Solved

Kepler-186f -- How can we deduce or learn so much with so little data?

Posted on 2014-04-19
14
446 Views
Last Modified: 2014-04-21
Kepler-186f, the first validated Earth-size planet to orbit another star in its habitable zone.

http://www.examiner.com/list/an-overview-of-the-kepler-186-system?cid=rss

What is the previous science and math that informs us so much about this newly discovered planet with so little specific data about it?

Thanks

WS
0
Comment
Question by:WaterStreet
  • 5
  • 4
  • 2
  • +2
14 Comments
 
LVL 83

Assisted Solution

by:Dave Baldwin
Dave Baldwin earned 80 total points
ID: 40010856
This article explains a lot about the circumstances: http://en.wikipedia.org/wiki/Kepler-186f   A key feature of the 'habitable zone' is that it is warm enough for liquid water and cool enough for it not to all boil off into space.
0
 
LVL 18

Author Comment

by:WaterStreet
ID: 40010892
Dave,

The link gives more statistics, but how do we get that with so little observable data.

I'll repeat my question:

"What is the previous science and math that informs us so much about this newly discovered planet with so little specific data [such as transit info] about it?"  In other words, how does transit data tell us so much?
0
 
LVL 83

Expert Comment

by:Dave Baldwin
ID: 40010899
I think it's just that what we see matches what we believe is needed.  http://en.wikipedia.org/wiki/Habitable_planets
0
 
LVL 27

Expert Comment

by:aburr
ID: 40010931
The key is the light received from the star. The receiver is VERY sensitive. The planet reduces the amount of light received. One can detect when the planet first blocks some of the light. One times how long before the light is unblocked. This gives info on the orbit.
The rate of change from blocked to unblock gives info about the size (and any atmosphere).
Change in amplitude of spectral lines from the sun gives info on composition of planet's atmosphere.
All these changes are very small so the uncertainty is large. So one makes the best guess and hopes that additional info reduces the uncertainty.
0
 
LVL 81

Assisted Solution

by:byundt
byundt earned 290 total points
ID: 40010997
The original article in Science lays out the evidence with clear discussion. The Wikipedia article links to it as reference 1, with a pdf available at http://www.nasa.gov/sites/default/files/files/kepler186_main_final.pdf 

From the distance of Earth to Kepler-186, its magnitude and its color temperature, you can make a very good estimate of the diameter and mass of the star.

From the repetitive changes in intensity, you can deduce the presence of planets passing between the Earth and the star. You distinguish planets from very dim companion stars in a binary system because it is unlikely (at the 0.5% level) that such a binary star would be at the right range of distances (1.4 to 4.2 AU). Any closer than 1.4 AU and the planetary orbits would be unstable. Any further than 4.2 AU and the companion star would be detectable.

The magnitude of the changes in light emitted by the star tells you the ratio of the cross-sectional areas of the planet and star. From that, you estimate the diameter of the planet.

Density of the planet is unknown. The article bracketed the density over a 12:1 range between a water/ice planet and one composed of pure iron.

After the bibliography, the pdf gives you the transit time data for each of the planets orbiting Kepler-186.
0
 
LVL 18

Author Comment

by:WaterStreet
ID: 40011434
Hi byundt.

I thought the pdf link you attached was exceptionally informative.

Please refer to your following quotation:
"The magnitude of the changes in light emitted by the star tells you the ratio of the cross-sectional areas of the planet and star. From that, you estimate the diameter of the planet."

However, I find it difficult to agree, because we are assuming that the Kepler telescope, the star (186) and the planet (186f) are all reasonably coplaner.  How can we assume that, and the so confidently diameter of the white dwarf (186)

WS
0
 
LVL 81

Assisted Solution

by:byundt
byundt earned 290 total points
ID: 40011585
The .pdf didn't say how the distance to Kepler-186 was determined, but I speculate that you calculate the distance (151±18 pc) by measuring its parallax compared to other stars at the extremes of the Earth's orbit. Closer stars exhibit more parallax shifting. More distant stars (the ones you want to use as a reference) exhibit less parallax.

No matter where Kepler-186 is, the Kepler telescope looks at it straight on. If you measure the spectra that it emits, you know its temperature (3788+/-54 K according to the .pdf). If you know the temperature, you can calculate the amount of light it emits according to Stefan Boltzmann law http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law.  From the measured light intensity and distance, you can thus determine the diameter of Kepler-186.

