gravity - Could I create a staircase to the moon and walk to it?

A space elevator has been proposed. With currently available technology it would be very expensive to build one, but not necessarily impossible.

Assuming the staircase or elevator being fixed at some point on Earth near the equator, the Earth's gravitational pull is compensated by centrifugal force at a height of about 36,000 km, in the geostationary orbit.

That's the height you would have to climb to, before you could slide "down" to the Moon along an even more expensive construction stepwise adjusting your velocity to that of the Moon. The total trip would be at least about 400,000 km, that's ten times around the Earth.

Your weight would decrease, the higher you climb from ground to the 36,000 km height. Assuming you make 3 km of height each day, it would take you about 34 years to reach geostationary orbit.

Then sliding down with 10 m/s would take at least one more year to reach the Moon. Hence, it's possible with a one-way trip of about 35 years, assumed the technical infrastructure is provided.

But it looks easier to take an appropriate vehicle.

n body simulations - How to scale down solar system data to simulatable values

Okay, I am seriously ashamed for asking this especialy when I geniualy study on physics but there is something that bugs me with the simulation I am working on.

I am re-creating the solar system in an n-body simulation that i programmed before. And I've a problem with scaling down the solar data. So even if I use kg and km for the metric units, the values are far bigger than the variables can hold in programming. Also as some of you know, bigger the value is, bigger the floating point error it makes. (error noise in data) Also it makes it slower to process.

I decided to scale down the data with a reference point, and for that, I took the earth's radius as 1 unit. And scaled down every other distance and radius according to it. (So a unit is 6371 km just to be clear)

But I am not sure whether if I should scale down the mass or not. My common sense says that I should scale down the mass so density of each body should remain same. So I took the density, and calculated a new mass value for each body, with the new scaled down radius.
But I am somehow couldn't conviced myself about If it's true or not. So here I am, asking to you :) Should I also scale down the mass?

PS.1: I used using F = GMm/r^2 equation for the calculactions as usual. (Iterating it through each body pairs)

If there are other programmers like me interested in making a simulation like this, how did you accomplish this data size problem? Are there any better solutions than scaling down the values?

PS. I have created an excel file that does the scale conversion. So I am sharing the sheet in OneDrive. (http://1drv.ms/1NIekGo) If you can check my calculations and values, that I'd also be really helpful to me. Thanks for any help.

meteorology - July/August hottest months (northern hemisphere): because of continued warming or other effects?

The amount of surface heating from the sun is a function of time of year and latitude. You are aware of the dependence on the time of year with the sun over the equator at the equinoxes and ~ $pm$23$^circ$ N at the solstices. The varying position of the sun overhead means the projection of a solid angle from the sun onto the surface of the earth will vary as well. This just means that when the sun is low, the energy in a "beam" of sunlight is spread over a very large area and warming will be reduced. This is why a bright sunny day in the winter cannot provide the same heating as the bright sunny day in the summer.

Surface energy loss is a function of the temperature of the surface, and atmospheric energy loss is a function of the local atmospheric temperature. There are some complications here, namely clouds and greenhouse gases that prevent some radiative losses.

At the summer solstice when days are long and the sun is at its highest it is easy to understand that we are getting more energy than we are losing, and the days should be warm. What isn't so apparent is that after the days start shortening and the sun lowering in the sky that the days continue to warm. This is simply because we are still getting more energy than we are losing. It isn't until the energy losses (which happen 24 hours a day, regardless of day/night) finally overcome our energy gains (which only happen when the sun is out) that the days will start to cool.

_{From www.atmos.washington.edu}

This image is from the northern hemisphere, but it is unclear what latitude it is for. The summer solstice is the maximum in solar insolation and the winter solstice the minimum. Radiative losses increase with temperature, and it is the point at which temperature has risen enough such that the loss is equal to the (diminishing) solar input. This is where the maximum temperature will occur. The lag between the solstice and the max temperature will vary with latitude.

As mentioned in a comment in the other answer, large bodies of water exaggerate this seasonal lag in temperature. This is for the same reasons as for land, with the added effect of the large heat capacity of water (about 3x that of land).

