Janus

The light's speed is not reduced, only it's apparent speed. As photons travel through the glass, They encounters molecules. These absorb the photons, increasing their own energy. After a slight delay, a new photon is released in the direction of the first and the molecule returns to its rest state. It is these repeated delays between absorption and emission that account for the increased time that it takes light to pass through the glass and the reduction in its apparent speed. While the light travels from molecule to molecule it still travels at c.

The body released later will run faster than the earlier one as it will always be at a high gravitational potential (even though both "feel" no gravity as they are in free fall) If the second was released from the same height as the first, it will run slower. Both from moving to a lower potential and from increasing its speed due to the fall. The sea level clock will run slower

No, it isn'tConsider two people holding on to a rope while swing in a circle. If one or the other let's up on the tension, you will separate some. This is because you reduced the centripetal force but not their mass and momentum. Now imagine an object in orbit around another. It releases electromagnetic waves. While the release of these waves will slightly reduce the mass of the object and thus its gravitational attraction to the other, However, at the same time you are reducing the mass and momentum of the object. This loss of momentum exactly compensates for the lessening of the gravitational attraction, and the orbit does not change. A case where "throwing weight overboard" would be on any help would be a rocket climbing out of a gravity well. Lessening the mass would decrease the needed thrust. But as I alluded to in an earlier post. Throwing that weight overboard in the form of billiard ball, would work just as well as throwing it overboard as electromagnetic radiation. (In fact, it would make more sense to throw out the billiard ball. To get the same effect from electromagnetic radiation you would have to convert a billiard ball's mass worth of matter to energy. So just toss out the mass without messing with the conversion.) But even then, it's a pointless exercise. If your rocket had extra mass that was there just for the purpose of being thrown overboard, you would be much better off just leaving it off the ship in the first place. No, it isn't. It is of no consequence at all. Once the light is emitted, how it interacts with the Earth via gravity has no bearing on the rocket. The only gravitational consequence is the loss of mass in the ship due to the mass equivalence of the emitted light. But you get that same effect from a standard chemical rocket. The mass of the exhaust gasses are removed from the ship, decreasing its mass as time goes on. IOW, there is nothing special about light's gravitational interaction that would affect this situation. The only advantage using light emission as a rocket has is that the exhaust speed is very high which increases efficiency. The downside is that because the momentum of a photon is very small, it is very hard to produce enough light to generate any significant thrust.

Absolute nonsense.

This makes absolutely no sense whatsoever. It does not follow that if electromagnetic waves are subject to gravity, then electromagnetic waves can control gravity. Billiard balls are subject to gravity, but a "billiard ball emitting apparatus" would not allow us to control gravity.

But light does have energy and momentum and in GR these also contribute to gravitational interaction.

The "change in distance" is not important for time dilation for a moving body. That part of what you see happening to the body's clock rate is accounted for by Doppler shift. If the body was moving towards you, you would "see" his clock running fast, but once you factored out the Doppler shift, his clock would be running slow according to you. Or you could have the body circling you at high speed, maintaining a constant distance, and his clock would still run slow according to you.

You wouldn't have to worry about the orbit if the impact just slowed the Earth's rotation so that it matched the period of the orbit.(of course, realistically, slowing the rotation by that much isn't much different from stopping it completely, and there is no practical way of doing it without cooking the Earth.) But even then, you wouldn't stop the Sun from rising and setting from some parts of the world. The Earth's orbit is elliptical, so the rotation and orbit will drift a bit out of sync, causing a East to West libration. If you were near the terminator, the Sun would bob up and down behind the horizon over the period of a year. Also, since the Earth is tilted to its orbit (assuming that the way in which the rotation was changed didn't alter this.), There will also be a North-South libration, and anyone closer to the poles than the arctic and antarctic circles will also see the Sun rise and set once a year.

The effects you mention are constant regardless of how long you keep the potatoes separated in the attic and cellar. IOW, as long as you keep the potatoes separated for long enough, any time difference caused by bringing them together or originally separating them will be inconsequential compared to the accumulated time difference due to their time spent separated.

How much fuel it would take depends of the exhaust velocity of the rocket. I could not find any numbers for the hybrid drive you mentioned, but it is estimated that a anti-matter drive could achieve exhaust velocities of 10,000,000 m/s . With this exhaust velocity, it would take 14 million times as much fuel as the mass of the space station itself to reach just half the speed of light. For a person traveling to Proxima C, the trip would take 7.27 years. But it's worse than that. The above gets you there, but you would just fly through the system at 0.5c. To stop, you would have to use just as much fuel as it takes to reach that speed. But since you don't use this fuel until you are ready to slow down, the mass of the fuel used to slow down has to be added to the mass accelerated up to 0.5c in the first place, this means that you would to start out with about 2 x 10^14 times as much fuel as space station. The space station masses 375,727 kg, so this means you would need about 7.4e+19 kg of fuel. This is about the mass of the Saturn moon Enceladus. ( And remember, a good portion of this would have to be anti-matter.)

