# Tag Info

2

Simply put, it is: how many aliens could we meet? More specifically: The Drake Equation is a way of predicting how many intelligent species there might be in the universe and the likelihood of them contacting us. There are a lot of things that can change the number of aliens we can expect to find. So we use what we know about the universe so far and ...

2

I will simplify this problem by assuming that the only forces come from Newtonian gravity and by limiting the masses of the moons to much smaller values than that of the planets, such that the moons exert a much smaller gravitational forces on all other bodies and therefore can be neglected; so the only sources of gravitational forces are the two planets. ...

0

I suspect that they were lucky that their predictions agreed with reality so closely, but any prediction was going to have Neptune roughly (perhaps very roughly) in the same direction as Uranus, during the times when it affects Uranus the most. So I suspect their calculations meaningfully ruled out large swathes of sky, which improved odds of finding it.

4

Although this may not be what you're looking for... They weren't "simply lucky." In fact, they didn't use Bode's law at all- they used calculations based on Neptune's supposed gravitational effect on Uranus. In fact, had the two used Bode's law, they would never have found Neptune, as the Bode "law" would predict a completely different location. (This is ...

52

I've made this into an answer because it's too long for a comment, and I really want to show the pictures. It is tempting to think of visible light as "close enough" to (near by wavelengths) and to conclude that "yes, actually, the yellow does affect it. I want a mirror without an obvious tint" However you are wrong, Physics will slap you down. Exhibit A ...

46

If you look at the reflectivity of gold (vs silver or aluminum) you can see a plateau at wavelengths below 500 nm source: If blue wavelengths are not reflected as well as other colors, the resulting image will look "more yellow" - which is what you see. At longer wavelengths, gold is a very good reflector (better than the other two above 600 nm). It also ...

2

There is a lot more to it than just astronomy. For example, the tide times inside Boston Harbor are significantly different from those on the southeast coast of Cape Cod. It is true that the primary force behind tides is the position of the Moon, but the macro tidal bulges take a long time to propagate around/across oceans, and then the shoreline shape ...

0

According to here, there was no precise definition before this group redefined what it meant for a group of galaxies to constitute a supercluster--before their redefinition, it seems it was just loosely defined as "extended regions with a high concentration of galaxies." They now define a supercluster to be a volume in which "the motions of galaxies are ...

1

You can check The Best Cheap Budget Telescopes Under $200; this presents the latest list of all affordable telescopes that are competent for beginners. Also you can check The cheap telescopes of 2014. Remember always that though magnification is good, but you must need to have a good apparture in your device & for that I prefer Newtonian telescopes or in ... 3 The current data release is DR12. As to how to interact with the data, it depends somewhat on what you want to do. If you have small queries, for instance if you want data for only a handful of objects, there are a couple of web interfaces. If you have larger queries in mind, CASJobs is a good place to start. You'll inevitably need the schema browser to know ... 4 Both From Wikipedia Luminance Versus Luminosity In astronomy, luminosity is the total amount of energy emitted by a star, galaxy, or other astronomical object per unit time. It is related to the brightness, which is the luminosity of an object in a given spectral region. Now your question: Let's assume we have two stars that have the same surface ... 1 Due to the force of gravity, which goes as the inverse of the square, planets trace out an ellipse in space as they orbit around the sun, which is located at a single focus. The other focus is unphysical. Actually, given two massive bodies, their "difference" vector will trace out an ellipse with the center of mass at the focus. Because the sun is so much ... 0 There is no physical object at the location of the second focus. Newton showed that an elliptical path was the consequence of an inverse square radial force from a fixed point. While you can identify the point that is the second focus, nothing associated with that point is required to create the elliptical motion. Deriving Kepler's Laws from ... 2 You will see Earth, Venus, JupiTer, Saturn. TheY are either bright enough, or come closer to Mars than Earth, that there are no complications/calculations necessary. Uranus is just a naked eye object from Earth (magnitude 5.3-5.9). Its closest approach to Earth, when it is brightest is 17.2 au distant. Mars' orbit takes it closer by about 0.5 au, so you ... 0 From numerical simulations, I would say that this is method (adjusting the nose to point to the star at all times): converges (you reach the star) if the star does not go too fast (the spaceship needs to go faster than the star on its orbit), does not look like it is the shortest path in general, because there is no anticipation of where the star will be, ... 0 Adding to the answer of the question 3 (How did arms of the spiral galaxies form?) one of the theories suggest an encounter between galaxies. This simulation I found in other answer "shows the influence of a nearby galaxy causing the imbalance which triggers arm formation" and could be interesting to see. 1 Even if your ship traveled at (almost) the speed of light, you would have to follow the trajectory of a ray of light traveling from your point of origin to the star; not from the star to your point of origin. As described, you keep following the trajectories of many different rays of light (emitted at different times) backwards, putting yourself on a curved ... 0 It's often not the quickest because you aren't taking into account how the star was moving when the light was sent. It is similar to a temperature controller. You can adjust the temperature based on the current temperature (point), the integral of the temperature (integral) and how the temperature is changing (derivative) and using all three (a P.I.D. ... 0 Due to the gravitational pull from other planets and suns it may be the shortest in terms of distance, but not necessarily the quickest. It would really depend on the star you are travelling to. 0 If the photon you described is going to move from A to B, there's no reason you can see it at position C. Because if you were able to see the photon, it had moved from A to C, where you are, not B. Therefore, you are not able to see the path light travels through, unless, as others said, something like an atom deflects it. In all the answers above, something ... 1 Under the conditions stipulated, the question is a false statement. We would, in fact, be able to "see" the light beam between points A & B. As the photons travel from point A to B, some photons will be deflected (by colliding with the dust particles) in our direction, allowing us to "see" the beam. Only in the absence of dust particles (or any other ... 0 To expand on this, you would point your telescope at an empty patch of sky (as mentioned). You expect the CMB to be thermal to a very high degree ($\sim 1$part in$10^5\$). Also, you can point your telescope to any direction on the sky and get (statistically) the same observation (if you have the resolution to pick up the primordial fluctuations that are ...

2

The instrument should be pointing at an empty ( no stars no galaxies) region of the sky and be able to record very low frequencies Other radiation comes in much higher frequencies from stars , and would not overlap with the low frequency part. >Discovered accidentally in 1964 by Penzias and Wilson (Nobel Prize, 1978), the CMB is a remnant of the hot, ...

10

The classic color mapping for Hubble is described in Flase-color astrophotography explained. What you have is (in the Hubble palette): Line Freq True False Ha (656.3 nm) Red -- Green S-II (672.4 nm) Red -- Red O-III (500.7 nm) Green -- Blue An example of this for true color from John Nassr at Stardust Observatory at Coming to Life ...

6

I find myself now answering my own question but only because the comment feature is not suited to this "comment". I have selected the answer by @HDE 226868 as my answer and primarily due to the linked Space.com reference. Very good answer to my question. In particular, I also thought this quote from the same page as being important as these reasons (below ...

24

Sort of. As Space.com writes, The raw Hubble images, as beamed down from the telescope itself, are black and white. But each image is captured using three different filters: red, green and blue. The Hubble imaging team combines those three images into one, in a Technicolor process pioneered in the 1930s. (The same process occurs in digital SLRs, except ...

0

It depends on, what consider we significant. On our current best hypothesis, the moon was created by a collision of the Earth and a Mars-sized planet. Thus, the core of the Moon contains much fewer heavy elements, and thus also much fewer Uranium and Thorium. This corresponds also the fact, that themean density of the Moon is only 3.3, while the Earths is ...

Top 50 recent answers are included