km/s = 426 seconds for the Earth to move one of its own diameters. So if the the inclination were 0 degrees, a putative alien would see an ingress/egress time of a little over seven minutes (and a total transit time of 1.4 MKm/29.8 km/s = 46979 seconds or 13 hours) . Assuming more likely angles would result in longer X-gress times than that, and with an ~ 1 minute short cadence, I think Kepler would easily be able to estimate the inclination angle and orbital velocity. I don't know if the orbital velocity at a point on the ellipse and the time for the orbit is enough to give you the precise eccentricity, but I think with a large sample you can begin to say some very interesting things about the eccentricity of the population. For example, if you are seeing numbers above and below the expected value for a circular orbit, then you can with some confidence say that X% are in non-circular orbits. Also, the highest and lowest numbers would give you information about the maximum eccentricity values in your population.

Interestingly, the only short cadence data available from the NStED website I gave upthread that had anything resembling planet data is this one, which seems to be pretty deep, (Nevermind, didn't see that the Y axis is linear and offset) but illustrates the number of points possible on the gresses. The paper mentioned above listed short cadence data being taken on three known transiting planets. I'll drop them a line and see if they can wring out the additional data.

TrES-02
Click to view attachment

HAT-P-07 (KIC 10666592) has long cadence but not short cadence data online

HAT-P-11 (KIC 10748390) isn't found in the database.

I submitted a ticket into their helpdesk for more information.

PS
Found a way to correlate names with Kepler Input Catalog numbers here:

http://nsted.ipac.caltech.edu/starsearch.html

Works for Kepler-1/2/3 and their original discovery names above.

I'm familiar with the 1-minute short cadence for Kepler, but in the very paper you mention.

Since only 512 of the targets are being observed at short cadence, and there's some 1,000 or so stars with planets, and it's unreasonable to expect each of the 512 short-cadence stars to be planet hosts, and since the long cadence targets represent the overwhelming majority of the target stars, it's those systems I was talking about.

I should have been more specific about which cadence I was talking about.
... for the overwhelming majority of stars since they're being observed at long cadence.

@Hungry Yes, time from first contact to second (call that t [small t]) call the time from first contact to mid-transit T (big t) so total transit time is 2T and time in full transit is 2(T-t). Assume you know R (radius of the star) hence you can compute r (radius of the planet) from the depth of the transit. Then (I claim) you can directly compute the velocity, v^2 = 4Rr/(T^2-t^2), from just a single transit. (As above, assuming a circular orbit.)

And I never said Kepler had the accuracy to do this, although Hendric's example suggests that it actually does under some circumstances.

--Greg

Very pretty visualization of Kepler data: (some apples to oranges comparisons in there I think, tho)

http://flowingdata.com/2011/02/09/a-view-o...to-kepler-data/

If I understand this correctly, some educated guesses could be made about the prevalence of longer-period planets even without observing more than one transit?

Correct. There have been (shhhhhhh) a few observations of such things. The single transit gives them information on approximate size and period of the apparent planet, but there's no solid statistical significance to it.

But, still, that those observations exist... all they need is time to remove the observation bias!

Shhh! They're right there in the released data! But don't tell anyone!

On a less snarky note, even absent the occasional long-period candidate transits, I imagine there are implications in the statistics that can be derived from the current Kepler release. For example, there does appear to be a hard edge as you get (very) close to a star, inside of which planets are extremely rare or absent. OTOH, no sign yet of any reduction in frequency for any size planets as you get farther from a star. In other words, it's not turtles all the way down, it's turtles all the way up!

Well, yes, I know that's overreaching. But that conclusion is at least not inconsistent with the data.

@hendric So for the planet you show, I figure it to be pretty big: 12% of the diameter of its star. And it seems to be moving pretty fast: 30.3 stellar radii/day or 175 mps if the star has the diameter of sol.

Any more info about it?

--Greg

The planet is TrES-2 b.
http://exoplanet.eu/star.php?st=TrES-2

Well, we're still early into the mission. I wouldn't assume that number is fixed, although it could be data bandwidth limited I suppose.