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You see the same effect in daytime maximum and minimum temperatures where (barring anything but solar influence) the daily max is often in the late afternoon and the daily min is often just before sunrise. This is the same reasoning as above, just on a daily scale rather than seasonal. The sun is overhead at solar noon (this varies with local noon based on longitude, time zone (daylight saving time)) but continues to provide solar heating after solar noon. It isn't until the radiative losses overcome the radiative gains (afternoon to late afternoon) that the temperature will drop. Likewise when the sun goes down temperatures will cool down until the sum comes back up.

_{From www.vsc.edu}

exoplanet - Conditions on Hot Jupiters

I think we can broadly distinguish two classes of effects here:

• Flares directly affecting the planet


• The by-products of the flare affecting the planet

I can loosely tell you what I know (although my knowledge here is far from being complete)

Direct effects include:

• Magnetic field geometries: Far-away from their host star (>0.05AU) planets usually have isolated dipoles. The solar wind then must penetrate the magnetosphere of a planet to erode its atmosphere. This is however much easier for the wind if due to small distance both star and planet-magnetospheres connect. What this will lead to, is quasi-free flow of a significant portion of the stars total ejected mass towards the planet and huge energy input into the planet's upper atmosphere.

Indirect physics can be:

An active young star can have frequent outbreaks of CMEs, those are accompanied by strong UV / X-ray emission and this will affect the planet:

• Increased photochemistry: This can generate a very active photochemistry in the upper atmospheres. Doubly ionized molecules like $O_2^{++}$ or $N_2^{++}$ can then be produced more efficiently and eject their constituents into space upon dissociation due to high dissociation energies


• UV-X-heating: Additional heat through thermalized high-energy photons can add to the already happening 'natural' atmospheric erosion.

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We don't know however, to which extent those effects play a role. I'd add tomorrow some references if you're interested.

There could be signs however of such activity in young stellar systems, as verious combinations or extensions of the above have been proposed to explain the radius anomaly in super-close extrasolar giant planets.

universe - Is it possible to be 'still' in space?

The so called expansion of the universe is not as trivial as most people think. What is happening, in fact, is that the distance between two points in space (note that I'm not talking about objects with velocities, but just coordinates in space) increases with time in a manner proportional to a given factor (in this case, the Hubble constant - which is actually not a constant in time, but let's ignore that for now).

This means that galaxies are not exactly moving away from a specific point with a particular velocity (i.e., a particular rest frame), but that the space between them is expanding. It is somewhat hard to wrap your head around the difference between these two situations, but this has to do with the geometry of space, and not with what's inside it.

It is widely accepted that the expansion of the universe is homogeneous and isotropic (the Cosmological Principle), which means that there is no "special" position (rest frame) on the universe, and whatever your velocity and your location, the universe will seem to expand the same way.

The best way to imagine this is to reduce to 2 dimensions expanding in a third dimension. For example, suppose that you live in a galaxy embedded on the surface (a 2D universe) of a balloon (completely unaware of the third dimension), with other galaxies distributed homogeneously. If, for some reason, the balloon is expanding isotropically and, every point of the surface will be moving away from every other point, so will every galaxy from each other. Think about this: is it possible to find a special, "still", point (without resorting to the elusive third dimension) where you will see galaxies moving from you in one side and galaxies moving towards you on the other side?

Our universe works similarly, but we're embedded on a three-dimensional space instead of a two-dimensional one.

the sun - Sun path at poles

At places other than the poles, the sun is seen to "rise" on the Eastern horizon and to "set" on the Western horizon. After a period of night, the sun "reappears" on the Eastern horizon.

At the poles, during the short periods of 24-hour daylight, what does the sun path look like? Where does the sun move after it has reached its Western most position? Back Eastwards along the same path in reverse?

Less stars in the night, compared to 15-20 years back

In my own case, if I see fewer stars than I did when I was younger, it's partly because my eyes have deteriorated with age. I don't know whether that applies to you.

