Easy physics puzzles
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9 members have voted
yes I think it is never really explained that if the 'falling' objects are really massive , everything changes
So it makes sense that the effect on even small objects is there even if tiny, it actually is a little unsettling to me to think the 'inertia' perfectly cancelled it all out
Quote: acesideDon’s consideration is very thorough; however, Earth is not a perfect sphere, so gravity at North Pole is very different from that at the Equator. Moreover, gravity is a field, so the equations you use are just a good approximation.
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Don's more massive ball still reaches the ground first (by a tiny fraction of a second) no matter where on Earth he stands. Einstein's general theory of relativity also has no relevance and no effect on the solution to this puzzle.
"gravity at North Pole is very different from that at the Equator" The accuracy of that statement would be improved by omitting the word "very".
I have climbed to the summit of Cotopaxi (almost 20,000 ft above sea level) in Ecuador (on the equator) and I can assure you that the weight of my backpack was not noticeably lighter than it was at home in Tennessee.
Quote: odiousgambit10^17 vs 10^18 is an order of magnitude so how is that "not by much"?...... and are you saying Galileo was wrong?
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"Not by much" in absolute terms.
Technically, Galileo was wrong, although in real world experiments, you have to take not only things like air resistance into account, but also, is the distance that each object falling the same? Put a grain of sand on the ground directly underneath the lighter object, and it will get to the ground first. Also, if the objects aren't the same shape, the centers of mass aren't the same distance from the center of Earth. In the general sense, Galileo was correct - as far as can be observed, two objects of different sizes will fall at the same velocity.
Here's a bonus problem that I have not worked out yet; again assuming the only forces are gravity between Earth and the objects, from how high would they have to be dropped before the heavier one lands one microsecond (0.000001 second) before the other? The problem here is, as the balls drop, r in the equation changes, so it's probably not a straightforward integral.
Quote: gordonm888
Don's more massive ball still reaches the ground first (by a tiny fraction of a second) no matter where on Earth he stands.
I just thought about this again. If the gravitational acceleration is the same everywhere for two objects of different masses, then the two objects arrive to the ground at exact the same time. I don’t understand what you and Don are questioning?
Quote: acesideQuote: gordonm888
Don's more massive ball still reaches the ground first (by a tiny fraction of a second) no matter where on Earth he stands.
I just thought about this again. If the gravitational acceleration is the same everywhere for two objects of different masses, then the two objects arrive to the ground at exact the same time. I don’t understand what you and Don are questioning?
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The "gravitational acceleration" of Earth pulling on objects is the same regardless of mass. However, you also have to take into account the gravitational acceleration of those objects pulling on Earth as well, and those aren't the same if the pulling objects have different masses.
Quote: acesideI briefly read it, but my understanding is that you are saying the gravitational acceleration is smaller at a higher altitude, but that does not change the result that two objects of different masses will reach the ground at exactly the same time.
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The Earth's gravity acts on the two balls with equal acceleration; but each of the balls exert gravity on the Earth and the more massive ball causes its acceleration to be slightly higher than the less massive ball.
Quote: acesideNo, I don’t think so. I believe the gravitational acceleration is the same, so the time it takes for an object to travel is exactly the same. Maybe you are not confident about the reference frame. Put it on Earth first.
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"On" Earth, each ball's movement is obstructed by Earth itself.
Like I said, the "gravitational force" that Earth exerts on each ball is proportional to their masses, so the acceleration of the balls toward Earth is the same.
However, you also have to take into account the "gravitational force" that each ball exerts on Earth. Each object pulls the other closer to each other; it's not just Earth that is doing the pulling. It only seems that way because of the relative sizes.
Quote: acesideI guess your question is not well stated. Why not you draw a picture and a reference frame to restate your question? After that, we continue on this.
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I don't think a picture is necessary.
Think of it this way: you have two rafts on a lake, each with a person on it. They are both holding ends of the same rope.
Both are pulling on the rope.
Each one is pulling the other one towards themself.
If the frame of reference is "person 1," then as far as person 1 is concerned, both person 1 pulling and person 2 pulling are moving person 2 closer to person 1.
