 # Janus

Resident Experts

2054

27

## Everything posted by Janus

1. Let's check that figure. The Sun converts 5' date='000,000 tons of mass to energy every second. This is considerably more mass than we could ever put in a bomb, so vaporizing 19% of the Solar system is out. In fact, considering that the gravitational binding energy of the Earth alone is in the order of [math']2 x 10^{32}[/math] joules, it would take the conversion of 2,444,444,444,444 tons of matter to energy to "vaporize" just the Earth alone.
2. ## Speed of Light - Time it take to...

For those in the ship the trip would take 395,967 years, compared to the 28 million years the trip would take fro those on the Earth.
3. A quick calculation shows that if 600,000,000 million people jumped at the same time, and all in the same direction, the Earth would move a maximium of 0.0000000000000009 of a meter. This maximum displacement would take place about 1/3 of a second after the Jump initiates. The Earth would then move back to its orginal position as the gravitational attraction btween it and the people bring them back together. No amount of jumping will cause any permanent displacement of the Earth. Considering that that Earth already changes its distance from the Sun by almost 12,000 km on a monthly cycle and 4,000,000 km on a yearly cycle, It seems like a lot of effort to go through for sush an unmeasurably small, temporary displacement.
4. ## Relationship between Planets size, mass, age and gravity?

The 'G' is the universal gravitational constant and is equal to $6.673 x 10^{-11}$N-m²/kg² Examples: mass of Earth : $5.97 x 10^{24}$kg radius of Earth : $6.378 x 10^{6}$ m acceleration of gravity at the Surface of the Earth: $\frac{(6.673 x 10^{-11})(5.97 x 10^{24})}{(6.378 x 10^{6})^2} = 9.792 m/sec^2$ mass of Moon : $7.35 x 10^{22}$kg radius of Moon : $1.738 x 10^{6}$ m acceleration of gravity at the Surface of the Moon: $\frac{(6.673 x 10^{-11})(7.35 x 10^{22})}{(1.738 x 10^{6})^2} = 1.623 m/sec^2$ mass of Uranus : $8.68 x 10^{25}$kg radius of Uranus : $2.556 x 10^{7}$ m Acceleration of gravity at the Surface of Uranus: $\frac{(6.673 x 10^{-11})(8.68 x 10^{25})}{(2.556 x 10^{7})^2} = 8.887 m/sec^2$
5. ## Who is More "Relative"?

6. ## Who is More "Relative"?

If that were true then both clocks would read the same as they experienced the same accelerations. But they don't' date=' they record different times. mass of Uranus: 8.68e25 kgRadius of Uranus 25,559,000 meters. Formula for acceleration due to gravity: $a_g = \frac{GM}{r^2}$ plug the numbers in and you get 8.87 m/sec² for Uranus comapred to 9.8m/sec² for the Earth. and you said Don't you se a contradiction ? No contradiction if you understand the difference between 'local force due to gravity' and 'gravitational potential'. If I drop the higher clock in the uniform field, it will fall, and as it falls it will gain speed. As it gains speed it gains kinetic energy. It can only gain kinetic energy by giving up potential energy. This potential energy is in the form of gravitational potential. Thus the two clocks are at different Gravitational potentials and according to Relativity, it is the this difference in potential that acounts for the time rate difference between the two clocks, not the gravitational force felt by the clocks.
7. ## Who is More "Relative"?

No' date=' because it would not be correct to say so.For example, I send two clocks off into space, accelerate them both up to the same speed with the same acceleration, let them coast for a while, and then bring them back, again with each clock experiencing the same acceleration. The difference is that I let one clock coast for a longer period before bringing it back. Then the clock that coasted for a longer period will read less when the clocks are brought back together While a clock on the surface of the Earth does run slower than one in space it is not due to the local strength of gravity. It is due to a difference of potential between the two clocks. For example, the Surface gravity of Uranus is less that that of the Earth's, but a clock on the surface of the Uranus will run slower than one on the surface of the Earth. Another example would be two clocks at different heights in a uniform gravity field(one that does not change in strength between the two heights). The higher of the two clocks will run faster, even though it feels the same gravtitational force as the lower one.
8. Fusion gives you more "bang for the buck" in terms of percentage of mass converted to energy.
9. ## determining semiminor axis in elliptic orbit

Sorry, not careful enough reading on my part. Okay then, to find the semi-minor[/b} axis you can use the formula: $b= a \sqrt{(1-e^2)}$ for which you need to know the eccentricity (e). You can find the eccentricity thusly: The Areal velocity can be found by $A = \frac{cos{\theta}vd}{2}$ or $A= \frac{\pi a^2 \sqrt{(1-e^2)}}{P}$ where: $\theta$ is the angle of the objects path relative to a line perpendicular to the radial vector (e.g. at perapis or apapis, or for a circular orbit $\theta=0]$) and $P$ is the period of the orbit and can be found by $P= 2 \pi \sqrt{\frac{a^3}{GM}}$ Again, if you equate the two equations for Areal velocity and solve for e, you will get the eccentricity.
10. ## determining semiminor axis in elliptic orbit