The most likely explanation is increased light pollution. If you can manage to get to an area far enough away from city lights, on a cloudless and moonless night, you should see just as many stars as you did when you were younger.

Stars are dead?

No. All the stars you can see in the night sky are within a few hundred light-years of Earth. That means that you're seeing them as they were no more than a few hundred years ago. Most stars live for billions of years; the Sun, for example, is about 5 billion years old and is expected to live for another 5 billion or so years in essentially its current form. Some of the brighter stars have shorter lifespans (because they're larger and "burn" their nuclear fuel more quickly), but even they have lifespans of at least millions of years.

No naked-eye visible stars have "died" in the last several centuries. We do sometimes see stars explode (as novas or, more rarely, as supernovas), but I don't believe any of the stars that have done this were naked-eye visible before they exploded.

Current position of the earth in galaxy

No. The stars do move relative to each other, but not quickly enough for the motion to be visible over a human lifetime. Barnard's Star has the fastest proper motion of any star in the sky, but it only moves at about 10 arcseconds per year; it would take it nearly 2 centuries to move the width of a full Moon. And Barnard's star isn't even naked-eye visible. The stars you see in the night sky are in very nearly the same apparent positions as they were thousands of years ago.

Is it possible for a moon to orbit a planet floating free in the galaxy rather than orbiting a star

The answer is Yes.. Planets that don't orbit around a star are known as Rogue Planets. There is nothing preventing a rogue planet from having one or many moons.

Not so long ago, the first candidate for a free-floating exoplanet-exomoon system was presented in this paper. It looks like a gas giant several times larger than Jupiter with a sub-Earth mass moon.

Another study calculated and simulated scenarios where planets where ejected from their orbits around a star and concluded that around five percent of Earth-sized planets who are accompanied by Moon-sized natural satellites would retain them after the event.

Why are radio telescopes shaped so differently than optical telescopes?

They're not different. Same principles do apply. You could have secondary, tertiary, quaternary, and so on, mirrors with instruments at any wavelength, either optical, or radio, or infrared, etc. You could also have instrumentation placed directly in prime focus (so no mirrors other than the primary) with any kind of instrument - radio or infrared or visible or whatever.

See this image of the 5 meter Hale telescope on Mt. Palomar - there is no secondary mirror in this case, the observer is sitting in a little cage at prime focus, using the primary mirror directly:

Of course, for other scenarios, the Hale telescope employs secondary and tertiary mirrors - it depends on the particulars of the telescope, the instrumentation, the experiment or research you're doing, etc.

One reason many of the large optical telescopes very often have at least a secondary mirror is that the architecture preferred in most of these cases is the Ritchey–Chrétien - chosen often for the largest professional telescopes because it eliminates coma, an aberration that is detrimental to astrometry (with coma, images of stars are not round, so it's hard to measure angular distances between them). You could use the primary mirror of such a telescope directly, sure, but being a concave hyperbolic mirror, it has strong aberrations of its own, and so requires the convex hyperbolic secondary (often a strong hyperbola, with a large eccentricity) to correct the aberrations.

The Hale telescope pictured above has a parabolic primary, so using it directly is not a problem.

Again, all of the above are not strict rules, just statistical observations.

Some radio telescopes have instrumentation at prime focus simply because it's convenient for that particular case. Other radio telescopes have secondary mirrors. Again, it all depends on what you try to achieve. E.g., the Arecibo radio telescope could be used either in prime focus, or with a secondary mirror in a Gregorian configuration - here's the image with the prime focus instrumentation and the Gregorian mirror to the left:

In the case of the Arecibo scope, the N-ary mirrors are sometimes used to correct the aberrationa of the spherical primary reflector, but that's not the only reason why they're used.

Here's a discussion comparing various architectures (classic Cassegrain versus Ritchey-Chrétien versus anastigmatic aplanat) for a large radio telescope, highlighting various design, performance and operational issues for each. TLDR: classic Cassegrain is traditional for radio telescopes, but the R-C architecture performs better and is not significantly more difficult to build; OTOH, with R-C you must always use the secondary.

