Janus

If the twin accelerates then S' is not an inertial reference frame and remain permanently attached to the twin. If it is an inertial refernce frame it cannot remained attached to twin. Now, the formula above does not involve the acceleration a of the ship, it only makes reference to the instantaneous relative speed v. So if you look at things from the point of view of the twin in frame S`, he sees his brother accelerate away from him at speed v as well. No, he can easily tell that he was the one who actually accelerated. He can feel the acceleration, he can fire a laser perpendicular to his path and notes that it curves, etc. And while the time dilation formula holds when used by either twin while they are at constant velocity, and holds for the twin that does not accelerate (remains in the same reference frame the whole time.), It is incomplete for the twin who undergoes acceleration while he is accelerating. For him he has to add an additional transformation that takes into account his acceleration and the distance between himself and his brother. No paradox as long as you take all the consequences of Relativity into account, including length contraction and Relativity of simulataneity. No, it is not. Again you must take all the effects involved into account to properly address this. In fact, when you do take them into account, the time dilation formula must be applied in [/i]all[/i] frames in order for there not to be a paradox. I would strongly urge you not to try and explain Relativity to others, you do not grasp it properly yourself.

No, that remark was not correct. Let's see if we can make things clearer. The force acting between two bodies is equal to the formula given already: [math]F= \frac{GMm}{d²}[/math] G is the universal gravitational constant. M is the mass of one body and m the mass of the other d is the distance between their centers. Using this we can calculate the force between the Earth and a 1kg mass sitting on its surface: The radius of the Earth is 6,378,000 m so this is d The mass of the Earth is 6 x 10^24 kg so we get: F= [math]F = \frac{((6.673 x 10-11)(1)(6 x 10^{24})}{6,378,000^2}=9.8N[/math] For the Moon, Mass = 7.35 x 10^23kg radius = 1,735,000m So the force acting on the same 1kg weight sitting on the Moon is: F= [math]F = \frac{((6.673 x 10-11)(1)(7.35 x 10^{22})}{1,735,000^2}=1.6N[/math] 1/6th that of if it was sitting on the Surface of the Earth. So we can see that even though the mass of the Moon is only 1/81 that of the Earth, an object sitting on the Moon's surface is is less than 1/3 as far from the Moon's center than an object sitting on the surface of the earth is from the Earth's center is. This is why the force of gravity on the surface of the Moon is 1/6 that of that on the surface of the Earth.

That's like asking: If Unicorns really existed, would their horns actually be an aphrodisiac?

Yes.

A magnet is a magnet because of two things: 1. the indivdual atoms/molecules have a magnetic moment(due to an imbalance of the spins of the electrons each atom has a net magnetic field. 2. These moments are aligned. (the magnetic fields of the atom all point in the same direction.) And thus re-enforce each other. If you strike a magnet sharply, you "dislodge" some of these atoms out of alignment and randomize them so that they no longer add to, and to some extent cancel out the total field of the magnet. The same thing happens if you heat a magnet. As you heat it, the atoms vibrate. Heat it enough and the atoms will vibrate right out of alignment, and you will demagnetize your magnet.

By "rotating at .25c" I'm goin to assume that you mean that the outer edge of the disk is moving a .25c. This is simply another example of the Relativistic addition of velocities, namely that velocities do not add by the relationship of [math]w=u+v[/math] but by [math]w=\frac{u+v}{1+\frac{uv}{c^2}}[/math] In this case, your second disk would rotate at [math]\frac{.25c+.25c}{1+\frac{(.25c)(.25c)}{c^2}} = .4c[/math] relative to you. Meaning that if the second disk was rotating at .25c relative to the first disk as measured from the first disk, then relative to you, it would move at .4c. if I add a third disk moving at .25c relative to the second it would be moving at [math]\frac{.25c+.4c}{1+\frac{(.25c)(.4c)}{c^2}} = .559c[/math] forth disk: [math]\frac{.25c+.559c}{1+\frac{(.25c)(.559c)}{c^2}} = .778c[/math] fifth disk: [math]\frac{.25c+.778c}{1+\frac{(.25c)(.778c)}{c^2}} = .988c[/math] sixth disk: [math]\frac{.25c+..988}{1+\frac{(.25c)(.988)}{c^2}} = .993c[/math] seventh disk: [math]\frac{.25c+.993c}{1+\frac{(.25c)(.993c)}{c^2}} = .996c[/math] eighth disk [math]\frac{.25c+.995}{1+\frac{(.25c)(.995c)}{c^2}} = .997c[/math] ninth disk [math]\frac{.25c+.997c}{1+\frac{(.25c)(.997c)}{c^2}} = .998c[/math] tenth disk [math]\frac{.25c+.998c}{1+\frac{(.25c)(.998c)}{c^2}} = .999c[/math] eleventh disk [math]\frac{.25c+.999c}{1+\frac{(.25c)(.999c)}{c^2}} = .9994c[/math] twelveth disk [math]\frac{.25c+.9994c}{1+\frac{(.25c)(.9994c)}{c^2}} = .9996c[/math] notice that each successive disk's velocity increases by a smaller and smaller amount. No matter how many disks you add the last disk will always move at less that c relative to you. If on the other hand you try to arrange it that each disks velocity increases by .25c as measured by you, then each disk will have to rotate faster with respect tot he last disk as measured by that disk. for instance if you want the second disk to rotate at .5c as measured by you then the second disk woud have to rotate at .286c realtive to the first disk as measured from the first disk. The third disk would have to rotate at .4c relative to the second in order for it rotate at .75 c as measured by you. And the fourth disk would have to rotate at 1c relative to the third to reach 1c as measured by you. But since the third disk cannot rotate at 1c relative to the third, this can never happen. In short, the only way for the last disk to have a velocity greater than c relative to you is that at least one of the disks to have a greater than c velocity relative to the disk it rests on. It doesn't matter is each disk rotates at .001c or .25c relative to the one before, the answer comes up the same; you can't acheive FTL speeds this way.