Launching from such a platform would only reduce the space vehicle's fuel requirements by ~1/2 of a percent, which would be wiped out by the fuel needed to fly the fuel and components to the platform. Whereas putting the same resources needed to build, maintain and keep aloft such a platform to increasing the exhaust velocity of the spacecraft by that same 0.5% will decrease your fuel needs by 0.8%. A much better payoff especially considering it wouldn't involve a nuclear reactor floating overhead.

Trying to lift slowly into orbit would be highly inefficient. You would be spending a great amount of energy/fuel just fighting gravity. A fast route serves two purposes: you spend less time fighting gravity until you reach orbit and you burn most of your fuel while you are close to the ground, meaning that you will not have to waste the energy it would take to lift that fuel.

For good reason, because there isn't. You are basing this on an insufficient understanding of both orbital mechanics and the structure of the atom. This is known as GIGO. No. Electron clouds are probability based. They just define where you are likely to find the electron. At a given instant, the electron might not even be in the area defined by the cloud. More GIGO. Forming an opinion from faulty premises.

On top of that, refraction and random scattering are frequency dependent. So you wouldn't get those nice clean spectrums where all the spectral lines are still properly spaced though shifted.

1. Electrons do not orbit the nucleus like planets orbit the Sun. This way of looking at atoms was discarded decades ago. 2. Electrons maintain discreet energy levels because there are no allowable energy levels in between' If you exert a force on a satellite such that it moves from a circular orbit to an elliptical one, it will stay in that elliptical one until something else acts on it to change that. It will not just adopt a new circular orbit on its own.

Well. in practical terms, you wouldn't have to go that far to escape the Earth. You only need to get to the edge of the gravitational sphere of influence. (the point where the Sun's gravity begins to dominate). For the Earth, this works out to to be ~925,000 km. It would still take 6.6 years to reach this distance at 10 mph.

Let's see: [math]v = c \frac{\sqrt{m'^2+2m'm}}{m'+m}[/math] and [math]m'= \left ( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} -1 \right ) m [/math] At first, to keep things from being unwieldy, we'll say that [math] \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math] So we can say that [math]m' = m \gamma - m [/math] Putting this into the first equation we get [math]v = c \frac{\sqrt{\left( m \gamma - m \right)^2+2\left (m \gamma - m \right ) m}}{m \gamma - m+m}[/math] [math]v = c \frac{\sqrt{\left( m \gamma - m \right)^2+2\left (m \gamma - m \right ) m}}{m \gamma }[/math] [math]v = c \frac{\sqrt{m^2\gamma^2-2m^2 \gamma + m^2 + 2m^2 \gamma -2m^2}}{m \gamma}[/math] [math]v = c \frac{\sqrt{m^2\gamma^2 - m^2 }}{m \gamma}[/math] [math]v^2 = c^2 \frac{m^2\gamma^2 - m^2 }{m^2 \gamma^2}[/math] [math]v^2 = c^2 \frac{\gamma^2 - 1 }{ \gamma^2}[/math] [math]\frac{v^2}{c^2} = \frac{\gamma^2 - 1 }{ \gamma^2}[/math] [math]\frac{v^2}{c^2} = 1-\frac{ 1 }{ \gamma^2}[/math] [math] \frac{v^2}{c^2} = 1- \left ( 1- \frac {v^2}{c^2} \right )[/math] [math] \frac{v^2}{c^2} = \frac {v^2}{c^2}[/math] [math] v^2 = v^2 [/math] [math]v = v[/math] Hardly an earth-shattering revelation. And not very surprising, considering that your initial equation solving for v relied on a term (m'), that itself relied on the value of v.

The units must belong to the same system, thus if the mass is in grams, c is in cm/sec and the answer is in ergs. If the mass in kg, c is in m/sec and the answer is in joules. A joule is a watt-second, so there are 3,600,000 joules to a kwh. There are also 1055 joules to a BTU If the mass is in pounds, c is in ft/sec and the energy is in ft-poundals.(the poundal being the unit of force in this system) If the mass in in slugs, c is in ft/sec and the energy is in ft-lbs. (there being 778 to a BTU) As will note, there are two systems that use pounds and feet; one in which the pound is a unit of force and the other where it is a unit of mass. They are often distinguished by designating them as lb(f) and lb(m). (there is also a third system that uses the lb for both force and mass, in which it is really important to keep them straight)

Saturn's rings are made of countless individual and separate particles of ice, each following it own individual orbit round Saturn. So while they consist of solid components, collectively these components do not act like a solid. Again, the Gas giant rings are made up of independent particles. The ring suggested is a single solid piece. As such,it is unstable, as James Maxwell was able to show in 1859.

