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For reference, Hubble's mirror is 2.4 meters wide, the upcoming James Webb's 6.5 meters, and the proposed ATLAST 8 or 16 meters. Let's assume a mirror nearly ten times Hubble's size, 21 meters, is sent up and benefits from a coronagraph. Could it detect the planets of stars like Proxima Centauri, at 4.2 light years, and likewise Epsilon Eridani, at ~10.5 light years?

By detect, not so much a pretty picture but enough light and data to confirm orbits and overall characteristics. Only other assumption would be that this telescope observes UV-O-NIR spectra, say between 200 to 1000 nanometer wavelengths.

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  • $\begingroup$ So in other words, you want to be able to resolve the planet? $\endgroup$ – Phiteros Dec 13 '17 at 0:49
  • $\begingroup$ Note that lens size might not be the most serious limitation. To observe a planet light-years away, you have to aim the telescope at exactly the right spot. That pretty much means you have to know the orbit beforehand. $\endgroup$ – Emilio M Bumachar Dec 16 '17 at 16:50
  • $\begingroup$ For those that didn't get the memo :) ATLAST (Advanced Technology Large Astronomical Space Telescope) has now morphed into LUVOIR (Large Ultra Violet/Optical/InfraRed Surveyor). $\endgroup$ – Vince 49 Jan 2 '18 at 22:23
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As of 2017, 22 exoplanets have been imaged directly. The most distant of those is 1200 ly away.

This shows 4 of them orbiting HR 8799, which is 128 ly away:

HR 8799

These observations are good enough to determine planet orbits, and to do spectroscopy.

So we can already observe large exoplanets at much longer distances, if they orbit far enough away from their star. Smaller planets closer to their star are more difficult to see.

In the image above there's a black disk at the center. This is used to block out the star's light. It covers the star, plus a radius of about 5 AU, so any Earth-like planets in the habitable zone are masked in this image.

This is necessary because a telescope's optics and imaging system aren't perfect: the star's light is not confined to the star's diameter, it gets spread out a bit. This spread-out starlight is brighter than planets in close orbits, so those planets are lost in the noise.

One way to combat this is an occulter, i.e. a physical object in front of the telescope that blocks out the star's light. Most of these are small disks at the front end of the telescope.

NASA is working on a starshade: a much larger disk placed 50,000 km in front of a space telescope. This would be able to block out the star's light more accurately, allowing the telescope to see exoplanets in the habitable zone (so at distances of ~1 AU for smallish stars). The plan was to use the starshade with WFIRST (a 2.4 m telescope), but recent concerns about costs could see the starshade removed from this mission.

So the solution for direct imaging of exoplanets may not have to be a larger mirror.

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  • $\begingroup$ The planets of HR8799 are giants of 5 or even 7 times the mass of Jupiter and a distance of several tenths of astronomical units. The smallest directly observed planet is less than 2 times the mass of Jupiter. Direct observation of extrasolar planets with a mass and distance like our Earth will take some time. $\endgroup$ – Uwe Dec 13 '17 at 13:57
  • $\begingroup$ @Uwe -- I believe it is "several tens" of AUs, not several tenths. $\endgroup$ – antlersoft Dec 13 '17 at 20:48
  • $\begingroup$ If there were an award fro the best use of GIFs in SE, this one would win. Beautiful! $\endgroup$ – uhoh Dec 14 '17 at 5:15
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The resolution of a telescope in radians is given (to a very good first-order approximation) by $\theta_{res}=1.22\frac{\lambda}{D}$ where $\lambda$ is the wavelength and $D$ is the diameter of the telescope aperture. Using the numbers you give, a 21 meter telescope at 200 nanometers would have a resolution of 0.002 arcseconds. Now, there are many other factors which will limit our resolution, but you can consider this, the diffraction-limited resolution, to be the highest possible resolution you can achieve. Anything with a smaller angular size than this will be indistinguishable.

So our next step is to figure out what angular size an exoplanet will subtend on the sky. Let's go with a Jupiter-size planet, to increase our chances. This means a planetary diameter of $~1.4*10^8$ meters. This object's angular size (in radians) will be given by $\theta_{size}=\frac{diameter}{distance}$. We can actually set this expression equal to our equation for resolution and solve for distance. This will give us an approximation for how far away a planet can be before we won't be able to resolve it. Solving for $distance$ yields:

$$ distance = \frac{diameter\cdot D}{1.22 \lambda} $$

Again, using our numbers, this gives us a distance of $1.2*10^{16}$ meters, or 1.3 light years.

