The original experiment using Cassini's onboard receivers would have had an accuracy of better than 1 m/sec and presumably similar positioning accuracy. Still, the probe accomplished a lot and was several different kinds of awesome.
If you have indeed done five minutes of research and still have doubt in your mind, you are a fucking idiot.
To deny the Apollo landings is a terrible insult to the tens of thousands of people who worked on what is arguably mankind's greatest achievement.
To take an agnostic position in the face of obvious historical record and no credible opposing evidence is nearly as bad, because it is giving credence to those who would deny it.
I say this as a subscriber to Skeptical Inquirer magazine.
"Maybe you should try being accepting of other people's views..."
"Keep an open mind, but not so open that your brains fall out."
I thought the grandparent was asking whether a black hold would have the same field strength as a star of the same mass. Also, aren't there a number of supermassive stars (up to 50 solar masses) that weigh more than conventional black-holes resulting from supernovae (~20 solar masses)?
I agree that in general most black holes are more massive than most stars.
So why would a black hole produce a greater gravitational force at a distance than any massive star?
It doesn't, but the black hole is very massive - considerably more massive than any star in the galaxy.
And why would a great force at the centre of the galaxy be inclined to spit out stars at huge velocities?
It's tricky to explain and not terribly easy to get your head around, but I think the principle is similar to this demonstration (check out the video). The grav. potential energy of the companion star due to its attraction to the black hole is transferred into kinetic energy in the ejected star.
Breeder reactors aren't perpetual motion machines.
There are three isotopes that are important when discussing fission reactors:
U-235 - 0.7% natural abundance. Rare and extremely difficult and expensive to extract from natural uranium. When used in concentrations >10% or so, makes an excellent fission fuel for a reactor. Very easy to use to make bombs but ONLY when at 95%+ concentration, and it takes a lot of effort to go from 10-20% conc. to 95% conc.
U-238 - 99.2% natural abundance. Relatively common, easy to refine and handle. Cannot be used as a fission fuel in any sort of reactor (excluding fission-fusion hybrids and things)
Pu-239 - does not exist naturally. Easy to use as a fission fuel. Also relatively easy to use to make nuclear bombs.
When people talk about breeder reactors as "producing more fuel than they burn", what they mean is that the reactor is run on either U-235 or Pu-239. It produces heat energy which is converted into electricity.
At the same time, excess neutrons from the reaction are reacted with an otherwise inert blanket of U-238 around the reactor, converting the U-238 into Pu-239 which can then be used to run the same reactor, or other reactors. It turns out that Pu-239 production is faster than Pu-239 or U-235 consumption.
It is relatively easy to use chemical methods to separate the produced Pu-239 from the leftover U-238 in the blanket, certainly MUCH easier than separating U-235 from natural uranium.
So it's not a perpetual motion machine because a resource is used up, i.e. the natural U-238, but that resource is plentiful and the overall process is easier than the conventional method of getting fissile fuel.
The reason that breeder reactors aren't widely used is partly technical, because they're fairly complex things to design and operate, but mostly political because the Pu-239 produced can relatively easily be used in bombs.
Breeder reactors aren't perpetual motion machines.
There are three isotopes that are important when discussing fission reactors:
U-235 - 0.7% natural abundance. Rare and extremely difficult and expensive to extract from natural uranium. When used in concentrations >10% or so, makes an excellent fission fuel for a reactor. Very easy to use to make bombs but ONLY when at 95%+ concentration, and it takes a lot of effort to go from 10-20% conc. to 95% conc.
U-238 - 99.2% natural abundance. Relatively common, easy to refine and handle. Cannot be used as a fission fuel in any sort of reactor (excluding fission-fusion hybrids and things)
Pu-239 - does not exist naturally. Easy to use as a fission fuel. Also relatively easy to use to make nuclear bombs.
When people talk about breeder reactors as "producing more fuel than they burn", what they mean is that the reactor is run on either U-235 or Pu-239. It produces heat energy which is converted into electricity. At the same time, excess neutrons from the reaction are reacted with an otherwise inert blanket of U-238 around the reactor, converting the U-238 into Pu-239 which can then be used to run the same reactor, or other reactors. It turns out that Pu-239 production is faster than Pu-239 or U-235 consumption.
It is relatively easy to use chemical methods to separate the produced Pu-239 from the leftover U-238 in the blanket, certainly MUCH easier than separating U-235 from natural uranium.
So it's not a perpetual motion machine because a resource is used up, i.e. the natural U-238, but that resource is plentiful and the overall process is easier than the conventional method of getting fissile fuel. The reason that breeder reactors aren't widely used is partly technical, because they're fairly complex things to design and operate, but mostly political because the Pu-239 produced can relatively easily be used in bombs.
IANAP (yet, starting my course in six months) but I think that the estimates for the amount of matter in the universe (dark or otherwise) are to do with gravitational attraction. Therefore only the rest mass of matter is important - if an object is accelerated close to c it has a large "relativistic mass" but that's mostly thought of as an outdated concept. Gravity only works on rest mass which is why if you travel close to c you wouldn't suddenly collapse into a black hole.
