Tag Info

1

Look at SETI's youtube channel. It's Life Jim, but Not as We Know It: The Prospects of Life in Titan's Seas, a rather informative talk by Jason Barnes about the prospects for life on Titan.

-1

As long as a neutron star emits any sort of elctromagnetic radiation, then it will emit radio waves, as the waves/particle gets 'redshifted' more and more.

3

I'm pretty sure it's restricted to the near-IR, with the shortest wavelengths being $Y$-band or $J$-band (i.e., 1 or 1.2 microns) and the longest being $L$-band (i.e., 3.8 microns). This is probably because 1) classical adaptive optics systems use the optical for corrections that are applied in the near-IR (for reasons I discussed in this answer); and 2) ...

3

This is a great question and sampling is always a little tricky. side note: It's important to make sure that no down-conversion has been done, that the " 1000-1400 MHz band" has not already been mixed with a 900 MHz local oscillator and shifted to 100-500 MHz before conversion. I had a hunch that you can get by with a lower frequency. I chose 1000 ...

6

While the statement in the block quote about the sphere is correct as far as it goes, was the shape of the correcting optics (secondary, etc) above also independent of where you look? Or ideally would you like a different secondary shape depending on how far off-vertical you look? If so, is this what happens? I will answer under the assumption this question ...

5

Below are some strengths of some radio sources. Let's choose 1000 Jy and 100 MHz bandwidth, a modest 1000 m^2 single dish and 8 hour observation. $$1000 \times 10^{-26} \ \text{W} \text{m}^{-2} \text{Hz}^{-1} \times 1000 \ \text{m}^2 \times 10^8 \ \text{Hz} = 10^{-12} \text{W}$$ and $$10^{-12} \text{W} \times 28800 \text{sec} = 3 \times 10^{-8} \text{J}$$ ...

4

Radio waves (approximately $\lambda=1\rm\,m$[1]) are very weak, about $E=\frac{hc}{\lambda}=\frac{6.63\cdot10^{-34}\cdot 3\cdot 10^8}{1}\rm\,J\approx2\cdot 10^{-25}\rm\,J$ every photon. An $1\rm\,mg=1\cdot10^{-6}\rm\,kg$[2] snowflake falling at $0.7\rm\,\frac{m}{s}$[3] has K=\frac{mv^2}{2}=\frac{1\cdot10^{-6}\cdot 0.7^2}{2}\rm\,J\approx2.5\cdot10^{-7}\rm\,...

Top 50 recent answers are included