This is what is happening with the ball and Earth. Earth is pulling the ball toward it, causing it to accelerate at about 9.81 m/sec2; this is regardless of the mass of the ball. At the same time, the ball is pulling Earth toward it, although the acceleration is much, much less - for the 5 kg ball, it is 8.2217 x 10-18 m / sec2. The height of the STRAT tower is about 350 m, so it would take about 8.5 seconds for the ball to reach Earth, during which time, the ball caused Earth to accelerate toward it by enough so it would have moved 3 x 10-16 m closer. For the 10 kg ball, Earth would have moved 6 x 10-16 m closer. The heavier ball will hit Earth while the ligher one is still 3 x 10-16 m away.
The main reason you don't notice this: this is 1/5 of the diameter of a proton.
Quote: acesideTo describe this problem more accurately, we must setup a reference frame first. Let’s put these three objects in the solar system and draw xyz coordinates on the Sun. It seems you are saying two heavier objects will collide more quickly than two lighter objects do. Is this true?
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Everything else being equal, yes, they will.
The gravitational force that every object pulls on every other object is G m M / r^2, as I said in my answer here. Since acceleration is force divided by mass, each object will cause another object to accelerate towards it by G M / r^2, where M is the mass of the pulling object, and r the distance between the objects' centers of mass.
Remember that each object is pulling the other one towards it simultaneously.
Let's go with your idea, and use the center of the Sun as the origin (x = y = z = 0). Also assume that the center of Earth and the center of the ball are on the positive x-axis (i.e. y = z = 0, and x > 0); however, for purposes of objects pulling each other, ignore everything except Earth and the ball.
The initial center of mass of Earth is at x = S, where S is the distance from the center of the Sun - about 150 million km.
The initial center of mass of the ball is at x = S + d, where d is the initial distance from the center of the ball to the center of Earth.
In the first second, Earth exerts a pull on the ball that causes it to accelerate at about 9.81 m/sec2, so after one second, its velocity is now 9.81 m/sec in the negative-x direction, and it has traveled about 4.905 m in that direction, so its location is x = S + d - 4.905. At the same time, the ball exerts a pull on Earth that causes Earth to accelerate at about 8.2217 x 10-18 m / sec2 (for the 5 kg ball), so after one second, its velocity towards the ball is 8.2217 x 10-18 m / sec in the positive-x direction, and it has traveled about 4.111 x 10-18 m in that direction, so its location is x = S + 4.111 x 10-18.
The distance between the two, if you ignore the rounding, is d - 4.904 999 999 999 999 995 889.
On the other hand, the 10 kg ball would have pulled Earth toward it with an acceleration of twice as much, or 16.4434 x 10-18 m / sec2, so Earth's velocity would now be 8.2217 x 10-18 m / sec, and its location x = S + 8.222 x 10-18.
In this case, the distance between the two is d - 4.904 999 999 999 999 991 778, which is very, very slightly closer.
Another problem, the Sun is moving all the time, so the reference frame is moving constantly. Is there any acceleration on this reference frame?
Quote: acesideI understand what you are saying, but every object is pulled by all other objects in the solar system, so all these objects will collide together someday. That’s going to be a doomsday.
Another problem, the Sun is moving all the time, so the reference frame is moving constantly. Is there any acceleration on this reference frame?
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The problem assumes that the only forces are the ball(s) and Earth. In real life, the Sun and Earth, as well as the Moon and Earth, definitely pull on each other; this is why tides exist.
The reason, say, Earth doesn't collide with the Sun - or the Moon isn't pulled into the Earth, for that matter - is the same reason neither the Moon nor Earth just fly off into deep space. Somewhere along the line, they had some "initial velocity" perpendicular to the object they are orbiting; when you apply the gravitational pull of the larger object to it, it bends the path so it becomes an orbit. There is a whole subset of physics about this - "orbital mechanics."
Yes, multiple objects pulling on each other causes problems. Even as few as three objects has been scratching heads for centuries - do a search on "three body problem."
Quote: acesideThis is a physics thread, so I am trying to get the physics out of this discussion. It’s not pure math like the other thread. Another thing I am trying to stress is, I just do not want anybody to question Galileo.
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It depends on what you mean by, "question Galileo." Nobody is saying that the Sun orbits Earth.