"I find it difficult to agree, because we are assuming that the Kepler telescope, the star (186) and the planet (186f) are all reasonably coplaner." Three points determine a plane, however. So I assume that you meant to argue that the Kepler telescope may not be on the plane of the planetary orbit around Kepler-186.

If the Earth is close enough (angular sense) to the plane of the ecliptic of the planets of Kepler-186, you will observe a planetary transit. If you observe the transit, then a certain fraction of the cross-sectional area of Kepler-186 as viewed from Earth (straight on) will be blocked. As a result, you can calculate the ratio of diameters of the planets to their sun.

If the Earth were quite far off from the plane of the ecliptic of the planets of Kepler-186, you would not observe a planetary transit. And we would not be having such a stimulating discussion.
0
Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

 
LVL 14

Assisted Solution

by:frankhelk
frankhelk earned 130 total points
ID: 40011711
Depending on the principle that Kepler used for detecting planets around distant stars, it is a absolute MUST that the earth is very close to be in the orbital plane of the planet. Kepler detects the planet by measuring the light emitted from the star. The intensity of it's light that raqches earth drops if the planet passes between it's star and earth. In that moment Earth, the planet and the distant star are positioned along a stright line that's part of the orbital plane of the planet.

To the things we derive out of the data ... the star's distance to earth could be measured i.e. by its red-shift or parallax measurements. The type of star - indicated by its spectral classification and the light intensity fives hints about mass, size and amount of light emitted. From mass and the planet's orbit data (orbiting time, shape of intensity drop) the distance to the star could be derived. The amount of light emitted gives hints about the amount of energy the planet receives form the star. When the star passes thru, its possible atmosphere causes slight shifts in the starlight's spectral pattern - that leads to some data about the atmosphere (amount, components). The analysis of atmosphere data from one transit against others gives hints about the weather ... and so on ....
0
 
LVL 81

Expert Comment

by:byundt
ID: 40011762
If the ratio of diameters of the planet and star is 0.021, then you can observe a transit if the Earth is within 0.39 degrees of the plane of the orbit of Kepler-186f. This is the "close enough (angular sense)" requirement that I stated in my previous Comment.

I got this by drawing the intersecting tangents between star and planet. The line connecting the center of the star and planet bisects these intersecting tangents.

By similar triangles, I know that:
star radius/L1 = planet radius/L2
L1+L2 = radius of planetary orbit

Since planet radius/star radius = 0.021 (from the paper), L2/L1 = 0.021 and L1 is roughly 0.98*orbital radius

The angle between the bisector and intersecting tangent is given by:
arcsin(star radius/L1)

Using values from the paper of:
L1 = 0.33 AU = 31 million miles
star radius = 0.47 radius of Sol = 203,000 miles

I get angle = (180/pi) * arcsin(203,000/31,000,000) = 0.39 degrees
0
 
LVL 18

Author Comment

by:WaterStreet
ID: 40012175
frankhelk,
1.  I understand the star is an "M dwarf" so that we have a good idea as to the average mass/size of one, but Kepler 186 could be an outlier in the tail of the bell curve.  So, we can only make a guess as to its probable mass -- the range of which could greatly alter our confidence in the estimated parameters for the planet 186f.  Do I understand this correctly?

2. Can you provide the next level of technical details as to how the "shape of intensity drop" relates to derivation of the star's distance to the planet, or provide a link?

Thanks.

byundt,
Nice application of math.

1. Did you really mean to say 0.39 degrees instead of 39 degrees?  If yes then that's seems to be a small window for expecting to see even a truncated (not full) transit.

In either case, that leads me to something that was bothering me when I opened this thread.

2. "If the Earth [angle] were quite far off from the plane of the ecliptic of the planets of Kepler-186, you would not observe a planetary transit."  In this case, aren't we just assuming a full, non-truncated, transit?  How would be know one way or the other?

Without informing me to the contrary it seems that we are using rough estimates for the star's mass and plugging that into calculations for  rough estimates of the planet's parameters.  So, the conclusions about the planet could be quite off.

Anyone,
What specific future planetary exploration plans are being made to confirm or expend what we learn from Kepler?
0
 
LVL 81

Accepted Solution

by:
byundt earned 290 total points
ID: 40012474
The .pdf said that the planetary period of Kepler-186f was 129.9 days, and that it had been observed for three years by the Kepler telescope. This means that 8 or 9 transits could have been observed. The repeatability of the data (not discussed in the paper) should give us confidence.