So, ultimately the answer is no. A 21 meter primary aperture will not be enough to resolve an exoplanet. So let's see how big our mirror would have to be, by rearranging our equation:

$$ D = \frac{1.22\cdot \lambda \cdot distance}{diameter} $$ If we want to resolve a Jupiter size exoplanet at the distance of Epsilon Eridani, it would need to be 173 meters in diameter.

Now, this is all assuming that we don't have to worry about other things, like glare from the star, which presents its own set of problems. But we can get around this by doing things like optical interferometry, which allows us to increase the effective size of our telescope without having to build bigger mirrors.

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  • $\begingroup$ Your answer seems to be about resolving planet as a disk (as opposed to seeing it as a point). I think we don't need that much resolution to confirm orbital parameters. $\endgroup$ – Alexander Dec 13 '17 at 1:27
  • $\begingroup$ I was basing my answer more off the "overall characteristics" part of the question. If all you care about is separating the exoplanet from the star, you could calculate the pixel scale of the telescope. But I don't think there's enough information in the question to answer that. You would need the focal ratio of the telescope (which you could make up whatever number for it), but you would also need to know the point-spread function of the parent star (unless it's blocked), and the separation between the star and the planet. $\endgroup$ – Phiteros Dec 13 '17 at 1:33
  • $\begingroup$ And in this case, you would still see the planet as a point, but it would be a point separate from other objects, so you could say that you were looking only at the planet and not other stuff as well. $\endgroup$ – Phiteros Dec 13 '17 at 1:36
  • $\begingroup$ that's fine, but I don't see why the planet's diameter should be part of the equation while the orbit's diameter (semimajor axis) is not. $\endgroup$ – Alexander Dec 13 '17 at 5:43
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    $\begingroup$ @uhoh that presentation is really great. But no matter what secondary optics you add, you can't get better than the diffraction limited resolution. That's the absolute best resolution you can accomplish. $\endgroup$ – Phiteros Dec 13 '17 at 19:50
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Yes. The ESO's VLT used the wobble method to detect Proxima Centauri b, a planet with a radius estimated at 0.8–1.5 R⊕ and a semi-major axis estimated at 0.0485 (+0.0041,−0.0051) AU, at a distance of 4.224 ly.

The Hubble Telescope can see the Proxima Centauri system's Sun.

The ground based VLT consists of four individual telescopes, each with a primary mirror 8.2 m across, they can achieve an angular resolution of about 0.001 arc-second. In single telescope mode of operation the angular resolution is about 0.05 arc-second.

Best case ground based conditions give a seeing disk diameter of ~0.4 arcseconds and are found at high-altitude observatories on small islands such as Mauna Kea or La Palma.

At the best high-altitude mountaintop observatories, the wind brings in stable air which has not previously been in contact with the ground, sometimes providing seeing as good as 0.4".

Under bad conditions a ground based telescope over 10 meters with poor seeing can limit the resolution to be about the same as given by a space-based 10–20 cm telescope.

Ground based telescopes must look through the atmosphere, which is opaque in many infrared bands (see figure of atmospheric transmission). Even where the atmosphere is transparent, many of the target chemical compounds, such as water, carbon dioxide, and methane, also exist in the Earth's atmosphere, vastly complicating analysis.

Existing space telescopes such as Hubble cannot study these bands since their mirrors are not cool enough (the Hubble mirror is maintained at about 15 degrees C) and hence the telescope itself radiates strongly in the IR bands.

The JWST telescope has an expected mass about half of Hubble Space Telescope's, but its primary mirror (a 6.5 meter diameter gold-coated beryllium reflector) will have a collecting area about five times as large (25 m^2 or 270 sq ft vs. 4.5 m^2 or 48 sq ft).

The JWST is oriented toward near-infrared astronomy, but can also see orange and red visible light, as well as the mid-infrared region, depending on the instrument.

JWST's primary mirror is a 6.5-meter-diameter gold-coated beryllium reflector with a collecting area of 25 m^2.


From the JWST FAQ:

  • At which wavelengths will Webb observe?

    Webb will work from 0.6 to 28 micrometers, ranging from visible gold-colored light through the invisible mid-infrared. The short wavelength end is set by the gold coating on the primary mirror. The long wavelength cut-off is set by the sensitivity of the detectors in the Mid-Infrared Instrument.

