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Science and Astronomy Questions
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| Watsisname | Date: Sunday, 23.10.2016, 12:12 | Message # 886 |
 Galaxy Architect
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| The (probably disappointing) answer is that you would have to simulate it with a model. This is what researchers actually do to study the atmospheric circulation on various exoplanets. You can read a bit into what this process involves on wikipedia:
https://en.wikipedia.org/wiki/General_circulation_model
And an example of this kind of work for a tidally-locked exoplanet can be found here:
https://arxiv.org/pdf/1105.4065v4.pdf
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| PlutonianEmpire | Date: Tuesday, 25.10.2016, 07:34 | Message # 887 |
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| Wow, that actually is a lot! 
Definitely not something I can do on a home PC now, isn't it? 
Specs: Dell Inspiron 5547 (Laptop); 8 gigabytes of RAM; Processor: Intel® Core™ i5-4210U CPU @ 1.70GHz (4 CPUs), ~2.4GHz; Operating System: Windows 7 Home Premium 64-bit; Graphics: Intel® HD Graphics 4400 (That's all there is :( )
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| Watsisname | Date: Wednesday, 26.10.2016, 07:32 | Message # 888 |
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Galaxy Architect
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| Yeah, anything involving the Navier–Stokes equations can get complicated in a hurry.
However, I think I might have been too quick to conclude that modelling it would be completely necessary. It is necessary to predict all the features of circulation with precision and confidence, but we can still get some insight to how many cells it is broken up into by the planet's rotation.
On a non-rotating planet (with equatorial heating, which is rather contradictory, but useful to visualize the physics), you get the classic Hadley cell. Warm air rises at the equator, which produces low pressure and a return flow from the poles.
The Coriolis force on a rotating planet causes these flows to be deflected. Following a simple argument of conservation of angular momentum (see §5.3 in Showman et. al -- this is also a really good paper to review for all sorts of information related to this topic) we can find an expression for the zonal (east-west) winds as a function of latitude:
Don't worry, I'll turn this into something sensible and usable.
This formula says the zonal winds on Earth at 30° latitude would be 134m/s, and even faster closer to the poles, which is unrealistic. The flow would break up into turbulence. In fact, we observe that the Hadley cell is broken up at approximately 30° intervals, so we have 3 cells in each hemisphere.
Let's use the predicted wind speed at the latitude for which we observe the cell structure get broken on Earth as a way to tune this model to predict the number of cells on other planets. This is an incredibly sloppy ad-hoc way to make a model, but as we will see, it happens to work reasonably well in many cases. There will definitely be exceptions though, so we'll need to make note of where and why it could fail.
Reworking the formula to find the number of cells (per hemisphere), I get the following:
where
N is the number of cells,
T is the planet's rotation period,
a is the planet's radius
u is the "critical zonal wind speed" that, for Earth, we computed to be 134m/s.
Okay, let's see how well this works. First, just to confirm it gives us 3 cells on Earth:
x(Earth) = (24 hours)*(3600s/hr)/(2*pi*6371000m)*134m/s = 0.2889.
Plugging 0.2889 in for x, we get N = 2.9989 (this is 3, we just got error from rounding).
Now for Jupiter:
Jupiter rotates once in 9.925 hours (35730 seconds) and has an average radius of 6.991*107 meters. Then x is 0.0109, and we get N = 15.06. So this predicts 15 belts and zones on each hemisphere of Jupiter.
Here's what we actually observe:
So this is not exact, but it seems pretty close. Definitely within a factor of 2.
Let's try Venus. We know Venus has a very slow rotation and it has only one Hadley cell.
x(Venus) = 74
N = 1.01
So far, so good! But we should take it only as a rule of thumb which is generally close to correct, not a precise universal fact. Indeed, it fails for Mars (it predicts two cells, but really there's just one.) This is because Mars has a very thin atmosphere which is quick to respond to changes in radiation, and a lot of the mass of the atmosphere is exchanged with the ice caps. And for other planets in general, the reality may differ for these and other reasons. This is why the global circulation model is really required to get a complete handle on what happens in the atmosphere.