How can I create tracks of the planet's orbits on my computer?

I'm wondering if there's any good computer software for tracking the orbits of planets over time. I like to see the orbit's trails against the sky background.

I'd prefer something that's free and of a good quality.

I've already tried Google Earth, but it doesn't seem to be able to do this.

star - What do we know about the characteristics of stellar objects, 13,3 billion light years away from us?

Holger,

I highly recommend you check out the NASA story on the discovery MACS0647-JD, which includes many dimensional facts (mass, distance, age, etc.) and comparisons:

http://hubblesite.org/newscenter/archive/releases/2012/36/full/

From the article:

The estimated total mass of the stars in this baby galaxy is roughly
equal to 100 million or a billion suns, or about 0.1 percent to 1
percent the mass of our Milky Way's stars.

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Measuring the mass of a galaxy is very tricky indeed (especially one as far as 13.3 billion light-years from earth!), but there are a couple different methods that can one can use, see here:

PSU Astro Measuring Mass of Clusters

And this Physics Stackexchange thread as well:

http://physics.stackexchange.com/questions/123/how-does-one-measure-the-mass-of-a-galaxy-and-other-such-large-quantities

habitable zone - Could there be Earth-like planets in binary or trinary systems?

Firstly, a few planets have been discovered around binary or trinary systems, mostly Neptune sized or larger. An example is are the two planets described in the NASA article "Kepler-47: Our First Binary Star 2-Planet System", where they are described as:

Now Kepler mission has discovered Kepler-47b and 47c, the first transiting circumbinary system — multiple planets orbiting two suns. To compound the excitement of the discovery, one of those planets is in the binary system's habitable zone (where liquid water may exist)

(A representation of the Kepler 47 system is pictured below)

The second planet discovered in the system, Kepler 47c, is within the binary stars' habitable zone, but is slightly larger than Neptune and is likely to be a gas giant - but that raises the potential for terrestrial satellites.

Another example is the Saturn sized Kepler 16b reported in the article "Weird Exoplanet Discovered Orbiting Two Stars" (Klotz, 2011), which orbits 2 stars that are smaller than our sun, thus has habitable zones closer to their parent stars, in this case, the equivalent of the Venus' orbit.

In terms of the habitable zones around multiple star systems, the article "Habitable Binary Star Systems" suggests a major advantage of a binary (indeed a trinary) system is that the combined energy between the stars could indeed extend the habitable zones, particularly,

Low-mass twins could make the best hosts, because their combined energy extends the habitable region farther away than would exist around a single star.

However, temperatures would vary significantly and the orbits would be irregular, and in simulations reported in "Double-Star Systems Can Be Dangerous for Exoplanets" (Wall, 2013) results in such disruptions that one or more of the planets therein could be ejected and sent hurtling into interstellar space. But such perturbations are likely to take a considerable amount of time (millions of years) to fully occur.

This is also discussed in "Exoplanets Bouncing Between Binary Stars" (Moeckel and Veras, 2012), where the planet in such as system is described as 'bouncing' between the binary stars, then resuming a relatively regular orbit.

The recent discovery of a terrestrial-like planet in the Alpha Centauri system, as reported in "Discovery! Earth-Size Alien Planet at Alpha Centauri Is Closest Ever Seen" (Wall, 2012) - despite this planet not being in the habitable zones of either star, this does raise the hopes for the possibility of Earth-like terrestrial planets existing in the habitable zones of other multiple star systems.

There is also the possibility that an Earth sized moon could orbit the larger planets such as those in the Kepler systems mentioned earlier.

infinite Universe - Astronomy

I think you're operating under a few false presumptions, the first being that that all infinities are instantaneous, which they are not!

Imagine I told you to start counting at 1:00 PM tomorrow, starting at the number one, and then to keep counting, once a second, until I say stop.

So at 1:00 PM you'd start: 1...2...3...4...

At 1:05 PM you'd be at: 300...301...302...

And at 1:10 PM you'd be at: 600...601..602...

