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Science and Astronomy Questions
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| spacer | Date: Saturday, 05.11.2016, 20:12 | Message # 916 |
 Star Engineer
Group: Users
Israel
Messages: 1258
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| i had some physics question and i got that the acceleration of a bullet is 1200000meters per second square.
is that Makes sense?
in the answers in the book for the question it was correct eventually when you use newton second law with ma
the mass of the bullet was 0.0035kg
so 0.0035x1200000=4200newtons=4200N
but accelration of 1200000? wow 
maybe it make sense because the accelration is happening for very short time. less than a second... 
also i think i understood now why we float in space even if earth gravity is still working.
the spaceship is going down in earth gravity acceleration...so we float,
the spaceship may going down in earth gravity acceleration but the speed is so high its never really goes down to earth.
its like an elevator when its goes down you float and feel easier.
i understood it to be something close to that...is that correct? and if so can you give farther information?
"we began as wanderers, and we are wanderers still"
-carl sagan
-space engine photographer
Edited by spacer - Saturday, 05.11.2016, 20:26 |
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| Watsisname | Date: Sunday, 06.11.2016, 01:13 | Message # 917 |
 Galaxy Architect
Group: Global Moderators
United States
Messages: 2613
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| Quote spacer (  ) i had some physics question and i got that the acceleration of a bullet is 1200000meters per second square.
Depends on the gun, but your result seems reasonable. Let's figure out a way to calculate the acceleration of a projectile (as it is fired) in general and see if we come up with a sensible answer that agrees with that.
In general we probably do not know the time it takes for the bullet to be accelerated. It is very fast and hard to measure. But we can usually know the muzzle velocity, and the length of the barrel. So we can use one of the basic kinematic equations to find the acceleration:
We know the initial velocity (v0 )of the bullet is 0, and it accelerates to vf which is the muzzle velocity. It does this over the length of the barrel, which is (s-s0). Let's redefine s-s0 as "L", and solve for a. We get
So the acceleration is the square of the muzzle velocity, divided by twice the barrel length. To get a ballpark figure, consider a muzzle velocity of 1000m/s and a barrel length of half a meter. Then the acceleration is exactly 106 m/s2. Makes sense!
Quote spacer ( ) maybe it make sense because the accelration is happening for very short time. less than a second...
Precisely. Big change in velocity over a very short interval of time. In fact, let's calculate that as well, using the above example of 1000m/s over 0.5 meters. How long did it take?
This time we'll use the kinematic equation
Solve for t, it is just v/a. 1000m/s divided by 106 m/s2 is 0.001s, or 1 millisecond. Pretty quick.
Quote spacer ( ) i understood it to be something close to that...is that correct? and if so can you give farther information?
Yeah, you pretty much have it. Not only is the spaceship being pulled down by gravity, but so is everything on it. And one of the funky things about gravity is that all objects in a gravitational field fall with the same acceleration, no matter what their mass is.
It's not obvious to us on Earth that gravity works that way, because air resistance adds an additional force which complicates it. But we can test it in vacuum chambers, or on the surface of the Moon:
So in space, the ship and everything on it are falling together, and to someone inside it seems like everything is weightless and freely floating. And remember, you don't feel gravity. What you feel are other forces which prevent you from being in freefall. This is just like the elevator example.
That the ship is falling down would definitely be a problem if it started out at rest somewhere above the Earth. It would fall down and into the atmosphere quite quickly, and then everyone on board would have a bad time. So the ship has to be moving sideways, really fast, so that it falls "around" the Earth. We can calculate the sideways speed that it needs to be able to circle the Earth, too. When you start learning about physics problems involving circular motion, this might be a fun thing to try to derive.
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| Alek | Date: Sunday, 06.11.2016, 07:58 | Message # 918 |
 Pioneer
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United States
Messages: 326
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| How much gravity would they feel if the station was somehow "locked" to the same place, the same distance from the Earth (basically, it would be locked in place in an "absolute space" that moves with Earth, though it still feels the Earth's gravitational field, it doesn't fall due to it.) Would they feel 1g, or less because they're farther from the ground, or more because the planet's gravity isn't pulling on it from 180 degrees like it does on the surface, and therefore is more concentrated in a downward pull?)
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: Sunday, 06.11.2016, 09:50 | Message # 919 |
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Galaxy Architect
Group: Global Moderators
United States
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| The ISS is currently at an altitude of 406.5km above the Earth. At that altitude, the acceleration due to gravity is 8.68m/s2, or about 88% the strength of gravity on Earth's surface. If you weigh 80kg here on Earth's surface, then on a stationary platform at that altitude you would weigh 70kg. (Of course, your mass is still 80kg.)
So yes, gravity is still pretty strong up there. That's because 400km altitude is not very much farther from the center of the Earth compared to being on the surface of the Earth.