I hope that's helpful!  For convenience, you can copy the following formulas into wolfram alpha:
To find x: (rotation period of planet)/(2*pi*(radius of planet))*134m/s
where rotation period should be in seconds and radius of planet should be in meters,
and for finding N: pi/(4arctan(sqrt((sqrt(x^2+4)-2)/(x)))) for x=[insert number]
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| Watsisname | Date: Thursday, 27.10.2016, 01:40 | Message # 889 |
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Galaxy Architect
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| Finally, an article (or at least a guest blog) which actually understood the significance of the recent paper about deriving cosmological parameters from supernovae data, unlike nearly every other pop-sci article on the internet which lead readers to think the universe isn't accelerating or that dark energy doesn't exist. Thumbs up to Scientific American for good science reporting. 
https://blogs.scientificamerican.com/guest-b....t-exist
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| Alek | Date: Thursday, 27.10.2016, 04:08 | Message # 890 |
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| Quote Watsisname (  ) Finally, an article (or at least a guest blog) which actually understood the significance of the recent paper about deriving cosmological parameters from supernovae data, unlike nearly every other pop-sci article on the internet which lead readers to think the universe isn't accelerating or that dark energy doesn't exist. Thumbs up to Scientific American for good science reporting.
I don't remember where, but I read an article that states how this "uncertainty" came about and did so in a non misleading way (aka not leading you to believe scientists have thrown out the accelerating universe model, since they explained that the observational data is technically within the margin of error but the measurements agree enough that the theory can't simply be considered rubbish and random chance/noise)
Living among the stars, I find my way. I grow in strength through knowledge of the space I occupy, until I become the ruler of my own interstellar empire of sorts. Though The world was made for the day, I was made for the night, and thus, the universe itself is within my destiny.
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| Watsisname | Date: Thursday, 27.10.2016, 23:02 | Message # 891 |
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| Yeah, that's quite a bit better than a large number of articles and online discussions that I have seen. But what I had not seen anywhere is a discussion of how the supernovae data fit in with the data from the other two methods (CMB and BAO). If we take the new supernovae data on their own, then we get 3 sigma [99.7%] confidence that there is dark energy and the universe's expansion is accelerating. With all methods combined, we get greater than 5 sigma confidence [99.99999%] .
Basically, BAO data do not constrain the amount of dark energy at all, but they do tell us that the matter density cannot be much less than 20% of the critical density, or much greater than 40%. CMB data tell us that the density of dark energy and matter together must add to about 100% of the critical density. (Equal to 100% if dark energy is 70%, a bit less than 100% if there is more dark energy, and a bit more than 100% if there is less dark energy).
Note: it is possible for a universe to have more or less than 100% of the critical density. Having a different density changes the spatial curvature, which means the rules of geometry are different from those of Euclid.
Put together, these data exclude all but a very small range of possibilities. Dark energy must be around 70% the mass density of the universe, and matter must be about 30%, to give a sum of 100% which makes the universe spatially flat. The regular matter that we can see is only about 5%, so that gives about 25% dark matter.
What's curious is how the new study arrived at uncertainty ellipses which are larger than previous studies, despite using more data. I'm not sure if this is an effect of their methods for error analysis, or a quality of the supernovae added to the sample. Even more curious is that their whole distribution is moved down and to the right by quite a bit. It disagrees with the overlap of CMB and BAO data by 2 sigma. Which isn't terrible, but it does beg for explanation. Is the new study off, or are CMB and/or BAO data off? Perhaps even all three! This is what cosmologists need to focus on.
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| PlutonianEmpire | Date: Saturday, 29.10.2016, 03:26 | Message # 892 |
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| Im watching this infotainment show depicts various doomsday scenarios, and this one is talking about GRB's. The event in question in this episode is a star-merger GRB. From the point of the merger event, how wide is a GRB ray, in terms of degrees in a circle? Is it always the exact same width for each merger GRB? And do supernova GRB's differ from merger GRB's in any way at all?