Also, as the pipeline is refined and additional processor power applied, it might be possible to download data, detect the start of an Earth-class transit, and enable short cadence for the egress portion of the transit. Given that the Kepler data should be very amenable to distributed computing, I don't see any major roadblocks to this. Perhaps during an extended mission, or Kepler II (if we decide a second mission to look in a different direction is worthwhile). Imagine an extended mission where the data is publicly available immediately, with BOINC clients to process it into lightcurves, and planethunters.org's thousands of eyeballs looking out for a sudden drop.

@Hungry Hmmm. I seem to have a factor of 2 error somewhere. I'd love to drop that factor of 4 from the numerator of the expression for the square of the velocity, but it's just as likely I screwed up calculating the velocity for TrES-2. I'll look at it again after work.

--Greg

@ Hendric,

The transits of these planets is measured in hours. If we were so unfortunate as to have only the ingress observed, then with the time it would take for Kepler to transmit the data, roll into its new position, and for the data to be received, processed, and the ingress discovered, I would think the transit would be over by then. How long does the whole transmission and roll take?

That I couldn't tell you. I assumed it had an articulated antenna and could transmit continuously. Still, nothing that couldn't be solved with enough ingenuity, no doubt.

Sounds like a great reason for a mission extension!

Are there any issues with the spacecraft if they do manage to extend the mission?

If they find a Jupiter-sized planet in a nice long orbit kinda like our Jupiter -- maybe there are some earthlings tucked away on a rocky earth-like planet to be subsequently discovered?

It seems that once there is a large enough database of candidate planets with lengthy transit times, they coulld statistically estimate how many of those are "center" crossings versus "side" crossings of the star. By some algebraic manipulation akin to deconvolution, they could then estimate counts for different orbital periods.

This could be very useful towards the end of the mission in order to infer really distant planets, such as Saturn with a 29 year orbit. I think we really would prefer not to wait for 3 transits for those ones!

@marsophile You can compute that from a single transit; you don't need statistics. If all you want to figure is how far off-center the transit was, you only need three things:

1) Duration of the entire transit--from first contact to fourth contact. (T)
2) Duration of the "deep" transit--from second contact to third contact. (t)
3) Square root of the amount the star dimmed during the deep transit. ®

Let h^2 = [T^2*(1-r)^2 - t^2*(1+r)^2]/(T^2 - t^2)

Then h is how off-center it was. h=0 means dead center and h=1 means all the way at the edge.

--Greg

@hungry and hendric Okay, I found my error. Applying my corrected formula to TrES-2b, I'm off by 0.9% on the velocity and 1.9% on the semimajor axis.

To make it simple, let T be the time from first to fourth contact and t be the time from second to third. (That's easier to estimate from the graph than the semitransits.) Use the published figure for R, the radius of the star, and estimate r based on the drop in brightness at the midpoint of the transit. Then

v^2 = 16*r*R/(T^2 - t^2)

For stars with large, close, transiting companions, we'll already have the doppler velocity and the period, so this will let us quite accurately estimate both the radius and the mass. To the extent we have accurate distance measures to these stars, I'm hopeful that'll let us produce models that predict radius and mass from spectrum, luminosity, and maybe other factors. There sure ought to be lots to study.

Then, for stars where we witness just a single transit of a large planet, once we have models to let us estimate radius and mass of the stars, we can use the formula above to get the semimajor axis, period, and inclination--all from a single transit!

--Greg

Nice, well done

Thanks. It turns out, someone else did the same thing (in 2003), but in more generality: http://iopscience.iop.org/0004-637X/585/2/1038/

In my own solution, I assume the planet is far enough from its star that I can treat its motion across the star as linear. These guys find that solution too, but they also find a more complex one that can deal with a planet that's so close that the transit is noticably "bent."

Separately, they report that for main-sequence stars you really can do a pretty good job of estimating the mass and radius. They talk a lot about estimating the period from a single transit.

Their big focus, of course, is ruling out "blended" stars. It gives me a better idea of what the Kepler guys are spending their time doing.

--Greg

Update on the short cadence data:

Hi Richard -

Indeed, the Kepler data made available by MAST in June of 2010 was not complete for TrES-2, HAT-P-07 and HAT-OP-11.