For all intents and purposes, Galileo is correct; things fall at the same rate regardless of their mass.
I was just pointing out that, under very controlled circumstances (which, technically, can't exist), a heavier object would fall faster than a lighter one.The problem was theoretical.
Quote: ThatDonGuy
The main reason you don't notice this: this is 1/5 of the diameter of a proton.
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I believe the engineers agree that 15 digits of pi is "enough", and to start walking down the hallway even if you can only cover one half of the remaining distance per day...
Quote: gordonm888ThatDonGuy has correctly explained the physical principles that were relevant to the stated problem. This is high-school level physics. The inertial reference frame of the problem was the Earth, however it happens to be moving or accelerating in the solar system and in the Milky Way galaxy. Sorry for being grumpy, but why is this being expanded to a discussion of celestial mechanics with solar-system-level non-inertial reference frames?
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Confusing all this, I believe, is the balls being dropped separately in two different trials. When we think about Galileo's tower drops we picture two objects being dropped together for comparison.
If the balls were dropped together, the torque on the surface of the earth by the moment between the two balls would surely produce an action of less than h/2pi so we can't say anything about which one lands first.
ah but they have figured it out for 1 million digits for [maybe only one] legit reason, if it had been shown that a pattern was repeating, unlikely as that was, then my understanding is it would prove pi is not actually an irrational numberQuote: Dieter(¡snip!)
I believe the engineers agree that 15 digits of pi is "enough", and to start walking down the hallway even if you can only cover one half of the remaining distance per day...
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Quote: Dieter(¡snip!)
Quote: ThatDonGuy
The main reason you don't notice this: this is 1/5 of the diameter of a proton.
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I believe the engineers agree that 15 digits of pi is "enough", and to start walking down the hallway even if you can only cover one half of the remaining distance per day...
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The Heisenberg uncertainty principle states (Δx Δp) is greater than or equal to hbar/2. While it doesn't set a strict "minimum distance" a particle can travel, measuring position (Δx) with extreme precision requires immense momentum (Δp). This creates a theoretical barrier around the Planck length (approx 1.6 x 10-35 meters). At this scale, the energy required to probe such a tiny distance becomes so concentrated that it theoretically creates a micro-black hole, preventing further observation. Many physicists use this to suggest that distances smaller than the Planck length may be fundamentally unmeasurable.
So theroetically it takes 35 digits of accuracy to become unmeasurable, not 15
Quote: gordonm888Quote: Dieter(¡snip!)
Quote: ThatDonGuy
The main reason you don't notice this: this is 1/5 of the diameter of a proton.
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I believe the engineers agree that 15 digits of pi is "enough", and to start walking down the hallway even if you can only cover one half of the remaining distance per day...
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The Heisenberg uncertainty principle states (Δx Δp) is greater than or equal to hbar/2. While it doesn't set a strict "minimum distance" a particle can travel, measuring position (Δx) with extreme precision requires immense momentum (Δp). This creates a theoretical barrier around the Planck length (approx 1.6 x 10-35 meters). At this scale, the energy required to probe such a tiny distance becomes so concentrated that it theoretically creates a micro-black hole, preventing further observation. Many physicists use this to suggest that distances smaller than the Planck length may be fundamentally unmeasurable.
So theroetically it takes 35 digits of accuracy to become unmeasurable, not 15
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Once the error sphere is down to an arm's length, I am comfortable asking for directions when I get close.
Stand on the middle of a disc that will rotate, and hold in your hand on a stick a disk that will rotate, if you hold the latter straight up and start spinning it, the disc under your feet will rotate in the opposite direction. It's conservation of angular momentum.
What happens if you start this process and without stopping the spin hold the disc in your hand straight out 90 degrees from the disc you are standing on?
What happens when, again changing nothing else, you hold it straight down 180 degrees?
Huh. I’m starting to wonder if this stumped folks. Lots of brilliant folks here, been some talk about torque too, so I predict responses soon.Quote: odiousgambitNew one
Stand on the middle of a disc that will rotate, and hold in your hand on a stick a disk that will rotate, if you hold the latter straight up and start spinning it, the disc under your feet will rotate in the opposite direction. It's conservation of angular momentum.
