Ad: "Black Holes in the Universe"
The next seminar of the Balseiro Institute in Bariloche Atomic Centre will be delivered by Dr. Felix Mirabel.
Saturday, October 31, 2009
Dr. Mirabel is senior researcher at CONICET, having received his Ph.D. in astrophysics from the University of La Plata, and one in philosophy from the University of Buenos Aires. For his discoveries and research in the area of \u200b\u200bblack holes has received several distinctions, including Doctorate Honoris Causa from the University of Barcelona, \u200b\u200bthe Bruno Rossi Prize for High Energy Astrophysics of the Astronomical Association of North America, and the Scientific Award of the Atomic Energy Commission of France. He has authored nearly 500 publications and has participated in the discovery of numerous novel astronomical phenomena, such as microcuásars, superluminal motion in the Galaxy, ultraluminous infrared galaxies and tidal dwarf galaxies. Service is a member of Astrophysics Atomic Energy Commission of France, and the prestigious European Observatory of the Southern Hemisphere and has been the leading representative in Chile for several years.
Friday November 6, 2009, 1430 Bariloche Atomic Center
Felix Mirabel
Institute of Astronomy and Space Physics, Buenos Aires
Summary: Black holes are the most enigmatic objects in astrophysics. Its gravitational pull is so intense that no light leak, staying dark. For more than two centuries its existence was cause of mere speculation, but for two decades convincing evidence has accumulated on the existence of two types of black holes in the universe: 1) as corpses of massive stars that cosmic dance devour stars produce light, and 2) as objects more individual mass of the universe, with masses equivalent to millions of stars, concentrated in regions as small as the solar system. We will review the properties of cosmic objects and their role in the evolution of the universe, according to recent research in the world's most advanced observatories. Demonstrate its effects with images and animations, reaching key conclusions about our knowledge of the cosmos. 
Sunday, October 25, 2009
Is Keri Hilsons Hair A Weave
a planet's surface temperature, surface temperature
comes the end of our trilogy on the orbital temperature. In the first part
calculate In this third part (last !) the equilibrium temperature. We have already determined that every second the earth receives $ 1.74 \\ times 10 ^ {17} $ J. What does the Earth with this energy? In short, it is heated. And we know that a hot body radiates energy according to the Stefan-Boltzmann (radiated energy increases with the fourth power of the temperature!). Suppose that the Earth has a temperature $ T_ \\ Oplus $ and behaves as a black body (a more reasonable assumption at this level), then the power radiated into space (usually using the letter $ L $ per light) is:
where $ \\ sigma $ is the constant Stefan-Boltzmann and $ \\ left (4 \\ pi R_ \\ Oplus ^ 2 \\ right) $ is the Earth's surface (here we consider the total surface of the Earth and not only its cross section as in the case of the energy absorbed .)
The equilibrium occurs when the energy absorbed is equal to the radiated energy. If there is an imbalance, the parameters are adjusted so as to reach a new condition. Suppose for example that for some reason increases the solar luminosity $ L_ \\ odot $. In this case, the energy that Earth receives will be greater and this leads to an increase in temperature, which implies an increase in the radiated energy. This equilibrium condition is therefore stable. Then:
\\ [E_ {abs} = E_ {emit} \\]
remember the expression for $ E_ {abs} $ obtained in the first part
and since brightness is energy per unit time,
\\ [L_ {abs} = \\ frac {L_ \\ odot} {4} \\ left (\\ frac {R_ \\ Oplus} {R_ {UA} \\ right) ^ 2} \\]
Therefore, our equilibrium condition occurs when:
\\ [L_ {abs} = L_ {emit} \\]
\\ [ \\ frac {L_ \\ odot} {4} \\ left (\\ frac {R_ \\ Oplus} {R_ {UA} \\ right) ^ 2} = \\ Left (4 \\ pi R_ \\ Oplus ^ 2 \\ right) \\ sigma T_ \\ Oplus ^ 4 \\]
and then clearing the temperature $ T_ \\ Oplus $ we have:
\\ [T_ \\ Oplus = \\ sqrt [4] {\\ frac {L_ \\ odot} {16 \\ pi \\ sigma} \\ frac {1} {R_ {UA} ^ 2}} \\]
Before numbers forward some conclusions:
The temperature is independent of the size of the planet (note that $ R_ \\ $ Oplus canceled) increases with the fourth root of the solar luminosity, and decreases with square root distance. 