Considering that the critical mass of Plutonium is only 2/3 lb, 2lbs can cause you a lot of damage as it goes BOOM!.( In reality the reaction will probably fizzle and not undergo complete fission, but even a fizzled chain reaction would not do you any good at close range.)

According to a solar calculator I found, the longest Solar day of 2005 falls on Dec 23 with a length of 24h 0m 29.9 s The shortest Solar day falls on Sept 17 with a length of 23h 59m 38.6s

I would say this. While the book is not spinning you are constantly having to exert effort in keeping the book balanced, by slightly moving your fingers as the book tends to topple to one side or the other. When the book is spinning there is a gyroscopic effect that tends to keep the book balanced for you. You don't have to do the work of keeping the book balanced, so the book seems "lighter".

This is too small by a factor of two.

Since nobody else is biting: The most elegant solution is to use the formula: [math]M = \frac{v^3t}{G}[/math] Where v is the orbital velocity and t is the orbital period. Converting to SI units this gives: [math]\frac{(2.98)^3 (6750)}{2 \pi 6.673x10^{-11}\frac} = 4.26x10^{14} kg[/math] The above formula is easily derived from the the fact that if we know the orbital velocity and we know the period, we can get the circumference of the orbit from [math]C = vt[/math] We can then get the radius by dividing by 2 pi: [math]r =\frac{vt}{2 \pi}[/math] This radius by the way turns out to be about 3.2 km. The equation for orbital velocity is [math]v = \sqrt{\frac{GM}{r}}[/math] we simply substitute for r: [math]v =\sqrt{\frac{2 \pi GM}{vt}}[/math] and rearrange to solve for M. We can now also solve for the density of the body. We know its mass and we can calculate its volume from its radius, giving us a volume of 1.37 x 10^11 m³. Dividing the mass by the volume we get the density of 3100 kg/m³. This is a density of a little less than that of our moon.

You can't incinerate nuclear waste. It remains radioactive no matter how hot you heat it.

If that is the case then why not just throw it out of the system entirely? As stated it takes a delta v of 30km/sec to drop something into the Sun. It only takes a little over 12 km/sec to attain escape velocity from the Sun. Use Jupiter as an assist and you can bring that down to under 6 km/sec. (You might not want to use the Galileo method, people screamed enough about a small nuclear power plant, imagine how they'd howl if it was a load of nuclear waste.)

Well, a single pass wouldn't be enough, and the problem with multiple passes is that the more passes you add, the smaller your initial launch window is and the less room for error you have. Add in the fact that your last pass has to put you on a trajectory that intersects Mercury, and I think you'll find that available launch windows are few and far between. Even if Mercury was dense enough to allow for a hairpin, 180° assist, a single pass would not be enough. Adding multiple Mercury passes on top of multiple Venus passes would even further restrict available launch windows. At some point you have to ask when do lack of launch windows outweigh the saving in fuel.

To drop it directly into the Sun you would need to change its velocity by nearly 30 km/sec. This works out to 450000000J per kilogram. If you tried to use Venus for a gravity assist, it would take an initial change of velocity of 4.8 km per sec to drop to Venus' orbit, but even with a perfect 180° assist (not possible) you would only make up about 6km per sec, leaving you about 19 km/sec short of falling into the Sun. I suppose you could use multiple passes in order to trim velocity at each pass, but this would take very good timing and would be a long drawn out process.

Light is not actually "slowed down" as it passes through a medium' date=' it is "delayed". As light travels through a medium the photons are constantly being absorbed and re-emitted by the atoms of the medium. There is a time lag between the absorption and emission(during this time the photons are stored as an increased energy state of the atom, and do not exist as photons), that leads to the apparent "slow down" of light. The photons still travel at c when between atoms. A loose analogy would be a car that travels at a constant 60 mph while moving. On a freeway such a car takes 1 min to cover 1 mile. Then the same car enters a city and has stop lights to deal with. The car still travels at 60 mph between lights, but has to spend a part of its time waiting at lights. Thus it will take more than 1 min for the car to travel 1 mile. Its [i']apparent[/i] speed drops below 60 mph, even though it never travels slower than 60 mph.