The landing was basically a bonus - it was designed as an atmospheric probe, and used a heatshield to decelerate to subsonic velocity, then a series of parachutes to float down relatively slowly over the course of two hours. It was calculated to impact the surface at 10m/s or so, around 20mph, so they designed the instruments to survive that sort of shock so that some data could be returned if a landing was successfully made.
Hey, quit the physics-bashing! Some of us are interested in working out how the world works.
("186,000 miles a second - it's not just a good idea, it's THE LAW")
"The Earth's mass
does affect it's orbit around the Sun. See Kepler's laws of planetary motion."
That's not correct.
Kepler's 1st Law: "The orbits of the planets are ellipses, with the Sun at one focus of the ellipse."
Kepler's 2nd Law: "The line joining the planet to the Sun sweeps out equal areas in equal times as the planet travels around the ellipse."
Kepler's 3rd Law: "The ratio of the squares of the revolutionary periods for two planets is equal to the ratio of the cubes of their semimajor axes."
You can prove it algebraically: Taking the simple case of a circular orbit, let r = orbital radius, M = mass of sun, m = mass of earth, G = gravitational constant, F = attractive force between sun and earth due to gravity, v = instantaneous linear velocity of earth
Since the orbit describes a circular path and no other forces act on the Earth, the centripetal force acting on it must be F. (Any object moving in a circle requires a centripetal force acting towards the center of the circle)
Centripetal force F = mv^2 / r
Since the forces are the same, equate them:
mv^2/r = GMm/r^2
Multiply both sides by r:
mv^2 = GMm/r
Divide both sides by m
v^2 = GM/r
As you can see, the orbital radius depends only on the Sun's mass (M) and the instantaneous velocity (v) since G is a universal constant. It is unaffected by the mass of the Earth.
The maths is a bit trickier for an elliptical orbit but the 'm's still cancel.
It is worth noting that in practice the Sun experiences the same force and is therefore displaced, but only by a tiny amount because it has a much greater mass than the Earth. I'm not sure if this is measurable but it has very little to do with the Earth's orbit.
On 15 January 2005 it will begin its descent into Titan's atmosphere, an event that might even be visible from Earth, provided you have a decent telescope, and are in the right place at the time.
Is this really possible? Considering this is all Hubble can make out of the whole moon, I find it hard to believe that the atmospheric entry could be bright enough to be seen from this far away. (The Cassini spacecraft itself is far far below the resolution of any Earth-based or orbital telecope)
Would be cool though if Cassini could photograph Huygens' descent, but I expect this will be precluded by the attitute necessary for proper radio communication.
Slashdot Mirror
The original experiment using Cassini's onboard receivers would have had an accuracy of better than 1 m/sec and presumably similar positioning accuracy. Still, the probe accomplished a lot and was several different kinds of awesome.
To deny the Apollo landings is a terrible insult to the tens of thousands of people who worked on what is arguably mankind's greatest achievement.
To take an agnostic position in the face of obvious historical record and no credible opposing evidence is nearly as bad, because it is giving credence to those who would deny it.
I say this as a subscriber to Skeptical Inquirer magazine.
"Maybe you should try being accepting of other people's views..."
"Keep an open mind, but not so open that your brains fall out."
You fucking idiot. Do five minutes of research. This is a fairly important thing to not have made up your mind about.
"Learn how Albert Einstein's Unified Field Theory was used in the Philadelphia Experiment and Nazi Bell Device!"
I agree that in general most black holes are more massive than most stars.
It doesn't, but the black hole is very massive - considerably more massive than any star in the galaxy.
And why would a great force at the centre of the galaxy be inclined to spit out stars at huge velocities?
It's tricky to explain and not terribly easy to get your head around, but I think the principle is similar to this demonstration (check out the video). The grav. potential energy of the companion star due to its attraction to the black hole is transferred into kinetic energy in the ejected star.
IANAPyet so please correct me if I'm wrong.
When people talk about breeder reactors as "producing more fuel than they burn", what they mean is that the reactor is run on either U-235 or Pu-239. It produces heat energy which is converted into electricity.
At the same time, excess neutrons from the reaction are reacted with an otherwise inert blanket of U-238 around the reactor, converting the U-238 into Pu-239 which can then be used to run the same reactor, or other reactors. It turns out that Pu-239 production is faster than Pu-239 or U-235 consumption.
It is relatively easy to use chemical methods to separate the produced Pu-239 from the leftover U-238 in the blanket, certainly MUCH easier than separating U-235 from natural uranium.
So it's not a perpetual motion machine because a resource is used up, i.e. the natural U-238, but that resource is plentiful and the overall process is easier than the conventional method of getting fissile fuel.
The reason that breeder reactors aren't widely used is partly technical, because they're fairly complex things to design and operate, but mostly political because the Pu-239 produced can relatively easily be used in bombs.