What happens if you start this process and without stopping the spin hold the disc in your hand straight out 90 degrees from the disc you are standing on?
What happens when, again changing nothing else, you hold it straight down 180 degrees?
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I saw a demonstration of this once, just like described, otherwise I would probably get the answer wrong. What I know about angular momentum isn’t that much… of course we all know that a spinning object requires force to stop the spinning. If it was not spinning in the first place, it required energy to get it going. That energy is not going to just disappear. What is not so obvious is that an object can become a system in which this momentum/energy has to follow a logic … ironically that can be counter-intuitive since it’s a logic that is invisible to us.
btw googlebot gave me the wrong answer!
Quote: ThatDonGuyQuote: acesideThis is a physics thread, so I am trying to get the physics out of this discussion. It’s not pure math like the other thread. Another thing I am trying to stress is, I just do not want anybody to question Galileo.
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It depends on what you mean by, "question Galileo." Nobody is saying that the Sun orbits Earth.
For all intents and purposes, Galileo is correct; things fall at the same rate regardless of their mass.
I was just pointing out that, under very controlled circumstances (which, technically, can't exist), a heavier object would fall faster than a lighter one.The problem was theoretical.
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I’ve thought about the two-body acceleration problem. Here is how I would handle it. To the first degree of approximation, we treat the smaller object as a free-falling mass in a stationary gravitational field, and the bigger one as an accelerating object in a weak free-falling gravitational field. Is this the method you used to calculate the moving of Earth?
Quote: odiousgambitah but they have figured it out for 1 million digits for [maybe only one] legit reason, if it had been shown that a pattern was repeating, unlikely as that was, then my understanding is it would prove pi is not actually an irrational numberQuote: Dieter(¡snip!)
I believe the engineers agree that 15 digits of pi is "enough", and to start walking down the hallway even if you can only cover one half of the remaining distance per day...
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A million? I think 300 trillion is the current record. Still irrational.
It leads me to believe you can never prove it's irrational or rational that way because you'd have to count to infinity to prove it never repeats or stops repeating.
Some of my conclusions may be incorrect, let me know, but I have seen this demonstration so I know what happensQuote: odiousgambitNew one
Stand on the middle of a disc that will rotate, and hold in your hand on a stick a disk that will rotate, if you hold the latter straight up and start spinning it, the disc under your feet will rotate in the opposite direction. It's conservation of angular momentum.
What happens if you start this process and without stopping the spin hold the disc in your hand straight out 90 degrees from the disc you are standing on?
What happens when, again changing nothing else, you hold it straight down 180 degrees?
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OK, maybe I don’t have any takers trying to guess. Angular Momentum does seem to be passed over pretty quickly in physics courses, which might be too bad because there are surprises within to keep your attention.
Intuition is a funny thing. The first time you ever saw a skater start to spin with arms etc held out, then bring it all in, causing the spin to increase dramatically, perhaps you were quite surprised. But now you’ve seen it so many times your intuition is set on the idea of what will happen, and you might laugh at a child who never saw it before.
The demonstration as described above has surprising elements as well. Unlike the skater, it shows things that you usually never saw before and will find counter-intuitive. What is unusual is that you have an isolated system that initially has no spin. Then you have two spinning objects within the isolated system, and this system must preserve the inserted momentum, redistributing the spin if necessary. Your intuition may try to tell you what happens in the second part is impossible! But what is actually impossible is that the spin would be redistributed in a way that the total amount of energy is increased or decreased, intuition be damned.
What happens if you start this process and without stopping the spin hold the disc in your hand straight out 90 degrees from the disc you are standing on?
In the demonstration I saw, the disc the man was standing on stopped spinning. A spinning object remembers it’s axis in a profound way and will not choose a different orientation! I think this means the spinning object the man held spun even faster to conserve the momentum, having to hold it all, but I don’t remember noticing that.
What happens when, again changing nothing else, you hold it straight down 180 degrees?
The disc the man is standing on rotates in the opposite direction from what it did in the beginning! This may not have the effect on you, reading this, that it did on the audience, which was astounded and gasped.
The held spinning object viewed from above was held upright and spun in a clockwise motion, say. When it was pointed down, viewed from above, that is now counter-clockwise [from the man’s perspective it doesn’t change]. The disc he is standing on must now change and rotate clockwise as surely as the skater doing the maneuver must spin faster.