  • How faint can Webb see?

    Webb is designed to discover and study the first stars and galaxies that formed in the early Universe. To see these faint objects, it must be able to detect things that are ten billion times as faint as the faintest stars visible without a telescope. This is 10 to 100 times fainter than Hubble can see.

  • What are the main science goals of Webb?

    Webb has four mission science goals:

    • Search for the first galaxies or luminous objects that formed after the Big Bang.
    • Determine how galaxies evolved from their formation until the present.
    • Observe the formation of stars from the first stages to the formation of planetary systems.
    • Measure the physical and chemical properties of planetary systems and investigate the potential for life in those systems.
  • How far will Webb look?

    One of the main goals of Webb is to detect some of the very first star formation in the Universe. This is thought to happen somewhere between redshift 15 and 30 (redshift explained in question 45). At those redshifts, the Universe was only one or two percent of its current age. The Universe is now 13.7 billion years old, and these redshifts correspond to 100 to 250 million years after the Big Bang. The light from the first galaxies has traveled for about 13.5 billion years, over a distance of 13.5 billion light-years.

  • Will Webb see planets around other stars?

    The Webb will be able to detect the presence of planetary systems around nearby stars from their infrared light (heat). It will be able to see directly the reflected light of large planets - the size of Jupiter - orbiting around nearby stars. It will also be possible to see very young planets in formation, while they are still hot. Webb will have coronagraphic capability, which blocks out the light of the parent star of the planets.

    This is needed, as the parent star will be millions of times brighter than the planets orbiting it. Webb will not have the resolution to see any details on the planets; it will only be able to detect a faint light speckle next to the bright parent star.

    Webb will also study planets that transit across their parent star. When the planet goes between the star and Webb, the total brightness will drop slightly. The amount that the brightness drops tells us the size of the planet. Webb can even see starlight that passes through the planet's atmosphere, measure its constituent gasses and determine whether the planet has liquid water on its surface. When the planet passes behind the star, the total brightness drops, and we can again determine more of the planet's characteristics.


Super short version: They're launching a slightly larger telescope than you have asked for that will reach ("detect", not provide close-up photos) virtually to the known edge of the universe.

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  • $\begingroup$ How deep into the infrared spectrum can a space telescope study without needing either coolant or extensive sun shielding? $\endgroup$ – Redliox Dec 15 '17 at 19:10
  • $\begingroup$ Some of the sensors on the JWST can see from 0.6 microns to 5 microns using the heatshield only (it's 5 sheets of plastic film); without using the jwst.nasa.gov/cryocooler.html - it uses helium gas, and very little, so not a coolant in the usual sense, or not much of it. It's designed to be light weight. More info about the Sunshield: jwst.nasa.gov/sunshield.html . $\endgroup$ – Rob Dec 15 '17 at 19:39
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At the present time, there is a plan to incorporate a planetary coronagraph on WFIRST based on the PISCES integral field spectrograph (IFS). The instrument will use a local occulter, rather than a starshade. The coronagraph will not resolve the planetary disk, but will be able to see a point image (PSF) separate from the host star. This will enable the IFS to extract a spectrum. The "smoking gun" for a habitable planet is a combination of water vapor, oxygen, and methane. [Methane can not endure over geologic time in the presence of free oxygen, so there must be a constant source (life?)].

Lessons learned from the coronagraph on WFIRST will be used to optimize the design of a revised planetary coronagraph on LUVOIR, or whatever the eventual name of the follow on to Hubble. A starshade is not only expensive (as Hobbs pointed out), but it also takes a long time to re-target. In an ideal world, and with sufficient funds, the local occulter would be used to get a first order spectrum of promising planets, then the most promising would be re-visited with a starshade to extract more detailed spectra.

Additionally, JPL is planning to propose the HabEx Mission which would incorporate a planetary coronagraph, also including an IFS. I believe it's important to point out that their site says: ... directly image planetary systems around Sun-like stars. This means they will be able to image luminous points (PSFs) that are distinguishable from the host star and adjacent planets--not planetary disks.

Of course, priorities change and funds come and go, so none of this is written in stone.

WFIRST/PISCES Integral Spectrograph: https://arxiv.org/abs/1707.07779

Proposed HabEx Mission: https://www.jpl.nasa.gov/habex/

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