Ok now I tell you to keep counting to infinity, so that you never stop counting! Eventually you'll reach every number you can think of (and even more, since you can always count another second after the one you just got to), but it might take you weeks, or months, or centuries to get to there!

So in that example we had a start, 1, and we're going all the way to infinity (i.e. we'll never stop). In the same way, the universe could start at one time and keep expanding forever (the only difference is that the universe "starts counting" at 0 instead of 1).

But why 0 and not 1? Or 2? Or -356? Or even -infinity? Why did the universe start 13.8 billion years ago? What about the time before the big bang?

This is a very, very tricky question (and one which may never be answered), but most cosmologists will tell you that there was no time before the Big Bang (since there was no space!), and even if there was, we probably wouldn't be able to probe it.

I will suggest you check out Sean Carroll's blog with a few great explanations to these kind of questions:

http://preposterousuniverse.com/writings/cosmologyprimer/faq.html

space - How do we get radio signals of the big bang?

The best estimates at the moment are that the Universe is "flat". By that we mean that a pair of parallel laser beams will travel as far as you like and they will always remain the same distance apart and parallel.

Ironically, the best evidence for this comes from detailed measurements of the cosmic microwave background by the WMAP and now Planck satellites.

The problem you are having is you (I think) are imagining that the big bang happened at a point in space 13.7 billion years ago, and that radiation has been travelling outwards from that point. This is incorrect. Every point in space that we see now was, about 13.7 billion years ago, part of the big bang. As the space in the Universe expanded, its contents cooled and due to the trapping of free electrons by positively charged protons to form hydrogen atoms, the light that was emitted by the hot gas everywhere in the Universe at a time about 400,000 years after the big-bang was able to travel unabsorbed and unscattered. This radiation has travelled (at the speed of light) in all directions.

Now, when we observed the microwave background, we are seeing light that has been travelling in a straight line for just short of 13.7 billion years. That light was in the near infrared and visible part of the spectrum when it was emitted, but because of the expansion of space, the wavelength has also been stretched by a factor of 1100 to the microwave/short-wave radio part of the spectrum.

How many earths fit in the observable universe?

Without checking the numbers in detail, according to Wikipedia, the volume of the observable universe is about $3.5cdot 10^{80} mbox{ m}^3$, and the volume of Earth is about
$1.08321cdot 10^{21} mbox{ m}^3$.

By dividing the two volumes we get a factor of $3.2cdot 10^{59}$, or written as decimal number: The observable comoving volume of the universe is about
320,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000-times the volume of Earth.

Why haven't Earth and Venus got any tiny moons? Or have they?

@adrianmcmenamin Has the right idea here. These objects he is referring to are called Trojans, and are defined to be:

a minor planet or natural satellite (moon) that shares an orbit with a planet or larger moon, but does not collide with it because it orbits around one of the two Lagrangian points of stability (trojan points), L4 and L5, which lie approximately 60° ahead of and behind the larger body, respectively.

Near to each massive body in the solar system (Earth included) are minima, maxima, and saddle points in the potential energy landscape due to the combination of said object and the Sun.

It is possible for objects to essentially get caught in the more stable of these extrema. Jupiter, being the most massive, has quite a number of these types of objects.

However, one such object has been discovered orbiting around a stable Lagrange point in 2010, called 2010 TK7.

star systems - Potential Re-Capture of Rogue Planets

For a start, the hot jupiters (HJ) are not similar to the planets of the Solar system (they are much hotter and very close to their host star). It's essentially impossible for HJs to be ejected away from their star's sphere of influence. The reason is that they are gravitationally rather tightly bound to their star, so that any gravitational interaction with a third body that would gain them enough energy for ejection needs to be closer than is possible without it leading to a collision with that third body.

So rogue planets that didn't form independent of a star must have been much less bound to their parent star, similar to the outer planets of the Solar system. Of course, once ejected (free-floating), such an object can in principle be re-captured into another planetary system or a binary system. However, this is quite unlikely, since the planet must move with a velocity very close to that of its capturer. Also, it must involve a third body (either another planet or star bound to its capturer) to absorb (into its orbit) the orbital energy (and angular momentum) released by the capture.