Edit: Watsisname, thank you for the formulas, I'll give them a try!
Specs: Dell Inspiron 5547 (Laptop); 8 gigabytes of RAM; Processor: Intel® Core™ i5-4210U CPU @ 1.70GHz (4 CPUs), ~2.4GHz; Operating System: Windows 7 Home Premium 64-bit; Graphics: Intel® HD Graphics 4400 (That's all there is :( )
Edited by PlutonianEmpire - Saturday, 29.10.2016, 03:56 |
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| Watsisname | Date: Saturday, 29.10.2016, 11:44 | Message # 893 |
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Galaxy Architect
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| Quote PlutonianEmpire ( ) From the point of the merger event, how wide is a GRB ray, in terms of degrees in a circle?
Probably not much less than a degree, or much more than ~20 degrees. Most likely toward the lower end of that range. The angle is very model dependent, and it almost certainly depends on what causes the GRB as well.
I should probably talk a little more about that, and the history of GRB research. GRBs were first detected in the 1960s (by classified surveillance against nuclear detonations), but it wasn't until the 90s with a dedicated gamma-ray observatory that their study really took off.
It did not take very long to realize that they were so far away -- and therefore so bright -- that they can't be radiating their energy isotropically (equally in all directions). To provide that much luminosity in all directions they would have to quickly convert several times the mass of the sun into energy, and we know of no plausible way for that to happen besides the merging of black holes (which don't emit that energy as light). So models began to be built under the premise that the energy was being radiated in a confined beam, and they appear bright if the beam is aimed in our direction, just like with pulsars. How confined, though? And what causes it?
That's what's really hard to answer, and it varies by the model. Much of this model variance is due to the variability of GRBs themselves. They have an astoundingly diverse range of behavior in terms of brightness over time. They're almost like snowflakes -- each one a bit different. This was a very tough puzzle to crack and a huge amount of astrophysical research was (and still is) focused on it. To emphasize it, here's a list of some of the studies on GRBs in a literature review. Give it a quick skim just to blow your mind.
http://www.astro.ku.dk/~malesani/GRB/papers.html
And one of my favorite individual papers is this one by Tsvi Piran, which gives a great treatment of many important aspects of GRB physics.
The most successful models are all based on creating an extremely fast jet, but they differ in what kind of engine powers it. We know that whatever it is, it must allow a lot of gravitational potential energy to be turned into kinetic energy, and a good way to do that is by merging massive objects like neutron stars together, or by the collapse of very large stars. We can run some simulations of these kinds of events to get some insight on the jet formation, but it's a really complex computational problem involving hydrodynamics and general relativity.
We can also get some insight into the dynamics of the jet by studying the after-glow of the bursts. This is the long-term, lower-energy emission which we think is caused by the jet slamming into the intervening medium, releasing energy through shock fronts. Unfortunately, this is also fraught with complexity and sometimes ambiguous results. But it's enough to give us some rough constraints on how narrow they are.
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| spacer | Date: Monday, 31.10.2016, 17:19 | Message # 894 |
 Star Engineer
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Israel
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| today we learned in physics how to caculate the density of a ball (like planets)
i caculated and got answer that earth density is 5.5tons (5500kg) per sqare meter. or m^3
or also 5.5 grams per square centimeter
Watsisname, is that correct? if so can you give another bodies to caculate? with given mass or density.
if i have density i will know how to caculate the mass.
today i caculate the mass of a boulder that is inside the western wall here.
if the boulder is 14meters length, 3.5meters wide and 3.5meters height.
14x3.5x3.5=171.5
now the known density of the boulder is 4.2 grams per square centimeter.
we want to know the boulder mass in kilograms.
