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The highest resolution 3d printers I know of are around 1600dpi, which is a resolution of about 15$\mu m$. Telescope mirrors have to be smooth to fractions of a wavelength of light, so the resolution of current printers is nowhere near good enough. Whether 3D printers could one day be good enough is a different question, but given that the improvement in ...

24

A typical giant galaxy, such as the one you've provided a picture of, has a radius of something like $10\;\rm kpc$ (kiloparsec - $1\;\rm pc \approx 3.2\;ly$). A supermassive black hole hosted in such a galaxy has a mass of something like $10^6-10^9\;\rm M_\odot$ (solar mass, $1\;\rm M_\odot \approx 2\times10^{30}\; kg$). The monstrous billion solar mass ...

24

One cannot tell by the light spectra. Hydrogen and antihydrogen would give the same lines in the spectrum. The prevalence of matter over antimatter from other evidence indicates matter is predominant in the observable universe, and here is a nice review. How do we really know that the universe is not matter-antimatter symmetric? The Moon: Neil ...

22

The moon does have a night and a day, but this isn't as fully connected to your question as you might think. The moon is tidally locked with the earth, meaning that the same side always faces earth. Since the moon also orbits around the earth (with a period of a lunar month), this means each side changes, over the course of a lunar month, between facing ...

16

You're right that the Sun being 4.5 billion years old makes observations difficult. The Sun goes around the Galaxy about once every 225 million years, so since the Sun formed it has gone around the Galaxy perhaps 20 times. The trouble is that the Galaxy is not like the Solar System: stars don't go around on nice nearly circular orbits, everything is a bit ...

15

Some numbers come from a review paper by Cullers (2000), who discusses the SETI Phoenix project. There, it is claimed that the Arecibo dish is capable of detecting a narrow band, coherent signal of $f=10^{-27}$ W/m$^2$ given a 1000 second observation. Assuming that this is an isotropic signal, then the implied power at distance $d$ is $p=4\pi d^2 f$, which ...

14

Martin Hoecker-Martinez's Answer is correct for perfectly noiseless observations of a two body Kepler system, i.e. the force between the bodies is directed along the vector linking them and the force magnitude follows and inverse square law with distance. An alternative to Martin's answer is that perfectly known position and velocity will determine all ...

12

With the atmosphere of the moon being $10^{14}$ times less dense than that of Earth, there is negligible scattering, so whereas on Earth, approximately 25% of direct solar radiation is scattered around (making the sky light up and appear blue), there is no mechanism for this on the moon, and all light from the sun travels (essentially) unaffected to the ...

12

Measuring $w$ is actually what I do for a living. The current best measurements put $w$ at $-1$ but with an uncertainty of $5\%$, so there's a little room for $w \ne -1$ models, but it's not big and getting smaller all the time. Indeed, we'd all be thrilled if, as measurements got more precise, $w \ne -1$ turns out to be the case because the $\Lambda$CDM ...

11

Generally speaking they refer to the distances from us when the light as emitted. No correction is usually made to say how far away the object is from us now, because this correction would be very small and inconsequential compared to the uncertainty in the original distance measurement. For instance, taking the Andromeda M31 galaxy as an example. Riess et ...

10

An ideal Kepplerian orbit is defined by six (6) parameters: angular momentum (3) total energy (1) Laplace–Runge–Lenz vector which is perpendicular to angular momentum (2) Therefore you need at least six (6) independent observations. Astronomical observations are direction (but not range usually) given by a pair (2) of angles therefore three (3) ...

8

To ask what do physicists expect to accomplish with gravitational lensing is nowadays somewhat like asking what do biologists expect to accomplish with looking at things with microscopes. Gravitational lensing is a well established method used across astronomy and the main challenges the field itself has to tackle are mainly technicalities. But I will try ...

7

To calibrate our expectations, consider the largest nuclear weapon ever detonated, the Tsar Bomba. It's yield was at most about $58$ megatons TNT equivalent, or about $2.43\times10^{24}$ erg. Now, let's consider a smallish star, something like Gliese 581, which is reasonably nearby, small and faint, and has a planetary system (of some sort: the number of ...

7

The atmosphere obscures data in three main ways; it absorbs light, it emits light in the infrared, and finally it diffracts light leading to distorted images. Observers have ways to deal with all three things, but I'll focus on the first two since they are more directly related to your question: 1) Atmospheric absorption. This plot gives a rough idea of ...

