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There are private space companies that will allow you to take a trip to space(it costs around $250,000). Take a trip and go in orbit around the earth or go to another space station. Another solution would be to go to the Zero-G place as Lasse said or fall farther as Rob Jeffries said which are both much less expensive.


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Establish a longer free fall time, for example in an aircraft or space station.


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The blue shift is actually necessary to conserve energy. The photon has energy, and therefore has potential energy relative to the neutron star. It loses gravitational energy as it approaches the neutron star, and it gains energy in the form of blue shift. Indeed, to the first order in GR, the change in energy of a photon is equal to the change in ...


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man made satellites uses gyroscope for orientation as they fall around earth. All these applications seem to be associated with gravity, therefore how can a gyroscope work in zero gravity? You are probably confusing or identifying the property of a gyroscope with the phenomenon of precession The bicycle wheel (gyroscope) doesn't fall down and ...


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Gyroscopes depend on the conservation of angular momentum. Orientation and navigation gyroscopes are finely balanced/symmetrized so that gravitational fields will not exert external torque and modify the angular momentum. As the container which holds the gyroscope moves, a gimbal mount allows the gyroscope to maintain a constant rotational axis ...


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Why does the mass of the orbitting object have no effect on its revolution at all? It does have an effect! However, the effect is immeasurably small if the orbiting object itself has a very small mass compared to the object it is orbiting. The most massive object we humans have put into orbit is the International Space Station, with a mass of 419.5 ...


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You are no doubt familiar with the apocryphal experiment of Galileo showing that falling bodies fall at a rate independent of their "weight". [ We should really say mass.] An orbiting body is just a special kind of falling body, albeit one that manages to miss the ground due its sideways motion. Hence the term "free fall" as used regarding astronauts or ...


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But can someone please explain why this is without using pure algebra? I will try without a single formula. In Newtonian gravity, the gravitational force on a particle is proportional to the particle's gravitational mass; the more gravitational mass, the more the gravitational force. In Newtonian mechanics, the acceleration of a particle, for a given ...


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Orbit is free-fall around a body, so it is for the same reason a feather falls as fast as a bowling ball (in a vacuum). In free-fall, $F = mg$ And we know that, $F = ma$ So we can substitute, $ma = mg$ And divide by $m$, $a = g$ Thus, no matter what mass is, acceleration equals $g$.


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You certainly know that all things fall at the same rate regardless of their mass (neglecting friction). An orbiting body is not different from a falling body in that the only force acting on it is the gravity of the thing it orbits, so there is no reason its mass should influence its orbit.


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Pete Brown wrote an essay which referred to this, see http://arxiv.org/abs/physics/0204044 and note the quote by Synge on page 20: "I have never been able to understand this principle. Does it mean that the effects of a gravitational field are indistinguishable from the effects of an observer’s acceleration? If so, it is false. In Einstein’s theory, either ...


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Why does an apple fall from a tree? Why do all objects accelerate towards earth at $9.8$ $m/s^2$? The 'out-of-the-box answer' is that the objects themselves don't move. It's the ground that rushes up! Regardless whether attached to the tree or not, Newton's apple is suspended motionless: it's earth's surface that accelerates up and meets the apple. This ...


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But doesn't a clock in an accelerating spaceship run at the same rate no matter where in the ship you put it? Remarkably, the answer is, even in the context of SR, no. It turns out that acceleration of an extended object is quite subtle. That is to say, we can't meaningfully speak of the acceleration of an extended object. Essentially, the 'front' ...



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