The recent MAST release of Kepler data on February 2nd, including new Kepler quarter 2 data, includes the short cadence data for these planets.

NStED is finalizing our release of these light curves, which we get from MAST when they become public. We expect them to be available to you on or about February 21st through NStED's website.

If you have any questions, or if you need the data urgently, please let us know. Thank you for using NStED.

PS

Can someone confirm my comment on the Systemic blog is correct? If the transit chord is on the trailing side of a star's proper motion, the chord should increase in length over time.

http://oklo.org/2011/01/01/a-safe-prediction/

Yes, that is correct.

I agree. I'm trying to decide whether study of the effect of proper motion over time can reveal anything other than \Omega;

--Greg

Greg,
Not sure what you mean by Omega, but if you had a long enough series, it could give another way to confirm the distance from the star during the transit.

Ω is the ascending node of the orbit, something that isn't actually known for the RV and Transiting planets.
Getting a long enough baseline could take centuries.

@Hungry I know it could take centuries, but it's an interesting thought experiment. Or if an ion-propelled telescope could be accelerated to (say) 10x the speed of the Sun, then perhaps a result could come in decades.

But it'd have to tell us a lot more than just Omega. :-)

--Greg

The latest data release was pretty spectacular in many ways. However, it did only cover a few months(~4) of observations. Is there any schedule for future releases or is it event dependent? Are we guaranteed a release every quarter or 6 months? If anyone has insight on the conditions or timetable for future Kepler data releases, please add your wisdom to this forum.

Adapted from the Kepler mission FAQ:

When does the proprietary period end for Kepler Data?
The table below lists the release dates for Kepler Key Project data, also known as the planetary search data. The data for each quarter (Q1-Q8) has a specific release date based on the Science Utilization Policy. Dates for quarters 7 and later will change if the Mission is extended.
Q1 = 6/15/10
Q2 = 2/01/11
Q3, Q4 = 6/18/12
Q5, Q6 = 6/18/13
Q7 = 9/18/13
Q8 and ff = 9/18/13
For Kepler non-Key Project data the nominal proprietary period is 12 months from ingest of the data into the archive by the DMC. For GO data, the proprietary period is 1 year from ingest of the data into the archive or 6 months after the last data for the proposal are ingested, whichever is later. From time to time, the Project may decide to extend these proprietary periods.

Of course that doesn't mean that you'll only get discovery announcements around those dates, as we've seen. If the current follow up season from ground telescopes provides confirmation of candidate planets, especially those based on data already released from the embargo, we'll hear about those before next summer, I'm sure.

It looks like the new Kepler data has arrived at the NStED database:
(short cadence data for example)
http://nsted.ipac.caltech.edu/cgi-bin/bgSe...splay_radius=on

http://www.spaceref.com/news/viewsr.html?pid=36329

JPL study based on released Kepler data.

Thanks for the link.

Summary:

- A planet with 0.95 <semi-major axis < 1.37 is defined as in the habitable zone.
- Additionally, 0.8 < radius < 2 defines an "Earth analog."
- Kepler's returned data is matched with predicted data.
- Conclusion: about 2.5% of stars have an "Earth analog" planet.

Comments:

- Not mentioned are stars harboring more than 1 Earth analog.
- Not mentioned are giant planets harboring Earth analogs. (Certainly anyone would just be guessing at this point.)

My thoughts:

- The papers uses the word "scarce" when describing the frequency of stars harboring Earth analogs. I don't think 2.5% is scarce at all. Holy cow. That's 1 in 40. Four times ten to the first power. Scarce would be like 1 in a 1000.

Agreed, 1 in 40 isn't that bad. And that's just with a baseline of a few months. The number will surely improve.

http://news.stanford.edu/news/2011/march/l...ets-030811.html



I just found out about this lecture about an hour ago. I hope to attend, but will have to cancel my existing plans. Wish I had known about this a few days earlier! How did this lecture sneak past us?

Their 1:40 number has predicted improvements and predicted future discoveries built in to it.

Still, if you had asked me three years ago if I'd like a 1:40 ratio, I'd have replied with an emphatic "Yes! Are you crazy?!!"