Breeder reactors aren't perpetual motion machines. There are three isotopes that are important when discussing fission reactors: U-235 - 0.7% natural abundance. Rare and extremely difficult and expensive to extract from natural uranium. When used in concentrations >10% or so, makes an excellent fission fuel for a reactor. Very easy to use to make bombs but ONLY when at 95%+ concentration, and it takes a lot of effort to go from 10-20% conc. to 95% conc. U-238 - 99.2% natural abundance. Relatively common, easy to refine and handle. Cannot be used as a fission fuel in any sort of reactor (excluding fission-fusion hybrids and things) Pu-239 - does not exist naturally. Easy to use as a fission fuel. Also relatively easy to use to make nuclear bombs. When people talk about breeder reactors as "producing more fuel than they burn", what they mean is that the reactor is run on either U-235 or Pu-239. It produces heat energy which is converted into electricity. At the same time, excess neutrons from the reaction are reacted with an otherwise inert blanket of U-238 around the reactor, converting the U-238 into Pu-239 which can then be used to run the same reactor, or other reactors. It turns out that Pu-239 production is faster than Pu-239 or U-235 consumption. It is relatively easy to use chemical methods to separate the produced Pu-239 from the leftover U-238 in the blanket, certainly MUCH easier than separating U-235 from natural uranium. So it's not a perpetual motion machine because a resource is used up, i.e. the natural U-238, but that resource is plentiful and the overall process is easier than the conventional method of getting fissile fuel. The reason that breeder reactors aren't widely used is partly technical, because they're fairly complex things to design and operate, but mostly political because the Pu-239 produced can relatively easily be used in bombs.
Actually that's even worse - the best you'll manage is to send information at the speed of sound in the string.
IANAP (yet, starting my course in six months) but I think that the estimates for the amount of matter in the universe (dark or otherwise) are to do with gravitational attraction. Therefore only the rest mass of matter is important - if an object is accelerated close to c it has a large "relativistic mass" but that's mostly thought of as an outdated concept. Gravity only works on rest mass which is why if you travel close to c you wouldn't suddenly collapse into a black hole.
FWIW: The highest energy particle ever detected carried an astonishing 50 Joules. That's equivalent to a 1kg object hitting you at 22mph. Ouchies.
Because you're returning to Earth. For (robotic) landers on other bodies e.g. Mars and Titan they just call it 'entry'.
FORTRAN???
The landing was basically a bonus - it was designed as an atmospheric probe, and used a heatshield to decelerate to subsonic velocity, then a series of parachutes to float down relatively slowly over the course of two hours. It was calculated to impact the surface at 10m/s or so, around 20mph, so they designed the instruments to survive that sort of shock so that some data could be returned if a landing was successfully made.
Photos 723 and 202 look like some evidence of splashdown(!)
Someone PLEASE mirror!
I have to agree with the other responses - I really like the live feeds, it's very atmospheric.
Flamebait? Parent was flamebait. Mine was a flame.
You're a fucking idiot. Learn some orbital mechanics.
Hey, quit the physics-bashing! Some of us are interested in working out how the world works. ("186,000 miles a second - it's not just a good idea, it's THE LAW")
That's not correct.
You can prove it algebraically: Taking the simple case of a circular orbit, let r = orbital radius, M = mass of sun, m = mass of earth, G = gravitational constant, F = attractive force between sun and earth due to gravity, v = instantaneous linear velocity of earth
From Newton's law of gravitation, F = GMm/r^2
Since the orbit describes a circular path and no other forces act on the Earth, the centripetal force acting on it must be F. (Any object moving in a circle requires a centripetal force acting towards the center of the circle)
Centripetal force F = mv^2 / r
Since the forces are the same, equate them:
mv^2/r = GMm/r^2
Multiply both sides by r:
mv^2 = GMm/r
Divide both sides by m
v^2 = GM/r
As you can see, the orbital radius depends only on the Sun's mass (M) and the instantaneous velocity (v) since G is a universal constant. It is unaffected by the mass of the Earth.
The maths is a bit trickier for an elliptical orbit but the 'm's still cancel.
It is worth noting that in practice the Sun experiences the same force and is therefore displaced, but only by a tiny amount because it has a much greater mass than the Earth. I'm not sure if this is measurable but it has very little to do with the Earth's orbit.
Doesn't matter if the Earth's mass changes. Only the Sun's mass affects our orbit.
On 15 January 2005 it will begin its descent into Titan's atmosphere, an event that might even be visible from Earth, provided you have a decent telescope, and are in the right place at the time.
Is this really possible? Considering this is all Hubble can make out of the whole moon, I find it hard to believe that the atmospheric entry could be bright enough to be seen from this far away. (The Cassini spacecraft itself is far far below the resolution of any Earth-based or orbital telecope)
Would be cool though if Cassini could photograph Huygens' descent, but I expect this will be precluded by the attitute necessary for proper radio communication.
Right.. now, a question I've never found an answer to: Why can't you do the same thing with a black hole? Henry
Here's a guy who built a software RAID using USB floppy drives on OSX.