PS: now finding more, using simply "angular momentum" for searching
Quote: odiousgambitDidn't think I would find anything on youtube after trying for a while. But finally found this, similar to what I saw.
PS: now finding more, using simply "angular momentum" for searching
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That is a good one! And of course the wheel slows down when he flips it, and regains its speed when he returns it to the original orientation.
Pretty cool if some figure skaters were to try something like this. Start with a little girl on skates, spinning as fast as she can. Then a lady skater comes up, picks her up and they're both spinning. Then her male partner picks them up and he's spinning. Finally a huge hockey player skates in and picks the lot of them up and they're still spinning, albeit very slowly now.
Quote: googlebot[the inventor] Bendik, an aerospace engineer, created the educational toy in the late 1980s and chose the name because Euler's equations formed the foundation for describing the complex motion of spinning and rolling rigid bodies.
If you got this going somewhere without an atmosphere how long would it keep going?
Another part of motion here is disc rolling.
you've got me wondering about thisQuote: acesidePeople often confused spinning and disc rotation. In my opinion, there is no spinning here. A disc spinning is the rotation of the disc around its central axis out of the disc, but here the rotation is around its diameter in the disc. Is this right?
Another part of motion here is disc rolling.
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I think what we see is the way a disc dissipates the energy by friction and air resistance if there is *some* spin. If there is no spin the coin might roll in a circle ... that would require setting it up too with a tilt?
Not sure about this
Quote: odiousgambityou've got me wondering about thisQuote: acesidePeople often confused spinning and disc rotation. In my opinion, there is no spinning here. A disc spinning is the rotation of the disc around its central axis out of the disc, but here the rotation is around its diameter in the disc. Is this right?
Another part of motion here is disc rolling.
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I think what we see is the way a disc dissipates the energy by friction and air resistance if there is *some* spin. If there is no spin the coin might roll in a circle ... that would require setting it up too with a tilt?
Not sure about this
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Physicists use polar, speherical or cylindrical coordinate systems and mathematical symbols to describe the motion. Words are less rigorous although certainly there are different meanings to words like rotation and revolution. In general 'spinning' is rotation around some particular axis in some particular
Quote: odiousgambitAccidentally came across Euler's Disc
Quote: googlebot[the inventor] Bendik, an aerospace engineer, created the educational toy in the late 1980s and chose the name because Euler's equations formed the foundation for describing the complex motion of spinning and rolling rigid bodies.
If you got this going somewhere without an atmosphere how long would it keep going?
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That would depend on the compressive strength and elasticity of the materials. That should be the only energy sink in vacuum.
Do it with a tub stopper on a cardboard plate, not very long.
Imagine there are three points in the universe Point A is at sea level, on Earth, at noontime during an equinox. Point B is what I call the gravitational point Nemo in the universe, which is the physical point in the known universe furthest away from all known sources of gravity and is the point where all gravitational forces add up to as close to zero as is possible. Point C is the asymptotic point just outside the event horizon of a black hole, where the gravitational force is so high the escape velocity is as close to c (light speed) as physically possible and still allow for escape.
What is the time difference between those three points? If one second ticks away at point A (Earth), how much time will elapse at point B with its lowest gravity in the universe and how much time will elapse at point C with the highest possible gravity that still allows for escape velocity?
Thanks!
Quote: GenoDRPhI have a physics puzzle, the may-or may not be easy. I legit don't know the answer,which is why I'm asking here. Has to do with gravitational time dilation.
Imagine there are three points in the universe Point A is at sea level, on Earth, at noontime during an equinox. Point B is what I call the gravitational point Nemo in the universe, which is the physical point in the known universe furthest away from all known sources of gravity and is the point where all gravitational forces add up to as close to zero as is possible. Point C is the asymptotic point just outside the event horizon of a black hole, where the gravitational force is so high the escape velocity is as close to c (light speed) as physically possible and still allow for escape.
What is the time difference between those three points? If one second ticks away at point A (Earth), how much time will elapse at point B with its lowest gravity in the universe and how much time will elapse at point C with the highest possible gravity that still allows for escape velocity?