To calculate, equate Newton's Law F=ma with the force of gravity F=GMm/r2. Solve for a. The little m, being the mass of the object, cancels (which is why all objects have the same acceleration due to gravity regardless of their mass). We get:
a=GM/r2
where G is the gravitational constant, M is the mass of the object you're gravitating towards (assuming it is a sphere and you're not inside of it), and r is your distance from its center.
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| apenpaap | Date: Monday, 07.11.2016, 15:15 | Message # 920 |
 World Builder
Group: Users
Antarctica
Messages: 1063
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| How did people come to realise the gas giants are gas giants and don't have a solid surface? I remember reading a popular theory for the great red spot used to be that it was a massive mountain poking out of Jupiter's atmosphere, so clearly there was a point where we thought they were solid worlds.
I occasionally stream at http://www.twitch.tv/magistermystax. Sometimes SE, sometimes other games.
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| Watsisname | Date: Tuesday, 08.11.2016, 01:40 | Message # 921 |
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Galaxy Architect
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United States
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| I hadn't heard that theory -- that's pretty interesting. I'm not sure of the history behind that, but I can propose one calculation which might have been part of what helped convince people that they are gas giants and cannot have solid surfaces. The calculation will be to determine the average density of the planet by observable quantities: the size of the planet, the distance of its moons, and the orbital period of the moons. Then the average density will tell us something about what the planet must be made out of.
Start with Newton's Law of gravity. F=ma, and the force of gravity is F=GMm/r2 (Aye, these two equations strike again! But we'll take them somewhere new.) With the moons being on essentially circular orbits, we can also equate the gravitational force of the planet on its moons to be a centripetal force of mv2/r
where r is the distance of the moon from the planet and v is the moon's orbital speed.
Let's do some algebra!
Relating these expressions for the force, we can write
We can cancel out the mass of the moon "m" from them, so that's nice.
The orbital speed of a moon around a planet is a bit inconvenient as a remote measurement. It would be nice if we could instead relate it to the orbital period and the orbital distance, which are easier to measure. For a circular orbit, that is:
where T is the orbital period.
Plug that expression in for v2 in the previous relation, and we get
Let's rearrange that slightly:
Does it look familiar to anyone? This is actually Kepler's Third Law, relating the square of the orbital period to the cube of the orbital distance.
Okay, one last variable which we want to replace. That's M, the mass of the planet, which we can't really measure directly. But we can relate it to the radius of the planet "R", and the average density "ρ". The average density is what we actually want to find at the end of the day, so that's great. Density is mass per volume, so mass is volume times density. And the volume of a sphere is 4/3*pi*R3. So we'll substitute in for M:
Which gives us
And finally, solving for the planet's density:
That's pretty neat. We can find the average density of a planet, provided that it has moons.
Consider Jupiter's moon Io. It orbital distance is about 422,000km, and its orbital period is 42 hours, 27 minutes. The (average) radius of Jupiter is about 69,900km. Plug these in and convert accordingly, and we derive an average density of Jupiter of 1.33 g/cm3. This is exactly the accepted value.
1.33 g/cm3 is very low for a rocky planet with a metal core. (Mercury, Venus, and Earth are all above 5, and Mars is 3.93) Even a purely rocky planet without a dense core should be at least 2 or 3. So Jupiter must be primarily made up of lighter materials. We might suppose water, but that doesn't really work too well on that scale. (And it doesn't work at all for Saturn -- it's average density is even less than water!). The only things that make sense are gaseous elements like hydrogen and helium. And if you have a planet made primarily of those, then it can't have a solid surface. Planets will differentiate with the denser stuff going to the center and the lighter stuff on top.
So we reach the conclusion that Jupiter and other "gas giants" must be made primarily of light elements, with thick gaseous atmospheres. Then, since the pressure goes up as you move in, there must be a smooth transition from gaseous to a fluid state, which we learn from studying physical chemistry and phase transitions.
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| steeljaw354 | Date: Tuesday, 08.11.2016, 11:13 | Message # 922 |
 World Builder
Group: Users
Pirate
Messages: 862
Status: Offline
| What if Pluto was the only object out beyond Neptune? What if Ceres successfully cleared out it's orbit? And by Pluto being the only object out beyond Neptune, I mean if it had cleared out the entire Kuiper belt, somehow.
Edited by steeljaw354 - Tuesday, 08.11.2016, 11:19 |
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| FastFourierTransform | Date: Tuesday, 08.11.2016, 12:38 | Message # 923 |
 Pioneer
Group: Local Moderators
Spain
Messages: 542
Status: Offline
| Quote steeljaw354 ( ) What if Pluto was the only object out beyond Neptune?
The New Horizons team would face unemployement by now instead of been active for several years more and we would be very enthusiastic about the contradictions this magic would generate with our current models for the Solar System formation.
Quote steeljaw354 ( ) What if Ceres successfully cleared out it's orbit?
