# Power of the James Webb Space Telescope [duplicate]

I have heard that the JWST will be capable of looking at the early stages of the universe. However I have also heard that it won't be capable of detecting the Oort Cloud that supposedly provides comets for the solar system.

How is the JWST able to see millions of light years away but cannot see a giant spherical cloud of rocks around the solar system?

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## marked as duplicate by Rob Jeffries, Hohmannfan, HDE 226868, Sir Cumference, TimtechMar 5 at 14:59

It all comes down to the brightness of objects (not their size). For all intents and purposes we can assume that the most distant galaxies and the small, but much closer, objects in the Oort clouds are unresolved point sources.The Oort cloud objects are too faint to see, with JWST, but it should be able to see bright galaxies and quasars even at 13 billion light years.

JWST will have a magnitude limit of something like 30th AB magnitude at red and near-infrared wavelengths. It turns out that quasars at Z~10 can have brightnesses that enable them to be detected at these magnitudes and brighter.

OK, but now work out how bright an Oort cloud object is seen in reflected light from the Sun. Taking a solar luminosity and calculating the flux at say the nearest hypothesised Oort cloud objects at about 2000 au from the Sun, we get $1.4\times 10^3/(2000^2) = 3.5\times10^{-4}$ W/m$^2$. Now let's be generous and say that as much as 10 per cent of this is reflected (a very high albedo) and that it is a big Oort object with a radius of 20 km. If this acts as a Lambertian reflector, then it will receive $3.5\times 10^{-4} \times \pi \times 20000^2 = 4.4\times 10^{5}$ W and it will radiate $4.4\times 10^{4}$ W isotropically back into a hemisphere towards the inner solar system.

Now reversing the calculation. After this light has travelled back 2000 au to the JWST, the flux received will be $4.4 \times 10^4 /(2\pi [2000 \times 1.5\times 10^{11}]^2) = 4.9 \times 10^{-25}$ W/m$^2$.

We can compare this with the original flux of the Sun at the Earth. This is $1.4\times 10^{3}$ W/m$^2$. Now assuming that the reflection doesn't change the solar spectrum, we can take the ratio of the solar flux at earth to the reflected light from the Oort cloud object to calculate that the Oort cloud object is 68.6 magnitudes fainter than the Sun as seen from the Earth (or close to the Earth - JWST will not be in orbit around the Earth) in any magnitude band you care to choose.

So in the near infrared, the Sun has an apparent magnitude of around -27.5. This means that the Oort cloud object would have a near infrared apparent magnitude of 41 and would thus be 11 magnitudes too faint to be seen by JWST. (And remember we chose a big object that was as close as an Oort cloud object is expected it to be and gave it a high albedo),

Now consider that 30th mag detection threshold. For a solar-type spectrum this equates to a flux threshold of $\sim 10^{-20}$ W/m$^2$. Ignoring the cosmological nuances of proper distance etc., then something at 13 billion light years would have to have a luminosity of a mere $10^7$ solar luminosities to produce this kind of flux - the brightness of a modest galaxy. Of course there is also the extreme redshift to consider, so this calculation is way too optimistic, but even allowing for an order of magnitude or two, one should easily be able to see the brightest galaxies and quasars at 30th magnitude.

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