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Can Bob be in the same reference frame as Alice while freefalling in gravitational fields of different strengths No. General relativity doesn't have global frames of reference. See How do frames of reference work in general relativity, and are they described by coordinate systems?


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Sanity check: Am i correct that at this point Bob should observe Alice's watch ticking slower than his own as a result of the difference in their accelerating reference frames? Yes. What relative rate will Bob observe Alice's watch ticking at now that they are both in free-fall? $$ \Delta\tau_B = \frac{\sqrt{\frac1{r_s} - \frac1{r_{B0}}} - \sqrt{\frac1{...


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The answer depends on what is meant by "frame of reference". The two falling bodies, in slightly different gravities and at different speeds, are slightly (minutely) out of reference. However, the two bodies under normal gravity could be carrying on a radio conversation together without misunderstanding, even if their watches are minutely different. ...


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Is an object that is neutral buoyant the equivalent of the same object in the microgravity (free fall) of the International Space Station? In some ways they are very similar, in other ways, not so much. In either case, the net weight is (approximately) $0$. This is the major similarity. Some differences would depend on the medium that you are buoyant in. ...


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One obvious difference is that the neutral buoyancy state of the ping pong ball on water is broken if it is moved vertically. If it is lifted and released it will fall back to the water surface; if it it pushed down and released it will rise back to the water surface. Whereas the ping pong ball in the ISS can be moved in any direction. Another difference is ...


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For an object to be in free fall there must not be any apparent weight.What means is that if you stand on a weighing scale it would give a reading of zero given that it has zero velocity relative to you. In free fall the reading will be zero, you must have seen astronauts floating in space? They would be weight less as they would not "push" the weighing ...


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This question is answered really well by the answers above. Of course, as explained in all answers the statement quoted from your textbook is incorrect. The phenomenon of weightlessness could be explained in two different ways: The object orbiting the planet is in free fall (as described in other answers). The centrifugal force counteracts the earth's ...


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A freely falling body is a body which moves in the presence of gravity without any other external force acting on it. A ball dropped on the surface of the earth is the example that you have in mind. Now let's give some initial tangential velocity to the object (wrt the direction of free fall). Now is this case any different than the ball falling without ...


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When people talk about "free fall," what they are referring to is the absence of other forces besides gravity. For example, there are no "normal forces" from the ground holding the object up. This is a correct description of how satellites move. It is a confusing term for the reasons you bring up. "Free fall" does give a sense that something is moving ...


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You are absolutely right, and the caption conveys a common error. The textbook is wrong! Congratulations on discovering this error. Indeed, gravity at typical spacewalk height is not all that different from gravity at the surface. Because we have that $$g \propto r^{-2}$$ where $r$ is the distance from the center of the planet, we can estimate (using an ...


1

A satellite is falling, it's just also moving forward fast enough that it is always falling around the curve of the Earth. So it is in freefall. If you throw a ball straight out in front of you, you will see it follow a curve as it goes forwards and falls downwards. If you could throw it fast enough, and there were no air resistance, it's forward speed could ...


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May be Newton's cannonball-thought-experiment can convince your intuition that there is no principal difference between a body falling down to the earth and a satellite circulating the earth. Consider a cannon located on top of a high mountain, shooting in horizontal direction. And let's neglegt air resistance. (image from Wikipedia - Newton's cannonball) ...


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Yes it is doing free fall. Consider a satellite at some moment in time. Say its velocity at that moment happens to be in the direction of the pole star. If the satellite were not being attracted to the Earth, it would move in a straight line and go off in that direction, moving further and further away from Earth. But, because of the attraction to the ...


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The astronaut can stay afloat because he/she is in free-fall around the earth. Picture a cannon on a cliff shooting a projectile with increasing speed. Eventually, with enough speed, the projectile will make a whole round trip. In space, with less air-resistance and friction, this process can go on for a much longer (in theory infinite, in reality finite, ...


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Whatever written in the book is wrong. But let us assume it is correct then had it been the case then the moon would have flown away as well as the ISS along with the astronaut.


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The statement in the image is shockingly wrong. Given that the Earth is visible behind the astronaut, the reduction of the gravitational force due to distance must be fairly small; on the ISS for example the force of gravity is about 88% the value at the Earth's surface - very far from being negligible! The reason why the astronaut appears weightless is that ...


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You are correct that the caption is wrong. The cause of weightlessness is that gravitational and inertial forces cancel, exactly according to the equivalence principle.


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I'm afraid your book is wrong. At distances from the Earth where it's still as big as in this picture, the gravitational field is still very relevant. As you said yourself, the astronaut is in orbit around the Earth, and thus effectively in free fall, leading to the illusion of an absence of gravity.


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The gravity here is emergent not only from the body they are falling towards but also each other. They will fall on different paths due to the relative masses. The force of gravitational attraction is the same so the speed of movement in the vacuum is the same but they will not fall towards the massive body at the same rate. Individual objects on tiny scale ...


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Sure. You just need two manoeuvring thrusters at equal distances from the spaceship's centre of mass and firing with equal thrust in opposite directions. These thrusters then exert a torque (turning force) on the spaceship but the net force on the spaceship is zero, so its linear momentum does not change.


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This is an interesting (and illuminating) question. The question to ask is as to what causes the stress in a body. The idea is that if different parts of a solid body are subjected to such external forces that they have a tendency to accelerate at different rates then the internal forces of the solid body will activate themselves so as to try to keep the ...


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