Thanks!
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NEVERMIND. I gave a response about special relativity and that's clearly not what you were asking about.
The event horizon of a black hole is the ultimate boundary, point of no return, and mathematical surface where the pull of gravity becomes so strong that nothing can escape it, not even light.
This is way too astronomical for an easy physics puzzle, I believe.
Quote: gordonm888Quote: GenoDRPhI have a physics puzzle, the may-or may not be easy. I legit don't know the answer,which is why I'm asking here. Has to do with gravitational time dilation.
Imagine there are three points in the universe Point A is at sea level, on Earth, at noontime during an equinox. Point B is what I call the gravitational point Nemo in the universe, which is the physical point in the known universe furthest away from all known sources of gravity and is the point where all gravitational forces add up to as close to zero as is possible. Point C is the asymptotic point just outside the event horizon of a black hole, where the gravitational force is so high the escape velocity is as close to c (light speed) as physically possible and still allow for escape.
What is the time difference between those three points? If one second ticks away at point A (Earth), how much time will elapse at point B with its lowest gravity in the universe and how much time will elapse at point C with the highest possible gravity that still allows for escape velocity?
Thanks!
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You appear to be confused about some aspects of physics that are, frankly, very easy to misunderstand.
Time dilation is observed when there is a large difference in relative velocity between two objects. That is Einstein's special theory of relativity.
Gravity bends space-time (Einstein's General theory of Relativity) but the rate at which time passes is not directly affected by gravity. However, gravity fields do affect how fast an object will accelerate which obviously can change the instantaneous velocity of an object. If you are on the surface of a planet you are not moving (relative to the planet) -no matter how strong the planet's gravity is. however, if you are falling into the planet and approaching its surface, you will have a finite (usually small) velocity relative to the planet and there will be some microscopic relative time dilation between you and someone on the surface of the planet.
A black hole has an event horizon - which is the distance from the black hole from which the escape velocity exceeds the speed of light (thus no information escapes from the black hole because no object can move faster than light.) If you threw a tennis ball towards a black hole it would continuously accelerate, go through the event horizon but its velocity relative to you would never equal or exceed the speed of light. The instantaneous velocity of the tennis ball at the event horizon would vary depending upon its history in arriving there.
So your question about time dilation relative to objects on the earth's surface, at point Nemo and at the escape horizon of a black hole depends upon the velocity of the objects that are present at those points. You could say the Earth is rotating around its axis, revolving around the sun and that the Sun is revolving around the center of the Milky Way galaxy. Whereas the object at Point Nemo is what?? - not moving at all? If it's not moving at all, then how did it get there? An object at the event horizon of a black hole is likely moving at a velocity that is a very high fraction of the speed of light - but the relative time dilation is very sensitive to the exact ratio of (v/c). And you can posit a weird theoretical case: a virtual particle could blink into existence at the event horizon of a black hole with a very low instantaneous velocity -say, lower than the Earth's velocity.
As for the Earth vs nemo: If the object at Nemo is defined to be stationary and relative to Nemo the earth is moving at one-millionth of the speed of light then time passes on Earth by one part in 1012 slower than it does at Nemo.
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Shouldn't we be analyzing this one in terms of general rather than special relativity, accelerations rather than velocities? In the sense of the Harvard experiment where an atomic clock on the top floor of the building runs faster than one at ground level, due to the different gravitational acceleration.
I think the idea of escape velocity is irrelevant, being acceleration is acceleration (the Equivalence Principle!) but more important is that fact that the time difference has to be measured by an observer who also has an acceleration. So an observer with an atomic clock at each point will notice nothing unusual about his, but the other two will be off, assuming he is able to get information from them.
Quote: AutomaticMonkey
Shouldn't we be analyzing this one in terms of general rather than special relativity, accelerations rather than velocities? In the sense of the Harvard experiment where an atomic clock on the top floor of the building runs faster than one at ground level, due to the different gravitational acceleration.
I think the idea of escape velocity is irrelevant, being acceleration is acceleration (the Equivalence Principle!) but more important is that fact that the time difference has to be measured by an observer who also has an acceleration. So an observer with an atomic clock at each point will notice nothing unusual about his, but the other two will be off, assuming he is able to get information from them.
