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Janus

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Everything posted by Janus

  1. It depends on how much matter you are talking about. For instance, if you had 9^16 joules to work with, you could have one electron moving at just slightly less than c, 1kg moving at .866c or 100 kg at .14c.
  2. When talking about the efficiency of a rocket, we are generally dealing with how much delta V we can get from our fuel/reaction mass. For this we use the rocket equation [math]\Delta V = V_e ln(MR)[/math] where Ve is the exhaust velocity and MR is the mass ratio, (the mass of the fully fueled rocket divided by the mass of the unfueled rocket) Note that for any given mass ratio, the dela v depends on the exhaust velocity. The above equation works well for every day velocities, but for relativistic velocities you need to use the Relativistic rocket equation: [math]\Delta V = c \tanh{\left (\frac{V_e}{c} \ln{(MR)}\right)}[/math] Again we note that the delta v depends on the exhaust velocity. We get the best efficiency if we have an exhaust velocity of c. With such as "photonic" drive one could expect the following maximum delta v: for a MR of 2, we can get .6c for 3, .8c 4, .882c 5, .923c 6, .946c 7, .96c ... 100, .9998c
  3. Yes it is. Or more correctly, if the mass of the planet is very small as compared to the mass of the Sun then the contribution of the planet's mass can be safely ignored. When the planet's mass becomes large enough you need to use the following formula to get an accurate answer. [math] P = 2\pi\sqrt{\frac{a^3}{G \left(M_1 + M_2\right)}} [/math] How large is "large enough" depends on what degree of accuracy you need. For instance, with the Earth's orbit, ignoring Earth's mass causes a difference of only about 70 seconds or about 0.00015%. If you needed to calculate Earth's orbital period finer than this then you need to include its mass. (However, in this case you also need to know the Earth's average dist from the Sun, the mass of the Sun, etc to equal accuracies first)
  4. What you are describing is known as a Doppler Shift. While light and sound both exhibit such shifts, it has nothing to do with the Relativistic effects described by Einstein. The effects of Relativity are what are left over after the Doppler shift has been accounted for. So, no, your method is not the easiest way to describe Relativity as it actually leads to an erroneous conclusion as to the nature of the theory.
  5. Actually it would be about 1.1547 years later on Earth.
  6. Er, that would have been a little difficult, since Kepler died in in 1630 and Newton wasn't even born until 1642.
  7. Kepker's Law actually were purely empirical. They were based on observation of the planets in their orbits. He observed that the planets followed ellipitical orbits and he observed that their periods and distances from the sun were related in a particular way. He then formulated these observations as laws. It wasn't until Newton that these laws were mathemathically derived from his laws of motion and universal gravity.
  8. Actually, you would expect it to work out that way. As has been pointed out, the force acting on the falling body decreases porportionally with the decrease in distance from the center. This implies harmonic motion on the part of the object. Harmonic motion can also be described as a projection of circular motion on the diameter of the circle. If you start with an object moving in a circle with a centripetal force of [math]f= \frac{mv^2}{R}[/math] You will find that the component of that force acting along the line of the diameter varies linearly. (as per harmonic motion). Since the centripetal force acting on a surface orbiting object and the intial restoring force of our falling object are the same, if you start the two objects at the same point at the same time, you will find that a line drawn through either object, and perpendicular to the path of the our falling object will always pass through both objects. Thus when the orbiting object reaches a point 90° from its starting point, the falling object has reached the center and both objects at that moment are heading in the same direction at the same speed.
  9. Just one more point. It turns out that the speed at the center (for uniform density) equals [math]\sqrt{\frac{GM}{R}}[/math] Where M is the mass of the Earth and R its radius. This is also equal to the orbital velocity for an object at the surface of the Earth (ignoring air resistance)
  10. No, CPL's figure for the speed is correct. It's what you get when you take the potential energy difference between the center of the Earth and the surface, and then figure out the speed needed to equal this in kinetic energy for a given mass.