It would have automatically to be considered a planet by the IAU definition. Since the mass of the entire Main Asteroid Belt is about 4% that of our Moon or 22% that of Pluto you would have a very very small planet.
Quote steeljaw354 ( ) Pluto being the only object out beyond Neptune, I mean if it had cleared out the entire Kuiper belt, somehow.
It would have automatically to be considered a planet by the IAU definition, also. Since the mass of the entire Kuiper Belt is 0,1 Earth's masses you would have a Mars sized Pluto. Would this generate disturbances? I guess not to much since Pluto is quite isolated: the closest aproach to another object of the Solar System would be Uranus (surprisingly!!!) and it would get at max to 11AU from it. Consider that a Mars-sized planet currently aproaches Earth as close as 0,36AU and nothing happens with our Moon for example, no disturbances at all. Nothing would change in the rest of the Solar System.
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| Watsisname | Date: Wednesday, 09.11.2016, 06:50 | Message # 924 |
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Galaxy Architect
Group: Global Moderators
United States
Messages: 2613
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| What would it take to break Washington, Oregon, and California off of the CONUS and move them between Hawaii and New Zealand?
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| JackDole | Date: Wednesday, 09.11.2016, 07:07 | Message # 925 |
 Star Engineer
Group: Local Moderators
Germany
Messages: 1742
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| Quote Watsisname ( ) What would it take to break Washington, Oregon, and California off of the CONUS and move them between Hawaii and New Zealand?
I think you would need a 'Dillon-Wagoner Graviton Polarity Generator', also known as the 'Spindizzy'.
Don't forget to look here.
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| midtskogen | Date: Wednesday, 09.11.2016, 11:19 | Message # 926 |
 Star Engineer
Group: Users
Norway
Messages: 1674
Status: Offline
| Quote Watsisname ( ) What would it take to break Washington, Oregon, and California off of the CONUS and move them between Hawaii and New Zealand?
Some nukes and Lex Luthor?
NIL DIFFICILE VOLENTI
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| Hornblower | Date: Wednesday, 09.11.2016, 11:52 | Message # 927 |
 World Builder
Group: Users
United States
Messages: 714
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| New question: Can we impeach him now?
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| PlutonianEmpire | Date: Thursday, 10.11.2016, 23:40 | Message # 928 |
 Pioneer
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United States
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| I was thinking about the whole terraformed Mars thing, because of a dream I had this morning, and I began to wonder about the common presumption. The common presumption is that terraformed Mars stereotypically assumes 1 atm of air with standard Earth composition. And my dream this morning reminded me of the Earthlike atmospheres thread. Looking at it again, we all know by now that 0.21 bars of 100% can be breathable.
So, lets apply that to Mars. We dont have the tech to add oceans, plain and simple. But, do we have the tech *today* to give mars 0.21 bars of at least 95% oxygen? How much co2 does mars have in its air do we need to compensate for? If we do have the tech, what is it? If were don't have it, what do we need to do to get it? And it won't be easy for sure, but how hard will all this be?
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 - Thursday, 10.11.2016, 23:41 |
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| Hornblower | Date: Friday, 11.11.2016, 01:54 | Message # 929 |
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World Builder
Group: Users
United States
Messages: 714
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| PlutonianEmpire, Well no matter how hard it is to get the oxygen there, it will be even harder to protect it. Mars has no magnetic field so the atmosphere it has will just be ripped away by the solar wind.
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| Watsisname | Date: Friday, 11.11.2016, 12:01 | Message # 930 |
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Galaxy Architect
Group: Global Moderators
United States
Messages: 2613
Status: Offline
| Hornblower:
That may actually not be as big of a problem as you think. There are many ways in which gases escape a planet's atmosphere. The one you're referring to is 'sputtering', where gas molecules get ejected due to collisions with solar wind particles, which a magnetic field protects against. However, this process is fairly slow -- it's important over million to billion year timescales. Assuming we can build a breathable atmosphere on century to millennia timescales, then that loss would be easy to manage.
There is also atmospheric loss by 'evaporation' (fastest molecules escaping), and there can even be atmosphere blown off of the planet by large impacts. All of these processes were important for the loss of Mars' atmosphere early in its history.
PlutonianEmpire,
It's a cool idea. One interesting effect would be the reduced boiling point of water on the surface -- about 60°C. Barely hot enough to cook an egg, after boiling it for at least 20 minutes. 
Quote PlutonianEmpire ( ) How much co2 does mars have in its air do we need to compensate for?
Mars' surface pressure is variable, especially by altitude, but let's take 11.5mbar at the lowest elevations. At 95.9% CO2, that's a partial pressure of about 0.011atm. This is like having 1% or 10,000ppm CO2 at Earth's atmospheric pressure.
If we kept this with the additional 0.21atm of oxygen, then it would cause some drowsiness and long-term CO2 adaptation.
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