Interesting contribution Hungry4Info and Syrinx. But, in France, thanks to Jacques Laskar's work at CNRS(*) who demonstrated that, thanks to our big Moon, the conditions on our homeworld seems to be very special indeed, we do NOT refer anymore to the words "Earth-like" or "Earth analog" when referring to those new discoveries, but to "Earth-sized" planets which is scientifically much more accurate. The French Press now also uses this term, not to give the Public falses hopes about the findings of new Earths. By looking to Jacques Laskar's conclusions, one may assume that the number for "Earth-like" worlds will surely be much lower, while the number of "Earth-sized" planets will surely grow in the near future.
(*) "Stabilization of the Earth s obliquity by the Moon, Nature, 361, 615-617, February 18, 1993."

I agree with your general thought. We need better terms. Everybody has their own, and what's worse sometimes the terms overlap with differing meanings.

But as per the paper under discussion these worlds are definitely more than just Earth-sized. They are in the habitable zone. The paper calls them Earth-analogs, but again you can call them anything you want. I prefer to call them Earth-like, but my "Earth-like" category is the first step on a 3-step ladder.

http://www.unmannedspaceflight.com/index.p...st&p=149307

I can't comment on a moon being a requirement for any category of Earth-likeness. Haven't read the paper.

We could take a cue from the Kepler team: they've got planet candidates, we could have Earth candidates. Meaning, the available data meet all criteria for Earth-likeness, but we lack sufficient data to say if a planet's Earthlike. (Of course, "Earthlike" opens a huge can of worms that I'm not going to touch here except to note that by present-day standards, Earth hasn't been Earthlike for most of its existence. But we already knew that.)

Also of note: "the full 3.5 to 6 year Kepler data set". I like that 6, but it's the first I've heard of it. I hope we keep hearing more!

I'm going to be lazy here since I've got work I'm supposed to be doing, but 2.5% of what, exactly? Surely not Kepler's entire sample of stars - 2.5% of 150,000 would give us 3,750 "Earth analogs" (of which Kepler would actually see - what? 375? that's still huge)! The original mission docs said they expected 50 - 60, IIRC.

EDIT:

OK, in the paper itself, we have this: "We found that 2% of stars have an Earth in the HZ; this means 3000 of the 150,000 Kepler stars have an Earth. Of this 3000, about 1 in 250, or 12 will be edge-on, producing a transit. The Kepler data set
contains 4 Earth and super-Earth planets colder than 300 K. This means that as of Feb 2011 Kepler found 30% of the transiting Earths and super-Earths that it will eventually find after 3~4 years."

So they're actually saying that Kepler will find <12 Earth candidates (since they've got "4 Earth and super-Earth planets").

Thing is, I thought the incidence of transiting planets was more like 1%, whereas if I'm reading it right they're saying it's .4%.

We could just call them all "Venus-like" until proven otherwise. That will keep the media off of them!

It's a function of stellar radius, planet radius, and orbital radius. For a giant Jupiter in a 0.1 AU orbit it'll be 10% (just guessing). For a tiny Earth 100 AU out it'll be just about 0%.

For an Earth-Sun relationship I'm remembering 0.5% but it very well could be 0.4%.

Thanks Syrinx : I feel good on your "3-step" ladder. Now you would need a "B prime" step : the stabilization of the Earth s obliquity by... (you name it, but it's the Moon for our home planet), because, as Jacques Laskar discovered, it helped stabilize also our climate over long periods of time (to match your "C" criterion) and be "Earth-survivable" as you say. Mars had not such chance...

A moon specificially isn't really needed. Planets in nearby (stable) orbits can assist in stabilising the obliquity as well. A paper on 55 Cancri found that it's planets could support an obliquity-stabilised planet in the HZ without the need for a moon. I'd have to find the paper, perhaps later. Furthermore, I'm not convinced the obliquity stabilisation is even a requirement for a planet being Earth-like. Is Myr-long climactic stability (on average, neglecting the kyr-long fluctuations) a requirement for a planet to be Earth-like? The deserts have to stay the deserts, the jungles have to stay the jungles?