 # Janus

Resident Experts

2039

26

## Everything posted by Janus

1. ## Will a Magnetic Induction Engine work?

2. 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 $w=u+v$ but by $w=\frac{u+v}{1+\frac{uv}{c^2}}$ In this case, your second disk would rotate at $\frac{.25c+.25c}{1+\frac{(.25c)(.25c)}{c^2}} = .4c$ 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 $\frac{.25c+.4c}{1+\frac{(.25c)(.4c)}{c^2}} = .559c$ forth disk: $\frac{.25c+.559c}{1+\frac{(.25c)(.559c)}{c^2}} = .778c$ fifth disk: $\frac{.25c+.778c}{1+\frac{(.25c)(.778c)}{c^2}} = .988c$ sixth disk: $\frac{.25c+..988}{1+\frac{(.25c)(.988)}{c^2}} = .993c$ seventh disk: $\frac{.25c+.993c}{1+\frac{(.25c)(.993c)}{c^2}} = .996c$ eighth disk $\frac{.25c+.995}{1+\frac{(.25c)(.995c)}{c^2}} = .997c$ ninth disk $\frac{.25c+.997c}{1+\frac{(.25c)(.997c)}{c^2}} = .998c$ tenth disk $\frac{.25c+.998c}{1+\frac{(.25c)(.998c)}{c^2}} = .999c$ eleventh disk $\frac{.25c+.999c}{1+\frac{(.25c)(.999c)}{c^2}} = .9994c$ twelveth disk $\frac{.25c+.9994c}{1+\frac{(.25c)(.9994c)}{c^2}} = .9996c$ 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.
3. 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.)
4. ## what if information can be transmitted faster than c

5. 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
6. 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".
7. This is too small by a factor of two.
8. ## Classical Mechanics Challenge

Since nobody else is biting: The most elegant solution is to use the formula: $M = \frac{v^3t}{G}$ Where v is the orbital velocity and t is the orbital period. Converting to SI units this gives: $\frac{(2.98)^3 (6750)}{2 \pi 6.673x10^{-11}\frac} = 4.26x10^{14} kg$ 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 $C = vt$ We can then get the radius by dividing by 2 pi: $r =\frac{vt}{2 \pi}$ This radius by the way turns out to be about 3.2 km. The equation for orbital velocity is $v = \sqrt{\frac{GM}{r}}$ we simply substitute for r: $v =\sqrt{\frac{2 \pi GM}{vt}}$ 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.
9. ## Min. Energy needed to launch to the sun?

You can't incinerate nuclear waste. It remains radioactive no matter how hot you heat it.
10. ## Min. Energy needed to launch to the sun?

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.)
11. ## Min. Energy needed to launch to the sun?

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.
12. ## Min. Energy needed to launch to the sun?

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.
13. ## what if information can be transmitted faster than c

14. ## What if Lightspeed were not constant.

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.
15. There is a natural process through which the moon could form a ring system. It is through the continued evolution of the same tidal action that is causing it to move away from the Earth as it is now. As the Moon moves into a slower, higher orbit the trade off is that the Earth slows down in its rotation. Given enough time the Earth will rotate at the Same rate as the Moon revolves around it, always presenting the same face towards the Moon. This is known as tidal lock. However, the Moon is not the only actor in slowing the Earth's rotation; the Sun also has an influence, as it tries to bring the the Earth into tidal lock with itself. The result would be that once The Earth and Moon have achieved tidal lock, the Sun will continue to try and slow the Earth's rotation. The Moon will try to keep the Earth locked with itself. As a result, the Monn will transfer angular momentum from itself to the Earth, in much the manner as the Earth now transfers angular momentum to the Moon. The loss of momentum will cause the moon to fall into a lower, faster orbit. It will continue to transfer angular momentum to the Earth in an attempt to speed up the Earth to match its new orbital period, but doing so just moves it into even a lower and faster orbit. The moon will continue to spiral in until it reaches its Roche limit and is torn apart by tidal forces and forms a ring around the Earth. The kicker is the phrase "given enough time". The above process takes so long that before then, our sun will expand into a red giant as a part of its evolution and very likely engulf and destroy the Earth and Moon.
16. All closed Earth orbits are elliptical (circle is just an ellipse with an eccentricity of 0). If you fire from a cannon a projectile that strikes the Earth. it also follows the curve of an ellipse, it just can't complete the ellipse because the Earth gets in the way. Over short distances, this curve follows closely that of a parabola, so we can get away with assuming that it is a parabola and still get an accurate enought answer. You can have parabolic orbits, but they would not be closed. Any object traveling at exactly escape velocity would follow such a path. if you fired a projectile at escape velocity, it would continue to fly away from the Earth following a parabola. (assuming that you didn't fire the cannon with a downward tilt.) If the object is fired at greater than escape velocity, it will follow a hyperbola.
17. ## What if Lightspeed were not constant.

No, c stands for the speed of light in vacuum. When you talk about the apparent speed of light due to refractive index, this can not be refered to as c.
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