You can get the semi-major axis from conservation of energy. The total energy of an orbiting body is given by $E= \frac{mv^2}{2}- \frac{GMm}{d}$ or $E= \frac{GMm}{2a}$ Where M is the mass of primary m is the mass of the orbiting object v is the velocity d is the distance to the primary a is the semi-major axis Just equate these two formulas and solve for a to get the semi-major axis when you know the velocity and distance of your object.
11. Two points: The solar wind is exceedingly thin. if you just consider the Earth's orbital path, the Earth collides with less than 6 grams of mass per sec. Gvien that the mass of the Earth itself is 6,000,000,000,000,000,000,000,000,000 grams, it would be a long time before this led to any significant slowing. The other point is that the Solar wind is "blowing" out away from the Sun. This tends to push the Earth out away from the Sun. Since the speed of the solar wind varies between 200 and 800 km/sec and the Earth's orbital velocity is 30 km/sec, this "push" exceeds any drag. But then again, this push is very, very, very small.
12. If anything, it is on the conservative side. For instance, I just calculated the energy needed to get up to a speed that will get you to Andromeda in 28 yrs. But that is coasting at that speed for the majority of the trip. If you are accelerating the whole way, the velocity I used would be just your average velocity. Your top velocity would be twice that, meaning you would need even more energy than I calculated. Edit: It turned out to be really conservative. Calculating for accelerating at 1g for the entire 28 yrs, it comes out to 2.5e41 joules of energy per kilogram. Thats equal to the entire energy output of the Sun for 21 million years!
13. The same direction. In fact, The Moon also orbits the Earth in the same direction. In addition, all the planets of the Solar system orbit in the same direction and the majority of the moons of those planets. It is so prevalent that it is called "direct" motion, while orbiting or rotating in the opposite direction is called "retrograde" motion. There are exceptions, but they are just that, exceptions, such as some moons that are captured asteroids (For complicated reasons it is easier for a planet to capture another body into a retrograde orbit), or due to some major event in the body's past.
14. ## Does Relativity affect Radioactivity?

15. ## Does Relativity affect Radioactivity?

take two identical candles in the exact same oxygen content, place them at different gravitational potentials and they will burn at different rate. Take two identical water clocks. Place one on the suface of Uranus(where the surface gravity is less than that on Earth). Put the other at an height above the Earth's surface so that the local gravity equals that of Uranus' surface, and you will find that the clock on Uranus' surface will run slower. Even though the gravitational force drivng each clock is equal This is because Uranus has a deeper gravitational field than the Earth does, and its clock will be at a different gravitational potential. A sun dial does not measure local time rate, but the "motion" of a Heavenly body. Since this motion varies over time due to different influences, such astronomical time keeping has been abandoned for any time measurements that require any high accuracy. What is weird is that you think the wording of the above makes any sense. This is gibberish Time to read what I posted above on this matter. Due to the Relative velocity the traveling twin has during the trip. This makes no sense. The Earth's orbital velocity has no effect on its gravity. And even if you were trying to invoke a relativistiv mass increase this is way off. Relativistic mass increase follows the rule of $M_r = \frac{M_0}{1-\frac{v^2}{c^2}}$ Earth's orbital velocity 30 km/sec. Assuming its rest mass is 1,000,000 "shuttles" its relativistic mass due to velocity is 1,000,000.005 "shuttles" . it only gains .005 of a shuttle in mass. Using the same formula we can find that the shuttle would have to travel at .943 of c to reach this same mass. ( Not the ten times the speed of light you mention below) See above for correct calculations. You also seem to be laboring under a false impression as to the effect a relativistic increase in mass has on the gravity of an object. (another common misconception) Which, as pointed out above, has a smaller influence on the shuttle's time rate than the relative velocity of the shuttle has. In the Twin Paradox gravitational effects are ignored, because due to the high Realtive velocity of the Traveling twin, they cause a negliable effect. Again, two different Relativistic effects in play, with the one causing a slower time rate overshadowing the other. And if you think that effect that velocity has is that it causes a mass increase, and the resulting gravity increase is what causes the clock to run slow, you are way off base. This is not what the theory predicts. Before you go around claiming to have proved Einstein wrong and proposing new theories, maybe you should actually learn what Einstein actually said
16. ## Hardest question in the World

It wouldn't be a spiral, but it would be curved, and your hole would have to be dug such that it misses the center. (Assuming that it isn't dug from pole to pole)
17. The point is that when Einstein made his imaginary ride with a beam of light he soon realised that it lead to non-sensical results. For one, the beam of light he was riding on would cease to exist.
18. We're not talking set theory, we are talking physics, and in physics, a 20% variance is huge and not considered "almost the same".
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