4.2x1000=4200kilograms per square meter.
so 4200x171.5=720300kilograms.
so the boulder mass is 720.3 tons. A LOT. how ppl moved it i dont really know...
is that correct?
also now i know to caculate your weight in an elevator if the acceleration of the elevator is known.
lets say your weight is 50 kilograms. and the elevator has acceleration of 5 (not really that high in real life i guess)
so a=5
now 50x(5+10) 5 is the accelration of the elevator while 10 is to do it in Newton units so 50x10 is to caculate the weight in Newton units.
now 50x5=250, 50x10=500
500+250=750
so given the acceleration of an elevator is 5meters per second^2
you will weight 750 newtons or 75 kilograms, that also can explain why when the elevator go up you feel heavier.
now what if the elevator goes down? lets say the acceleration is a=-2meters per second^2
now 50x10=500
50x-2=-100
500-100=400
you will weight 400 newtons or 40 kilograms
that why when the elevator goes down, you feel less heavy or even float abit given your weight is very low or the accelration is pretty high.
 Watsisname, is that correct?
"we began as wanderers, and we are wanderers still"
-carl sagan
-space engine photographer
Edited by spacer - Monday, 31.10.2016, 17:25 |
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| midtskogen | Date: Monday, 31.10.2016, 20:58 | Message # 895 |
 Star Engineer
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Norway
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| Quote spacer ( ) square
You mean "cubic".
NIL DIFFICILE VOLENTI
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| Watsisname | Date: Monday, 31.10.2016, 23:49 | Message # 896 |
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Galaxy Architect
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| Yep, density is mass per volume, and volume has units of length cubed. Length times length times length.
Mass per length squared is a 'surface density', (think the density of a thin sheet of foil or something like that). There is also a 'linear density', which is just mass per length (think the density of a long straight wire). But for your purposes here you want mass per volume.
Your calculation for Earth's density is correct. The idea is to take the mass of the Earth (about 6x1024kg), and divide by the volume of the Earth (4/3*pi*(radius)3), where Earth's average radius is about 6371km. This gives about 5500kg/m3. Does the result make sense? Well, this value is somewhere between the density of rock and the density of solid iron. So yes, it makes sense for a planet which is made up of a metal core with a rocky mantle.
Your result for the mass of the boulder is correct, too. Nice work on the unit conversions. That can be a tricky part of this kind of calculation. Do you feel comfortable with it? If not, I can give you some further help -- there is an easy method for writing it out so that it makes sense and you'll be less likely to make mistakes.
And finally the elevator. You are 3 for 3 -- correct answers all across the board! Well done!
The way I solve or explain the physics behind the elevator problem (and being thorough) is with a free body diagram. This is using the ideas of Newton's Laws, forces and vectors, which you should either be developing now or very soon in your physics. If you haven't seen any of this yet, then don't worry about it.
Newton's Second Law says that the sum of the forces acting on an object is equal to the object's mass times its acceleration. The forces and acceleration are both vectors, meaning they have a magnitude and a direction. We can draw them as arrows.
As a picture, the second law is telling us that adding the arrows for all the forces on the object together should give us a new arrow whose length and direction represents the acceleration. The elevator is accelerating up, and since you are accelerating with it (you neither fall through the floor or fly up into the air), your acceleration vector must be pointed up.
You know the force of gravity is pulling you downward, and its strength is mg (your mass times 9.8m/s2). For your acceleration to be upward, the sum of the forces acting on you must also be upward. So there must be another force, upward, with a strength greater than mg. The only force this could be is a "normal force" -- the force of the elevator floor pushing up on you. We'll draw it as an upward arrow labelled "FN", with a length longer than mg.
Now apply Newton's 2nd Law symbolically:
Sum of forces = ma
= FN - mg
Solve for the normal force that the elevator applies on you:
FN = ma + mg = m(a+g)
So the force that you actually feel is greater than the gravitational force (mg). It is mg plus your mass times the elevator's acceleration. A similar argument with the elevator accelerating downward will give us mg-ma. If the elevator were to accelerate downward with a=g, then the normal force on you is zero. You are in free-fall and seem weightless.
A neat lesson here is that you do not feel gravity! Even in freefall, or in orbit, there is still a gravitational force mg acting on you. What you actually feel are normal forces. The force of the elevator, or the ground, pushing you up.