The only possibility is that the planet was ejected very early on, when the stellar association / group / open cluster within which it was born had not yet been dissolved. Then it may find another home with one of the other members of that association.

I don't think that anybody has estimated the actual probabilities for this to happen, but why are you interested?

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As a side remark, it seems clear that orbiting planets form from / in gas discs surrounding protostars, but it's not clear whether brown dwarves cannot also form this way. Whether or not an object formed this way becomes a gas planet, a brown dwarf, or even a star depends entirely on its mass and not the formation process.

What limits the usable focal length of telescopes currently?

Comparing telescopes that observe the visible spectrum to the radio spectrum, radio astronomers have been able to create telescopes with apertures of the order of kms, thanks to aperture synthesis. This is extremely hard in optical telescopes and the only telescope (afaik) that does so is the Large Binocular Telescope. The reason this is possible in radio astronomy is because we can measure the phase of the incoming wave using radio telescopes where as information about the phase is not captured by optical telescopes. Maybe in the future, technology will help us make optical detectors that can measure the phase of the wave.

coming to the size of the aperture itself, larger and larger sizes doesn't help as long as we don't account for the atmospheric seeing. the reason stars twinkle is because of atmospheric seeing. effects seeing can be negated using adaptive and active optics and advancement of these technologies will help astronomy move ahead.

coming to the actual detectors, the intrinsic noise from radio detectors (eg. bolometers) is much smaller than that in optical detectors (eg. CCDs). so again, maybe in the future, we have better detectors with extremely low noise.

(sorry coulnd't add more links. needed more rep :D)

earth - Could the Philae comet lander be recharged by laser?

I only have time for a partial answer, so I'll discuss the bit about rotating the comet.

We have to do some calculations with radiation pressure, the force on an object from light.$^1$ This pdf has some good starting calculations, although they assume that the source is spherical - the Sun. Our laser clearly is not. However, the beam will diverge over such a long distance, so I'll have to use a substitution (given by CountIblis).

The force due to radiation pressure is
$$F_{text{rad}}=frac{2I}{c}A$$
where $I$ is the intensity, $c$ is the ubiquitous speed of light, and $A$ is the area onto which the force is applied. However,
$$I=frac{text{power}}{4 pi r^2}$$
so we get
$$F_{text{rad}}=frac{text{power}times A}{2 pi r^2c}$$

The $4 pi$ has to be replaced (by CountIblis' substitution) with $$Omega=pi left(frac{alpha}{2}right)^2$$
For an ideal laser, $$alpha=2.44frac{lambda}{d}$$
where $lambda$ is the wavelength and $d$ is the initial beam diameter. Let's assume that our laser has a power output of one megawatt - the COIL laser, from a Boeing YAL-1, as per Randall Munroe's idea. $1,000,000 text{ watts}$ is a lot from a laser. Let's also assume that no power is lost.

For COIL, $lambda=1.315 times 10^{-6}$ and $d=0.1016 text{ meters}$. This gives an $alpha$ of $3.158070866 times 10^{-5}$, and thus an $Omega$ of $7.83309915 times 10^{-10}$.

Philae is currently on comet 67P/Churyumov-Gerasimenko, which has, on its longest side, a surface area (combined lobes) of about $19,370,000 text{ meters}$. However, it's about $310,000,000 text{ miles}$ from Earth, which is $4.9879 times 10^{11} text{ meters}$. Knowing that, to rotate the comet, we have to apply force on only half the area of one side, we find that this laser would apply a force of
$$F_{text{rad}}=frac{1,000,000 text{ watts}times 9,685,000text{ meters}^2}{7.83309915 times 10^{-10} times 2.487914641 times 10^{23} text{ meters} times 299,792,458 text{ meters/second}}$$
$$=1.65771483 times 10^{-10} text{ Newtons}$$

Yep, we're not rotating a comet any time soon. I'll try to add in the bit about recharging Philae by laser sometime later. I do think that will be a lot more plausible.

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$^1$ Unfortunately, I can't find any calculations on laser propulsion.