This Is The First Photo of an Exoplanet Candidate 1,200 Light-Years Away

ScienceAlert ^ | June 17, 2016 | Josh Hrala

[layman]

Oh .... so THAT’S what it looks like!

[Captain Rhino]

Hmmm...this is very curious, indeed.

At this point, I’m struggling with the implications that, at 1200 light years, one pixel in the image = 17 AU (if taken by the HST). The “tiny tiny brown dot” in the article image is what? ... two pixels wide and two pixels high?

Stars can, of course, be of enormous sizes.

(https://en.wikipedia.org/wiki/List_of_largest_stars)

It’s unfathomable to think of a planet 34 AU in diameter. Is that even possible geologically?

Yet the only variable at play in the equation you provided is the size of the aperture (diameter) . Hubble’s is 2.4 meters. How much larger are the mirrors on the telescopes at the participating observatories? How much larger would they have to be to get the planet down to say, ... Jupiter size or even 1 AU?

On the other hand, this “planet” began as a pixilated image that subsequently has been reproduced digitally and manipulated digitally (perhaps many times). The image size could be an observational approximation being provided by the processing algorithms; a mathematical way of saying “I saw something here.” Or it could be some text editor’s attempt to assist the general reader.

Either way (or perhaps some other way), its true size and composition is for some very distant future generation to know. File it away in the box marked “Unanswerable Questions.”

Thanks for entertaining my comments and questions and providing informative answers.

Be well.

[Lonesome in Massachussets]

A planet 34 AU in diameter would collapse under the weight of its own gravity. The 17 AU would be the “beamwidth”, think of it as a bell curve, containing about half the energy, the rest would be smeared out, so a 17 AU planet would just fill about one (well matched) pixel.

I blew up the photo as big as I could and the planet just spans about 4 mm on my screen. Then I measured from the planet to star, about 152 mm. The planet is 660 AU from the star. 660 AU / 152 mm = 4.38 AU per mm, so 4.38 x 4 = 17.4 AU. The planet spans about 17 AU on the scale of that photograph. That does not mean that that planet is 17 AU across, it means that the optical resolution of the camera taking the picture is about 17 AU at that distance. It cannot distinguish a 1000 km diameter planet from a 1,000,000 km diameter planet.

The number of pixels used in displaying an unresolved target in a JPEG has nothing to do with the target’s actual diameter. The enormous size of the star itself is also almost certainly an artifact. It is how large the image of the star bloomed during the long exposure. The far tails of the diffraction bell curve, not the size of the star.

[Captain Rhino]

I must admit that I was surprised that the prospective discovery of a PLANET that large was not creating a bit more excitement. But your evaluation makes it clear that saying much more than “we can see directly that some sizeable object is at this location” exceeds what can be stated with certainty.

It is still a remarkable achievement.

Thanks for taking the time and applying your expertise to tease out these details from the image.