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1

It's a pretty broad topic. It depends a bit on the type of DSO. Speaking in general, most DSOs lack brightness, and so to observe them you need two things: lack of light pollution lots of aperture Light pollution Observing from the city, DSOs are challenging. The further away you are from city lights, the better your views. Deserts, national parks, ...


3

The angular resolution of the telescope really has no direct bearing on our ability to detect Oort cloud objects beyond how that angular resolution affects the depth to which one can detect the light from faint objects. The detection of an Oort cloud is simply a question of detecting the (unresolved) reflected light in exactly the same way that one detects ...


1

The main problem with determining an object's orbit is we only know the position with certainty in two dimensions. The distance to the object is largely unknown. This accounts for the large uncertainty in the period of newly discovered TNOs. Many possible orbits could fit the early observations, and therefore the uncertainty is large. As time goes on, ...


1

Some can, some can't. Viewing really is determined by ability, experience and technique when you leave viewing conditions aside. A lot of DSOs have a lower surface brightness, one thing I always notice is when an observer continuously increases their magnification to see the object larger. Surface brightness is proportional to magnification, the higher the ...


0

Some filters also help improve the view as seen through the telescope. Even a simple and cheap red filter would help,as it will remove most part of the light pollution spectrum and will let Ha reach your eye :)


1

Just because you can keep your eyelids open for $x$ seconds doesn't mean you are collecting light for $x$ seconds and using it to form a single image in your brain. How would you "save" the photo? How would you decide when to end light collection? You know as well as I do that you can't simply lift your finger off your brain's shutter release! And that's on ...


3

Few photons -- You have tiny pupils. Only photons that manage to travel that far over that much distance along a path that manages to intersect with your tiny pupils will have a chance of being seen. And only some photons that reach your retina actually interact with molecules that register their arrival. Interference -- The molecules of the atmosphere, ...


17

Surely if you stared long enough, the light from them would eventually hit your eye? Collecting light over a long span of time is how telescopes can see very dim objects. The human visual system doesn't work that way. For one thing, even when you think you are staring at something, your eyes still dance around a bit. It's a built-in response called ...


12

Not at all a dumb question, but actually you can see distant galaxies with the naked eye. From the northern hemisphere, the Andromeda Galaxy, our biggest neighboring galaxy, is visible if you know where to look, and is at a reasonably dark place. From the southern hemisphere, the two smaller, but nearer, irregular galaxies called the Small and Large ...


4

Your reasoning would be valid not only for galaxies, but also for stars and anything it shines in the Universe, but there is an important effect which invalidates it: absorption of light. Intergalactic and interstellar medium is filled with dust and gas, which contributes to absorb and scatter the light from distant objects. Especially on the plane of our ...


6

Yes, indeed! Many nebulae are visible from Earth in a small and cheap telescope, and even to the naked eye (if you are standing in a sufficiently dark place). In fact, yesterday I was watching the Orion Nebula with my 4.5" telescope (which is worth $200 or so) from my apartment in the middle of Copenhagen. The term "nebula" is a bit of a… well, nebulous ...


1

Approximating the Earth as a sphere with radius $R$, then when viewing from a height $h$ above the surface, the Earth blocks out a cone of some opening angle $2\vartheta$, where $\csc\vartheta = 1+\frac{h}{R}$. Thus, the visible portion has a solid angle of $$\Omega = 2\pi\left(1+\cos\vartheta\right) = 2\pi\left(1+\frac{\sqrt{h^2+2Rh}}{R+h}\right)$$ ...


0

For the area of sky as defined by "areas visible above your head inside Earth's atmosphere" then the answer depends on the height of the objects that you would like to observe. I like to observe airplanes, so without loss of generality I'll use ~10,000 meters as my "sky". We can see airplanes up to about 40 KM away. π * (40)^2 ~= 5000 KM^2 The ...


1

Arguably, "rogue planets" have already been discovered by direct imaging. Giant planets when first formed are big and hot. They radiate their own light, mostly in the infrared. So young isolated planets can be seen directly. There have been various claims in the literature that objects as small as a few Jupiter masses have been identified in young star ...



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