1. Newly-built planet-finding instrument installed on Very Large Telescope aims to be first to directly image a habitable exoplanet. The instrument, called NEAR (Near Earths in the AlphaCen Region), is designed to hunt for exoplanets in our neighbouring star system, Alpha Centauri, within the “habitable zones” of its two Sun-like stars, where water could potentially exist in liquid form. It has been developed over the last three years and was built in collaboration with the University of Uppsala in Sweden, the University of Liège in Belgium, the California Institute of Technology in the US, and Kampf Telescope Optics in Munich, Germany.
Since 23 May ESO’s astronomers at ESO’s Very Large Telescope (VLT) have been conducting a ten-day observing run to establish the presence or absence of one or more planets in the star system. Observations will conclude tomorrow, 11 June. Planets in the system (twice the size of Earth or bigger), would be detectable with the upgraded instrumentation. The near- to thermal-infrared range is significant as it corresponds to the heat emitted by a candidate planet, and so enables astronomers to determine whether the planet’s temperature allows liquid water.
2. On June 13, 2019 NASA’s planet counter ticked up to 4,000, then passed it by three, with 31 new discoveries.
NASA’s Exoplanet Archive announces 31 newly confirmed exoplanets – planets beyond our solar system – discovered by ground and space-based telescopes. Five were detected by the recently launched TESS space telescope.
3. While we now know of thousands of exoplanets — planets around other stars — the vast majority of our knowledge is indirect. We can only see these worlds as points of light or see the effect of gravity on stars. However, the number of exoplanets that have been directly imaged is growing over time. When NASA’s James Webb Space Telescope launches in 2021, it will open a new window on these exoplanets, observing them in wavelengths at which they have never been seen before and gaining new insights about their nature.
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6 thoughts on “Improved Telescope to Directly Image Alpha Centauri Exoplanets and Other Exoplanet Updates”
Agree. Though it will come down to a signal to noise problem, where even the alignment precision will have an impact. The question is if the telescope at hand has the necessary light gathering capacity….
Back in “my time” it was believed we would never image a star due to issues with
the atmosphere, sagging mirrors etc.. And to see a star’s planet was out of the question.
And then one day there’s this Time magazine with a photo of an exo-planet on the front
Next great prediction – we won’t see a computer as complex as the human brain, due to
the reliability of vacuum tubes, cooling issues etc.. I love these great predictions. But oh, we won’t go faster than light (?)
Though… if I can record the light I am getting from a dot, over time, as said dot is rotating I can make a 2 dimensional map of how the colour and albedo changes with longitude.
If I can also have some changes in the latitude that I am observing, say the planet’s ecliptic is not edge on, then the same technique produces a 2 dimensional map of how the colour and albedo changes with latitude.
Combine those and I get a rough map of the surface. At least to the point of continents are blobs.
Hmmm… let me try those calculations again
| Earth-to-Sol distance = d = 1 AU = 149.5×10⁹ m
| Sols-illumination-at-Earth = S = 1 Sol = 1363 W/m² at 1 AU
| Area-of–1AU-sphere = A = 2.809×10²³ m²
| Total-power-of-Sol = Ptot = SA = 3.828×10²⁶ W
| Lambertian power = Ptot/4π = 3.04×10²⁵ W/sr
| Earth-radius = R = 40,000 km / (2π) × 1000 m/km = 6,366,000 m
| Earth-face-area = f = πR² = 127×10¹² m²
| Earth-visual-albedo = j = 0.33
| Total-visual-power = jSf = 5.72×10¹⁶ W (max);
| Lambertian power = jSf/2π = 9.115×10¹⁵ W/sr
| RATIO = 3.04×10²⁵ ÷ 9.115×10¹⁵ = 3,340,000,000 = 3.34×10⁹x
Looking back on my original calculations, by the pattern of the coffee stains and scratch marks, it is clear that I had a brain malfunction, and was working with volumes, not areas. If we exclude the albedo thing, then it is in absolute agreement with your correction … about 10,000,000,000x
Still … for real-world null-cancelling interferometers (which is what this cobbled together thing is), the amount of nulling is rare to be greater than 99.9%.
The conclusion remains the same: the “image” will be a DOT.
Note that there are two different lambertian total-to-steradian divisors. A planet, unlike a star, only emits essentially from ½ its facial area at a go.
Hmm, last time I worked through the calculations, IIRC, the reflected light from an earth 2.0 around a sol 2.0 was “only” ~10^10 below the light from the sun. 10^10 for the sun-to-planet light ratio is pretty daunting, but it’s ten billion times better than the 10^20 ratio that you give above.
I don’t think near IR helps much, since planets with earth-like surface temperatures don’t emit significantly at near IR. But perhaps it helps for an earth-based telescope. Less small non-diffraction scattering of light from the central star?
Just to be crystal clear, the “direct imaging of an exoplanet” isn’t quite taking a full-face-on picture like an upside-down Earth with clouds, oceans and crazy assortment of island continents (or lava fields, or swirling gas bands, or metallic cratered surfaces, etc.)
It is at most … a DOT.
The telescopes will be used in a mode that coherently blocks most of the light of the gravitational center star; the word MOST is important… blocking 99.999999999% of the light form a star which is nominally 100×10¹⁸ (100,000,000,000,000,000,000×) brighter than the planet STILL lets thru 10×10⁹ times as much stellar light. Way too much to get meaningful image of a planet with any detail. However, great optics might be able to make out a single fuzzy pixel’s worth of reflection from the planet.
So, while it ain’t much, its still worth performing.
… for our CLOSEST neighboring star
… which just happens to be of a type similar to Sol.
At the edge of optical physics.
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