The equation is valid for any star-planet system, and is used inter alia to determine the "
\\ [T_ \\ Oplus = \\ sqrt [4] {\\ frac {3.85 \\ times 10 ^ {26} \\ mathrm {W}} {(16 \\ pi) (5.67 \\ times 10 ^ {-8} \\ mathrm {W} \\ mathrm {m} ^ {-2} \\ mathrm {K} ^ {-4}) (1.5 \\ times 10 ^ {11} \\ mathrm {m}) ^ 2}} \\] \\ [T_ \\ Oplus = 278 \\ mathrm {K} = 5 ^ \\ mathrm {o} \\ mathrm {C} \\]
comes the end of our trilogy on the orbital temperature. In the first part
, we calculate the solar energy that Earth receives, and the second part
, which would estimate the temperature of the Earth if the Sun was not where $ \\ sigma $ is the constant Stefan-Boltzmann and $ \\ left (4 \\ pi R_ \\ Oplus ^ 2 \\ right) $ is the Earth's surface (here we consider the total surface of the Earth and not only its cross section as in the case of the energy absorbed .)
The equilibrium occurs when the energy absorbed is equal to the radiated energy. If there is an imbalance, the parameters are adjusted so as to reach a new condition. Suppose for example that for some reason increases the solar luminosity $ L_ \\ odot $. In this case, the energy that Earth receives will be greater and this leads to an increase in temperature, which implies an increase in the radiated energy. This equilibrium condition is therefore stable. Then:
\\ [E_ {abs} = E_ {emit} \\]
remember the expression for $ E_ {abs} $ obtained in the first part
:
\\ [E_ {abs} = \\ frac {L_ \\ odot t} {4} \\ left (\\ frac {R_ \\ Oplus} {R_ {UA} \\ right) ^ 2} \\]
and since brightness is energy per unit time,
\\ [L_ {abs} = \\ frac {L_ \\ odot} {4} \\ left (\\ frac {R_ \\ Oplus} {R_ {UA} \\ right) ^ 2} \\]
Therefore, our equilibrium condition occurs when:
\\ [L_ {abs} = L_ {emit} \\]
\\ [ \\ frac {L_ \\ odot} {4} \\ left (\\ frac {R_ \\ Oplus} {R_ {UA} \\ right) ^ 2} = \\ Left (4 \\ pi R_ \\ Oplus ^ 2 \\ right) \\ sigma T_ \\ Oplus ^ 4 \\]
and then clearing the temperature $ T_ \\ Oplus $ we have:
\\ [T_ \\ Oplus = \\ sqrt [4] {\\ frac {L_ \\ odot} {16 \\ pi \\ sigma} \\ frac {1} {R_ {UA} ^ 2}} \\]
The temperature is independent of the size of the planet (note that $ R_ \\ $ Oplus canceled)
The equation is valid for any star-planet system, and is used inter alia to determine the "\\ [T_ \\ Oplus = \\ sqrt [4] {\\ frac {3.85 \\ times 10 ^ {26} \\ mathrm {W}} {(16 \\ pi) (5.67 \\ times 10 ^ {-8} \\ mathrm {W} \\ mathrm {m} ^ {-2} \\ mathrm {K} ^ {-4}) (1.5 \\ times 10 ^ {11} \\ mathrm {m}) ^ 2}} \\]
more impressive results, is not it?
Thursday, October 22, 2009
Franklin Mint Vase Butterflies
3rd part of a planet, part 2
say that sequels are never good, but we will make the attempt ...
In the first part say that sequels are never good, but we will make the attempt ...
calculate the amount of energy the Earth receives from the sun But what about that energy? In a nutshell, is used to warm the Earth. Which brings us to another question before continuing: What temperature is the empty space? 
course to answer this question we must first define what it means "Empty space." Strictly speaking, the empty space there. And for various reasons: cosmology, quantum and relativistic. Re-ask: At what temperature is a region of space with no stars or other energy sources nearby?
(yes, as the oven!), Which extend in the electromagnetic spectrum between 300 MHz and 300 GHz
Alpher and R.
Hermann
, and was discovered (by accident) in 1965 by Hermann
A. Penzias and R. Wilson (Nobel Prize winners in 1978 for this work) is actually a relic of the Big Bang to the time when the universe became transparent to radiation, or put another way, when finally electrons and photons allowed to interact and electrons began to form atoms with the baryons. From that moment (about 300,000 years after the Big Bang), the photons propagate through the expanding Universe with almost no interaction (the current density is 410 photons per cubic cm.)
And since the universe expands adiabatically (by definition, the universe is a closed system), the temperature of this background radiation decreases as time progresses. The current temperature of the background radiation was measured with extreme accuracy (parts per million) and is 2.725 K, and its distribution corresponds to that of a black body at that temperature.
We finally have our answer: if the sun suddenly disappeared, and if the Earth had no nuclear processes inside, the temperature of the Earth gradually fall to reach thermal equilibrium with the background, about 2.7 K (about -270 ° C).
continued ...