  11. Here's the problem: Recent evidence points to Mars still being geologically active. Any past civlization on Mars would have likely died out before all the water left Mars. erosion and geological activity mean a rock cycle, which means that Mars' crust could have gone through a complete recycle since said possible civilization thrived, leaving no traces.
  12. To answer that, just make u = .1c and v = c in the equation given, like thus: [math]u'= \frac{.1c+c}{1+\frac{c(.1c)}{c^2}}[/math] [math]u'= \frac{1.1c}{1+\frac{.1c^2}{c^2}}[/math] [math]u'= \frac{1.1c}{1+.1}[/math] [math]u'= \frac{1.1c}{1.1}[/math] [math]u'= c[/math] Thus for you, the light travels at c, and for the "static" observer, the light travels at c.
  13. Which does absolutely nothing towards supporting your interpretation of Relativity. The transforms between frames in relative motion explain these experiments quite nicely. Once again, you are just taking the results of these experiments, and trying to fill in the causes by yourself, rather than actually learning the theory and what it says on the matter.
  14. Sorry' date=' but it [i']you[/i] that doesn't understand SR. Kinetic energy is relative not absolute. Which object has more kinetic energy depends on who you ask. For it to be otherwise would require a prefered Absolute reference frame from which all objects' kinetic energy shall be measured. Relativity denys the existance of any such frame. All this proves is that you have never bother to study SR deeply enough to learn that in Relativity, the kinetic energy of a mass is [math]KE= mc^2 \left ( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1 \right )[/math]not [math]KE= \frac{mv^2}{2}[/math] These accelerator experiments actual disprove your interpretation of SR. The Earth orbits the Sun at 30Km/sec, relative to the Sun. If your interpretation were right you would have to take this velocity into account while determining the Relativistic effects on a particle being accelerated. If the particle is being accelerated in the same direction as the Earth is moving, then this 30km/sec is added to its velocity, and if it is acclerated in the other direction it will be subtracted. This would mean the particles moving in one direction relative to the Earth would have more kinetic energy than particles moving in the Other and thus a greater Relativistic effect. Our accelerators are to the point that they can get close enough to the speed of light for this difference to result in a noticeable effect. For instance, the difference between .99c and .999c equate to a time dialtion factor difference of 7 to 22. For example , it would take make more energy to accelerate the same particle up to .999c relative to Earth in one direction than in another. One particle would also show a greater time dilation than the other. Now, we routinely accelerate particles in opposite directions (Many accelerator experiment consist of coliding accelerated particles), and in all sorts of realtive directions to Earth orbital path, and we have never seen this type of effect. It always takes the same amount of energy to accelerate up to any given speed relative to the Earth and said particles always show the same time dilation no matter what direction they are moving relative to the Earth's orbital vector. You really need to thoroughly learn about a subject before spouting off about it, rather than just picking up bits and pieces and then trying to fill in the missing parts yourself.
  15. 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.
  16. 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.
  17. 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.
  18. The 'G' is the universal gravitational constant and is equal to [math]6.673 x 10^{-11}[/math]N-m²/kg² Examples: mass of Earth : [math]5.97 x 10^{24}[/math]kg radius of Earth : [math]6.378 x 10^{6}[/math] m acceleration of gravity at the Surface of the Earth: [math]\frac{(6.673 x 10^{-11})(5.97 x 10^{24})}{(6.378 x 10^{6})^2} = 9.792 m/sec^2[/math] mass of Moon : [math]7.35 x 10^{22}[/math]kg radius of Moon : [math]1.738 x 10^{6}[/math] m acceleration of gravity at the Surface of the Moon: [math]\frac{(6.673 x 10^{-11})(7.35 x 10^{22})}{(1.738 x 10^{6})^2} = 1.623 m/sec^2[/math] mass of Uranus : [math]8.68 x 10^{25}[/math]kg radius of Uranus : [math]2.556 x 10^{7}[/math] m Acceleration of gravity at the Surface of Uranus: [math]\frac{(6.673 x 10^{-11})(8.68 x 10^{25})}{(2.556 x 10^{7})^2} = 8.887 m/sec^2[/math]
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