Planet Hunters Talk

New planet here

  • JenniferConnors by JenniferConnors

    Seems to be a planet with a period around 4 days

    Posted

  • Dolorous_Edd by Dolorous_Edd

    UKIRT image seems fine, however AFAIK K2 uses much larger pixel mask so perhaps needs TPF analysis

    and transits are rather V-shaped

    enter image description here

    Posted

  • Shellface by Shellface

    Again, the transits are rather deep (~2%). However, this star's colours indicate it is late G-early K, and the transit length (~3 hours) suggests it is a dwarf. For a typical radius of a K0 dwarf, this suggests a companion radius of ~1.3 Rjup, which is pretty typical for a Hot Jupiter.

    Looking at the the Kepler image, all three bright components in the UKIRT image fall within the telescope's aperture. Though the radius of the companion is probably approximately equal regardless of which star it is orbiting due to their similar brightnesses, the same thing also implies a considerably larger radius for the companion. Further modelling is required to say much more, but it is quite possible that the companion is stellar.

    Posted

  • ajamyajax by ajamyajax

    I was just looking at this one also. The fit for this LC does looks stellar to me, and there might also be an interesting oscillating TTV or ETV pattern that suggests a binary with one companion being a larger mass star as Shellface suggests, I believe. Just not sure how a hot Jupiter would move like this, but maybe it's possible. And maybe a third unseen companion is in play also. Hope to come back to this LC again, when there is more data to look for a L/P cycle. Also note I didn't fine-tune the O-C plot for MMR etc, just plugged in my observations.

    s1=1940.58 p1=3.9285 d1=0.27 (6.48 hours)

    ttv1 = [1940.58,1944.564,1948.416,1952.41,1956.262,1960.2455,1964.087,1968.081,1971.9225]

    f1
    oc1

    Posted

  • Shellface by Shellface in response to ajamyajax's comment.

    Hm. I'd say that a more plausible explanation for that TTV diagram is that the odd transits are due to a different object than the even ones, so that the system is actually an approximately equal binary with a slightly eccentric orbit with p~7.8 days. That would be mechanically easier, but requires a great deal of tertiary flux due to the shallow transits - something that this object has a good deal of.

    aja, what do you get out of this data when you analyse the odd transits and the even transits independently (but keep the periods equal)? Does that suppress the TTVs, and are the depths for the two sets of transits significantly different?

    (Also, I love those diagrams! How do you make them?)

    Posted

  • Dolorous_Edd by Dolorous_Edd

    (Also, I love those diagrams! How do you make them?)

    I will second that!

    Posted

  • ajamyajax by ajamyajax

    SF: you know what, I didn't think those plots would be statistically significant with the limited K2 data we so far so I didn't even run those options before... But looks like they might support your explanation of this system better than mine. Thank you for the correction.

    And all the plots you see here were written in python by myself using "getting starting" examples off the internet. (I worked for years as a computer programmer.) Libraries include matplotlib, numpy, scipy, pillow, six, pyparsing, python-dateutil, and pyfits. Anybody can do the same with a little time and effort as I like to say. But glad you and DE like them. And once in a while, I hope they are even a bit useful. 😃

    f1ab
    oc2
    f1a
    f1b

    K2 corrected light curve data credit: Vanderburg & Johnson (2014)

    Posted

  • Shellface by Shellface in response to ajamyajax's comment.

    Aha, those look fantastic! Though it's difficult to judge whether the odd/even transit depths are statistically different, the eradication of those TTVs decisively shows that this is an equal EB (for the level of TTV precision one would expect from K2, TTVs of ~5 minutes for 1.8% transits are probably consistent with measurement errors).

    With this in hand, the open questions are:

    • Which star in the aperture is the binary?
    • What is the component spectral type?
    • Why is the system eccentric?

    The last question is potentially the most interesting. Stars with non-negligible convective zones (for the main sequence, spectral types less than ~F7) can efficiently tidally circularise companions with periods of less than a week or two on timescales similar to their pre-main-sequence lifetime. Thus, if the the component spectral type is early, then the system may be primordially slightly eccentric, but if the component spectral type is late, then something may be pumping the binary eccentricity.

    And yes, your diagrams are very good to look at! Their design is clean and clear, and they make visually pleasing renditions of the relevant data. And hey, they beat excel!

    Posted

  • andrew418 by andrew418

    Hi all,

    I just checked data I reduced using other different apertures, and I can answer the question about which star is the binary -- it is the brightest one in the aperture, slightly above the intended target in the K2 image. (see the aperture on this page: https://www.cfa.harvard.edu/~avanderb/k2c0/ep202065879.html).

    I can see that the giant apertures that are often being selected are causing a lot of headaches -- in future data releases, I will try to come up with a better way of choosing apertures that favors smaller ones, at the slight expense of photometric precision.

    Best,
    Andrew

    Posted

  • JenniferConnors by JenniferConnors

    At first sight I thought it looked like an Eclipsing Binary too because of the transit depth, yet I had never seen an Eclipsing binary of such a perfect balance to appear like this.
    Interesting system!

    JC

    Posted

  • mschwamb by mschwamb scientist in response to Dolorous Edd's comment.

    question about the UKiRT image did you just query UKIRT to see if they had a previous observation?

    Cheers,

    ~Meg

    Posted

  • Dolorous_Edd by Dolorous_Edd in response to mschwamb's comment.

    Yep, this is exactly what I did

    Posted