Tuesday, October 13, 2009
Contemporary Costume For Sale
surface temperature of a planet, 1st part
We will try to estimate what the temperature of the earth's surface if it had no atmosphere. This is generally known as orbital temperature. mass $ M_ \\ Oplus $ and radius $ R_ \\ Oplus $ that orbits the Sun at a distance $ R_ {AU} $. 
On the other hand, the Sun Let's start by determining what fraction of the energy emitted by the Sun is received on Earth at a given time $ t $. Recalling that the luminosity $ L_ \\ odot $ is the energy emitted by a star per unit time, the energy radiated in time $ t $ is just $ L_ \\ odot \\ times t $.
Since the solar emission is isotropic (equal in all directions), at a distance $ R_ {AU} $ del Sol solar energy will distributed evenly over the surface of a sphere of radius $ r = R_ {AU} $ and therefore the energy density on the surface of the sphere is:
\\ [E_ {R_ {UA} } = \\ frac {L_ \\ odot t} {4 \\ pi R_ {UA} ^ 2} \\] The fraction of this energy that reaches Earth only depends on the relationship between the surface of the sphere and the exposed surface, ie the cross section of the Earth: $ \\ pi R_ \\ Oplus} ^ 2 $. To better understand this result, imagine that far off lantern illuminate a ball, on a wall form a circular shadow is equal to the radius of the ball and the surface of that circle is equal to the area of \u200b\u200blight absorption for the ball.
Therefore, the energy reaching the Earth at the same time $ t $ is:
\\ [E_ {tierrras} = E_ {R_ {UA}} \\ times \\ pi R_ \\ ^ 2 Oplus = \\ frac {L_ \\ odot t} {4} \\ left (\\ frac {R_ \\ Oplus} {R_ {UA} \\ right) ^ 2} \\]
Take some numbers: the solar luminosity $ L_ \\ odot = 3.85 \\ times 10 ^ {26} $ W, the Earth-Sun distance (also known as Astronomical Unit): $ R_ {AU} = 1.5 \\ times 10 ^ {11} $ m (150 million kilometers), the Earth radius $ R_ \\ Oplus = $ 6371 km and take $ t = $ 1 second. Then:
\\ [E_ {earth} = \\ frac {3.85 \\ times 10 ^ {26} \\ mathrm {W} \\ times 1 \\ mathrm {s}} {4} \\ left (\\ frac {6371 \\ × 10 ^ 6 \\ mathrm {m}} {1.5 \\ times 10 ^ {11} \\ mathrm {m}} \\ right) ^ 2 \\]
Every second the Earth receives from the Sun $ 1.74 \\ times 10 ^ {17} $ joules, that is: 174,000,000,000,000,000 joules!! and this is every second. To put things in perspective, the average
2005 was 16 TW (16 Terawatts = $ 1.6 \\ times 10 ^ {13} $ watts): the illuminated surface of the Earth receives from the Sun about 11,000 times the world energy requirement!
Monday, October 12, 2009
Milena Velba Airport Movie
free Stone Blaise
Pressure is a physical quantity that measures the force per unit area. Because it relates two vectors by a scalar product
, it is a scalar (a number!), Defined as: \\ [P = \\ frac {d {\\ bf {F}} } {dA} \\ cdot {\\ bf n}. \\] 
Where $ $ {\\ bf n} $ $ is versor This complicated expression is greatly simplified when force is normal to the surface: if on a flat surface area $ $ $ $ A uniform force acting F $ $ $ $ perpendicular to the surface, the expression for the pressure $ $ P $ $ is: , Pa, which corresponds to 1 newton per square meter: \\ [[P] = \\ mathrm {P} \\ equiv \\ frac {\\ mathrm {N}} {\\ mathrm {m} ^ 2}, \\] and was named after Blaise Pascal, French mathematician, physicist seventeenth century.
And now, a study to determine concepts: The draw will start counting touched Einstein . The idea, of course, was that he counted to 100, while all were hidden and then dedicate them down.
Einstein Beginning your account they all ran in different directions looking for a hideout.
Einstein finished his account: ....... 97, 98, 99, 100. He opened his eyes, turned around, and found Newton standing right in front of him.
Then Einstein said (as usual):
- "Stone Free to Newton "
, Stone free to Newton! "
, Stone free to Newton! "
- No, no, I have to disagree: I was not found, because I am not Newton.
Before the stunned Albert, who looked seriously Isaac, all the rest of the scientists went one by one from their hiding places, between intrigued and surprised to finally hear an explanation of Newton with which forced to match.
Newton said
- As see, I'm standing in an area of \u200b\u200b1 square meter, therefore, I am a Newton per square meter, in short .... I am Pascal.
"
Our thanks go to
Analía
for this wonderful piece of humor;)
for this wonderful piece of humor;)
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