# Tag Info

1

All the answers serve very good explaining why all the telescopes are synced. The only question which remains is that why we are using 4 telescopes or any number of telescopes for that matter. Actually, the quality of data collected increases with the diameter of the telescope. So, they place many telescopes in an array to increase the diameter of the ...

2

What you are seeing is the ALMA http://www.eso.org/public/teles-instr/alma/ which is an array of radio telescopes whose data can be combined to simulate the performance of a much bigger radio telescope. So the reason they're moving in synchrony is that they are all looking at the same target.

2

The video is of the Atacama Large Millimeter Array, and this is an atronomical interferometer. Interferometers work by combining the signal recorded by many individual dishes to reconstruct the original image. To make this work all the dishes have to be pointed towards the same object, and that's why the video shows them moving in sync.

5

Using a camera that can capture "Motion at a Trillion Frames Per Second", this can be done at the laboratory scale. The technique used has been called femto-photography. (Image credit to Ramesh Raskar, Associate Professor, MIT Media Lab) More info from MIT here.

4

Let's say you build a ping pong ball counter. It increments the count every time a ping pong ball hits the sensor. You throw a ball, and it hits the sensor: Detected! You throw a ball across the sensor from left to right... no detection, because you didn't hit the sensor. Your eyeball is a light sensor, which creates pictures from the light that hits ...

7

If some of the light is reflected off the dust at such an angle that it is diverted to reach the observer, the observer will see that light. However, those specific photons reaching the observer will not reach B (unless they are reflected there by the observer). Similarly, unless the observer is at point B (which is not the case in the question as asked), ...

3

My masters project was on something like this (though with hydrogen alpha emission lines for gas clouds between galaxies rather than dust between stars) and the answer in that case (and almost certainly in this one too) is that you can't see it because it's just too dim, though using hundreds of telescope hours can get you close (maybe). Again, this is not ...

20

It would be possible to see the progress of photons through space if the light pulse were exceedingly intense, and if the dust cloud from which they reflect were positioned and shaped to reflect the light toward us. Rather than shooting a beam from Point A to Point B, it would be better if the light source were between us and the dust cloud, as light ...

42

Sometimes we do, and the phenomenon is called a light echo. What you're looking at there is NOT moving gas. It's an "echo" exactly as you describe. The problem is that you need a pulse of light. If you have a constant stream of light, the "light echos" will be exactly like what you see in fog on earth.

1

In the early days of the solar system, it is generally believed , as the comment above implies, that it was a place of constant collisions. The moon may have been part of the Earth, there is strong evidence for that, and an impact is the most probable way that it would be have been able to became a separate orbiting body. For the giant planets, Jupiter ...

5

The image is that of the entire sky above the telescope taken with some kind of fish-eye lens. It is quite common at observatories to use such arrangements to monitor for cloud cover. However you seem to have found a particularly poor example - possibly using a CCD imager with very few pixels. It is also possible that there is some light cirrus over the ...

1

As you might know, we use RA (Right Ascension) and DEC(declination) to locate the stars in the sky. Now the RA and DEC co-rodinates of any star is fixed. However, locally to locate a star we use ALT / AZ co-ordinate system. The ALT and AZ of a star depends upon the LAT/LONG of the place you are observing from and the time. There are many online converters ...

1

The answer is a qualified yes, depending on your definition of fusion. If you collect together a mass of iron less than about $1.2M_{\odot}$ it is possible for that to form a stable configuration, a little smaller than the size of the Earth, supported by electron degeneracy pressure. Such a star would just sit there and cool forever and all that would ...

0

Iron won't fuse into heavier elements. It's a question of nuclear physics. Iron is the most stable form of nuclear matter. In other words, iron has the lowest energy configuration of all nuclear matter. Fusion can appear in the cores of stars because it's an exothermic process, that is, fusing nuclei lighter than iron can lower the nuclear matter's energy ...

0

Let's say a field of stars all die within a short amount of time. Just for argument's sake they produce a debris field of iron ( or any other heavy element). Provided that there is enough time the debris will agglomerate, we know this. My question: Given enough mass, will this agglomeration of heavy elements fuse into even heavier elements? ...

1

As I understand it regular fusion in a star takes light elements as input and the output is heavier elements and energy. There are several potential steps in the regular process, e.g: Hydrogen fusing to helium and producing energy which keeps the star from collapsing. After a lot of hydrygen is spent and helium has collected in the center of the star the ...

0

As I understand it, iron will not be able to fuse with iron no matter how much of it you have gravitationally bound. Instead to fuse iron atoms together requires a supernova.

3

Rigid bodies with three distinct moments of inertia have two stable rotation axes, the axes with the greatest and least moments of inertia (typically the shortest and longest axes). Non-rigid bodies have but one stable rotation axis, the axis with the greatest moment of inertia. The axis with the least moment of inertia becomes unstable thanks to entropy. ...

3

the flux drops off as the square of the distance, but the solid angle subtended by the source drops off the same way, so surface brightness is constant, right? Right. But what happens when you can no longer resolve the source? Then the "solid angle subtended by the source" stops dropping, and only the reduced flux of the entire source can be ...

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