By a truncated transit, I assume that you mean that the Kepler-186f planetary cross-section is never fully "over" the Kepler-186 star cross-section as viewed by Kepler telescope. If the ratio of diameters of planet and star is 0.021 (as presented in the .pdf), then only the extreme 0.01 degree of the 0.39 degree allowable angle between the Kepler telescope and Kepler-186f planetary ecliptic would exhibit truncation.

If an observed transit occurs closer to the "top" or "bottom" of the star cross-section, then the duration of the transit would approach 0. Such was not the case with the actual data presented at the end of the .pdf. The duration of the transit could even be used to estimate the angle that the Kepler telescope makes with the planetary ecliptic.

If an observed transit were being truncated, then you would expect a continuous change in the light intensity during the transit. An untruncated transit would have a "flat" part in the middle in which complete shadowing (according to the ratio of diameters) is occurring. The data presented in the .pdf for each of the planets of Kepler-186 showed such "flat" parts.

I agree that it is fortunate that the Kepler telescope lies within the allowable 0.39 degree angle compared to the Kepler-186f planetary ecliptic. But there are many, many stars in the Milky Way galaxy, and we should expect to see a certain portion of them with favorable alignment of their planetary ecliptics so that we might observe them.
0
 
LVL 14

Assisted Solution

by:frankhelk
frankhelk earned 130 total points
ID: 40013225
about "1": OK - many of the findings from the data available are more or less "good guesses", but they rely on data and they mark the most probable scenario that fits the data. There's a (more or less broad) space for variations, but the gross direction fits. The more assumtions are made to explain the measured date, the broader is the range of possible deviation. That's science, especially if outer space.is involved.

about "2": It's like an eclipse ... the planet enters the visible disk of the star, passes over it, and leaves. At both ends, the light curve forms a "ramp", in between, the drop forms a "constant" floor. If the ramp is steep, and the floor is long compared with that, the planet's disc is small measuerd by the star's disk. If the floor is short, the phase where the the planet is fully inside the star's disc is short. The depth of the dip (along with its orbital parameters) gives a hint about the size of the planet compared to the size of the star. Or it might help to make the orbital data more precise.

About the .39 degrees ... I think that's not completely wrong. The angle differs with the distance from the star and the planet size. If I calculate that for the earth, I come to about 0,005 degrees (arctan(eart diameter/1 AE)).

Given the mentioned 0.39 degrees, there's another intersting value - if you divide a full circle of 360 degrees by the 0.39, you learn that for every Kepler 186f there are  roughly another 1000 planets of that constellation around, where the orbit is not suitable for Kepler to observe the transits.
0
 
LVL 81

Assisted Solution

by:byundt
byundt earned 290 total points
ID: 40013304
The observed transit time is approximately 6 hours.

If the transit occurs across the diameter of Kepler-186, then the planet must travel approximately 406,000 miles in its orbit (twice the radius of the star).

Since the planet is orbiting at 0.33 AU radius, the orbital path is 2*pi*0.33*93,000,000 miles = 195,000,000 miles.

The orbital period is 129.9 days, so a diametral transit will take 129.9*24*406,000/195,000,000 = 6.49 hours

If the transit path takes 6 hours and a diametral path takes 6.49 hours, you can calculate the angle between star center, diameter and endpoint of the transit path as:
arcos(6/6.49) = 23 degrees

The angle between the Kepler telescope, Kepler-186 star center and Kepler-186f planetary ecliptic plane would then be determined as:
0.39 degrees * sin(23 degrees) = 0.15 degrees
0
 
LVL 18

Author Closing Comment

by:WaterStreet
ID: 40013772
Very informative.

Thank you all.

WS
0

Featured Post

Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
Q1. Magnets and Electromagnetism 33 105
Graph function 4 72
How to access ANSI/IEEE Std 754 or equivalent information ? 3 45
Does a rhombus have 2 pairs of parallel sides 5 38
Article by: Nicole
This is a research brief on the potential colonization of humans on Mars.
Lithium-ion batteries area cornerstone of today's portable electronic devices, and even though they are relied upon heavily, their chemistry and origin are not of common knowledge. This article is about a device on which every smartphone, laptop, an…
This is used to tweak the memory usage for your computer, it is used for servers more so than workstations but just be careful editing registry settings as it may cause irreversible results. I hold no responsibility for anything you do to the regist…
This is a video describing the growing solar energy use in Utah. This is a topic that greatly interests me and so I decided to produce a video about it.

910 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

23 Experts available now in Live!

Get 1:1 Help Now