Nope, and here's why. First off, we'll replace the keys and moon with balls of equal size, but of the same respective masses of each. This makes it a fair race, as both the centers of the objects and their surfaces start at an equal distance from the surface of the Earth. It is true that if you drop the moon-massed sphere from the tower of Pisa, it will hit the ground sooner than if you separately drop the key-massed sphere from the same height. However, this is because the closing acceleration is equal to the sum of the falling acceleration of the sphere towards the Earth and the falling acceleration of the Earth towards the sphere. Since the Earth will fall towards the moon-massed sphere faster, they will hit sooner, even though both spheres fall towards the Earth at the same speed. However, when you drop them together, side by side, the Earth can't fall towards the moon-massed sphere without also falling towards the keys-massed sphere at the same rate. In fact, what happens is that the Earth falls toward the pair in response to the sum of their masses and the pair will hit the ground together just a hair sooner than the moon-massed sphere did by itself.

It doesn't matter what you "believe", what matters is where the abundance of the evidence points.The only thing dark matter and dark energy have in common is the word "dark". It was first applied to DM, for the obvious reason that it emitted no light. When it was learned that the expansion of the universe was accelerating, they needed a term to apply to whatever was causing it. Even though "dark" really carried no meaning in this case, they decided on "dark energy" just for the verbal symmetry and nothing else. The other side of what? Again, this statement make no sense when applied to DM. DM is needed to explain the rotation curves of galaxies. They rotate too fast and differently than they should according to the visible mass distribution. It is also needed to explain the mass discrepancy in galaxy clusters, neither of which can related to any type of mass imbalance in the Universe.

The reason that you most likely haven't heard of the alternatives is that, to date, all of them have failed to match observation(they make predictions contrary to what we see happening. DM and DE are separate issues are are not related to each other. Since we do not know just what DM or DE consist of, of course it is going to be something yet undiscovered. The alternatives have been explored. As I stated above, they have all come up short in one way or the other. It is pretty easy to discount additional stars and planets. Even if we can't make out individual star in a galaxy, they would still contribute to the over all brightness of a galaxy. Even planets will radiate at some temp above background temp. Since we explore galaxies with radio telescopes that could see this radiation, we should be able to detect them, especially considering how much mass they would have to account for. Black holes could only make up a small percentage of the needed mass. Black holes are formed from stars going supernova. Supernovas spread heavy elements throughout space. If enough black holes had formed to account for DM, the elemental make up of the universe would be a lot different than what we measure. Why should it be important that we be able to see it? Just because we find it handy to explore the universe using the electromagnetic spectrum, doesn't mean that the universe has to oblige us. Besides we already know of one type of matter that matches the properties that DM would have in the neutrino. Sorry, but this makes no sense in addressing either the DM or DE issue. Galaxy rotation curves.Gravitational lensing surrounding galaxies It is the one model that explains all the observations, including the Bullet cluster(The Bullet cluster is the instance where two galaxy clusters have collided. By examining the results of that collision, we can compare the what we see to what the DM model predicts, vs, other models.) The only alternative is that gravity doesn't behave like we thought it did. The problem is that all attempts to come up with a new model of gravity have failed. In addition, the Bullet cluster observations have shown that no new model of gravity, unless it includes DM can work. For the same reason that it is not visible; It does not interact via electromagnetically. It does not interact or emit electromagnetic radiation, nor does it form the electromagnetic bonds need to make atoms or elements. This also explains why it does not clump like baryonic matter and remains more or less evenly distributed. IOW, its properties naturally lead to the type of distribution we see in our observations. The non-interaction means that, just like the neutrino, it can pass right through matter and itself as if it wasn't even there. This becomes important in the case of the Bullet Cluster. The visible matter is slowed by the collision, but the DM is not. Thus the DM halos should separate from their respective clusters. This is what we see in the aftermath of this collision. By mapping the gravitational lensing caused by the DM, we can see that it is located away from the visible matter.

Yes. its gravity would disturb the orbits of the other planets, as a result, even when those planets where on the the same side of the Sun as we are, they would be out of position. In addition, the other planets would disturb this planet's orbit and it would drift out of it's orbit until we could see it. The Sun only blocks an area of the sky about 1/2 degree wide. Any comets would follow a hyperbolic or parabolic orbit, and we would see them at other points of the orbit. I really don't see any advantage to a solar orbiting station like the the one you suggest.

Somebody has already done the math. For Doppler shift where Relativistic effects are ignored, you use the expression: [math]\left( 1-\frac{v}{c} \right )[/math] Relativistic Doppler shift (after taking all relativistic effects into account) is found by: [math]\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}[/math] Where v is positive when the source is moving away from you. This formula always gives a blueshift when the source is moving towards you.

Actually, antimatter, is the form of positrons is emitted in the radioactive decay of some isotopes. This is the beta+ emission and occurs when a proton within the nucleus changes into a neutron and positron. Such beta+ emitters are used in PET scans. The agent injected into the body contains such a isotope. The isotope emits positrons which then annihilate with electrons, emitting a gamma ray. The gamma rays are then mapped by detectors outside of the body.