6

I've added this because I don't think the accepted answer is very clear. Estimating the number of stars in the Galaxy relies mostly on two things. We estimate the present day mass function (that is the number of stars that exist per unit mass per unit volume) in the solar neighbourhood. We construct a model for the overall density distribution of the ...

6

Here is an update on (my) answer that your refer to. I have changed the visual threshold to V<6.5 mag (which is what Sky and Telescope have done) and I have used the revised Hipparcos reduction from van Leeuwen (2007) to obtain a (almost) complete catalogue of stars with their trigonometric parallaxes. It contains 7892 stars. I am not going to investigate ...

6

Yes, it's possible. Take a look at the diagram below. You're looking at the Earth from above the north pole. The yellow arrows indicate the direction of light coming from the Sun. The times on Earth show approximate local time; see how the Sun is most "direct" at noon but is below the horizon late in the night? (I think those times are most accurate during ...

6

The practical detection limit for HST is about a visual magnitude of 30 - that sort of number was reached in the ultra-deep field. Assuming that the solar sail kept reasonably stationary for the 100 hours or so of required exposure then we could do a calculation based on that. There is absolutely no need to resolve the object in order to detect it. If you ...

6

This would ultimately be more a problem of signal processing than physics. The situation is detecting a signal at a very low signal to noise ratio. At the broadband level, the noise (starlight) is several orders of magnitude more intense than the signal (the explosion). The only hope would be some sort of spectral technique , taking advantage of spectral ...

6

That's not quite correct. You may have noticed that during summer the days are longer (and the nights shorter) than during winter. That is because the earth's axis is tilted about $23^o$ from the plane of it's orbit around the sun. With this tilt, as the earth travels around the sun the northern hemisphere gets longer days is the north pole is tilted ...

6

You are neglecting two important facts. The first one is that stars, toward the end of their lives, return to the interstellar medium (ISM) a lot of their initial mass, but now enriched with heavy elements produced by nuclear reactions inside the stars themselves. In this way, younger stars which form from the ISM begin their life with a larger fraction ...

6

In short, yes it completely makes sense to keep searching for Dark Matter using Earth-based direct-detection equipment. Even in the case that there is no significant amount of dark matter in the Solar vicinity, that is the only area we are currently able to search using direct-detection. So it makes sense that if we search for dark matter (and we should ...

5

A brief overview of stellar evolution can be depicted in the following image: (From here which says it is originally from an encyclopedia; click here for larger image). The heavier stars (top track) have very short life times (a few million years) because they run through hydrogen, helium, carbon+oxygen, ..., iron fusion in the core. Once a particular ...

5

Have a look at this article. It gives the number as $10^{24}$ rather than $10^{23}$, but it's such a vague estimate that a factor of ten is within the expected error. The number is the number of stars in the observable universe i.e. within 13.7 billion light years of Earth at the time the light we see today was emitted. Note that visible means visible to a ...

5

I did some research and I think I can answer my question myself now, after all. I hope you find it interesting. As it turns out, the technology of the Kepler space telescope would indeed allow detection of all Solar system planets except Mercury and probably Mars, i.e. all of them are big enough to be seen by it from a distance of about 2,000 ly. However, ...

5

SSC: synchrotron self-Compton BBC: Compton upscattered blackbody radiation; Compton upscattering of stellar blackbody photons XC: upscattering of photons emitted by the accretion flow; accretion flow photons From: http://arxiv.org/abs/1307.1309 and http://arxiv.org/abs/1403.4768

5

This paper contains an important analysis of the different trade-off between bandwidth and energy efficiency. The interesting conclusion from that paper is that the most energy-efficient way to send and receive interstellar messages (over flat spacetime) that maximise the bit-rate requires making the bandwidth of transmission very large. In particular, this ...

5

The angular resolution of a telescope is given approximately by $1.22 \lambda/D$ in radians, where $\lambda$ is the wavelength of observation and $D$ is the diameter of a circular mirror. Say we study a star that is 10pc away with a telescope working in the optical band (you didn't specify) at 500nm, then the spatial resolution of a 100m telescope at the ...

5

In addition to the lack of an atmosphere (the primary reason, already addressed in other answers), something that compounds the effect even further is that the lunar surface is quite reflective, and because the people who are taking the images of lunar surfaces often want to have Earth and/or astronauts in the shots and also be able to make out detail in the ...

5

$G$ was historically calculated from the Cavendish experiment, involving balls and a torsion balance. The earth's mass was actually calculated before the sun's mass. Using the assumption that the earth was a sphere, its circumference and thus its radius could be determined through geodesy, as was done historically even before Newton. The acceleration of an ...

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