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| spacer | Date: Tuesday, 01.11.2016, 04:55 | Message # 897 |
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Star Engineer
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Israel
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| Quote midtskogen ( ) You mean "cubic".
yes sorry
Watsisname, i am familiar with all of the equations but still take me time to fully understand it.
can you explain to me how to know in which diraction the Net force is?
like in the second law questions
when i draw it like in what you did i have hard time to find the net force. the mg and normal force are ok but the net force i always stop there.
"we began as wanderers, and we are wanderers still"
-carl sagan
-space engine photographer
Edited by spacer - Tuesday, 01.11.2016, 04:55 |
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| Watsisname | Date: Tuesday, 01.11.2016, 06:51 | Message # 898 |
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Galaxy Architect
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| Quote spacer ( ) can you explain to me how to know in which diraction the Net force is?
like in the second law questions
when i draw it like in what you did i have hard time to find the net force. the mg and normal force are ok but the net force i always stop there.
Ah, well there's a few things to say about that. First, have you been introduced to vectors and how to add them? If not, then you don't need to worry about the drawing.
How to know the direction of the net force.
In this case, we know the direction that you are accelerating. It must be upward, same as the elevator. Then we know the direction of the net force (the combination of all the forces acting together) on you as well. Newton's Second Law says the direction of the net force on an object is the same direction as its acceleration.
Now what if we are trying to find the net force, but don't know the acceleration? What if all we know are the directions and magnitudes of each of (possibly several) forces acting on it? In that case, we can use vector addition. The idea is to represent each force with an arrow, whose direction corresponds to the direction of the force, and its length corresponds to the magnitude of the force. Then the net force is found by drawing all the vectors together, tip to tail. A youtube video explaining this can be found here (particularly from 3:35 to 7:55).
Also, I forgot you asked for further examples to practice on with density. So I'll give you one. Hopefully not too challenging, but I also don't want to make it trivially easy.
Consider the Earth. Our beautiful blue planet. 
Let's crush it until it becomes a black hole.
My question for you is:
How dense would the Earth be, right at the moment we crushed it enough to form a black hole?
Got it? Okay, here is some helpful info:
Einstein's theory of general relativity tells us that any object will become a black hole, if it is compressed within a sphere of a certain critical size. The radius of that sphere (which we call the "Schwarzschild radius") depends only on the mass of the object itself, and a few constants of nature. The formula is:
where:
G is the gravitational constant: 6.67x10-11 m^3/kg/s^2 (meters-cubed per kilogram per second-squared),
M is the mass of the object (Earth = 5.97x1024 kg)
c is the speed of light (3.00x108 m/s)
Good luck!
Added: Oh, and let's say you have until November 13 and I'll post the solution then. I don't want you to feel pressured to work on it too quickly.
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| spacer | Date: Tuesday, 01.11.2016, 15:41 | Message # 899 |
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Star Engineer
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Israel
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| Watsisname, mmm didnt understand much how to do the massxgravitational constant.
6.67x10-11 m^3/kg/s^2x5.97x10^24 kg?
"we began as wanderers, and we are wanderers still"
-carl sagan
-space engine photographer
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| Watsisname | Date: Tuesday, 01.11.2016, 17:03 | Message # 900 |
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Galaxy Architect
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United States
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| Yes, the way you wrote it is exactly right.
The mass and the gravitational constant are numbers, but both numbers have units associated with them. In the MKS ("meters-kilograms-seconds") metric system, mass is measured in kilograms, and the gravitational constant is measured in cubic meters per kilogram per second-squared.
If we multiply G by M, then the units will be (m^3)/(kg*s^2) multiplied by (kg). The kilograms will cancel, leaving (m^3)/(s^2). Then you'll get new units again after you divide by c^2. Indeed, it should simplify to just meters, since the units on the right hand side of the formula must be the same as the units on the left hand side. The left hand side is R (radius), which is in meters.
This technique of working with the units, or "dimensions" of the quantities in a calculation is known as "dimensional analysis", and is a very useful tool.
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