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You're correct. I overlooked gravitational time dilation, which is clearly and explicitly what GenoDrPh was asking about. I'm deleting my answer, but it will live on in your response.
The gravitational time dilation ratio for an object in a gravitational field is sqrt(1-2GM/rc2) where r is the radial coordinate of the observer in the gravitational field and M is the mass of the sphere causing the gravitational field. In principle you could use this equation to get a solution to the question that has been posed. I think that for the event horizon of a black hole that r is taken to be essentially zero, so its a singularity and cannot be properly calculated.
Quote: gordonm888Quote: AutomaticMonkey
Shouldn't we be analyzing this one in terms of general rather than special relativity, accelerations rather than velocities? In the sense of the Harvard experiment where an atomic clock on the top floor of the building runs faster than one at ground level, due to the different gravitational acceleration.
I think the idea of escape velocity is irrelevant, being acceleration is acceleration (the Equivalence Principle!) but more important is that fact that the time difference has to be measured by an observer who also has an acceleration. So an observer with an atomic clock at each point will notice nothing unusual about his, but the other two will be off, assuming he is able to get information from them.
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You're correct. I overlooked gravitational time dilation, which is clearly and explicitly what GenoDrPh was asking about. I'm deleting my answer, but it will live on in your response.
The gravitational time dilation ratio for an object in a gravitational field is sqrt(1-2GM/rc2) where r is the radial coordinate of the observer in the gravitational field and M is the mass of the sphere causing the gravitational field. In principle you could use this equation to get a solution to the question that has been posed. I think that for the event horizon of a black hole that r is taken to be essentially zero, so its a singularity and cannot be properly calculated.
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As I understand it there is a nonzero r that is proportional to the mass, and for a very large black hole the da/dr as you cross the event horizon is small. So for something like the supermassive black hole at the center of a galaxy an astronaut crossing the event horizon wouldn't notice anything at all, there would be no "spaghettification" or anything like that. But time would stop for him relative to everything outside the black hole, so if he looked behind him he'd see all the stars burn out, and what happens after that he'll know but won't be able to tell us. People on the outside watching him fall in would stop seeing him fall, or stop seeing him do anything at all because the light from it would be so redshifted a photon of it would have insufficient energy to produce any action. That's assuming he's freefalling and there would be a mix of general and special relativistic effects. If he had a physically impossible rocket capable of holding him steady at the event horizon he could stop and there would only be the GR effect due to gravity, but that rocket would be releasing so much energy its mass would rapidly be depleted also due to the famous E=mc2 and that would be a more complicated GR problem, what an outside observer sees.
Being a black hole can have charge, that's another thought experiment I've thought of- what would happen if a charged particle approached a black hole of the same charge? The charge of the black hole seems like it has to be on the surface, and we get that division-by-zero problem if we try to make the charged particle touch the surface, but there should be some point where the gravitational attraction and electrostatic repulsion cancel each other out and the charged particle just hovers there, unless there is some mechanism by which the charge and mass of a fundamental particle can be separated.
See, one semester I had spent a weekend in the apartment of two freshman coeds. When we got back to campus, they gossiped, and that week I was called into the Dean of Faculty's office to explain myself.
He apparently did not like my response. I told him that in my position as a professor of physics, it was perfectly acceptable for me to be performing the... performing the...
Quote: AutomaticMonkeyJust full disclosure here- I actually was a professor of physics. Until they fired me. "Moral turpitude," they called it.
See, one semester I had spent a weekend in the apartment of two freshman coeds. When we got back to campus, they gossiped, and that week I was called into the Dean of Faculty's office to explain myself.
He apparently did not like my response. I told him that in my position as a professor of physics, it was perfectly acceptable for me to be performing the... performing the...Young Double-Slit Experiment!
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LMFAO.
Well, I'll defer to you. I earned a BS in physics long ago and completed courses for a PhD in Physics but couldn't take the time off from my job to do a PhD thesis.. I was more interested in atomic and nuclear physics than in cosmology. My professional reputation was made in Nuclear Engineering and Energy Technology R&D, not gravitation.
Over the total time-span of Earth (4.6 billion years), a clock set in a geostationary position at an altitude of 9,000 meters above sea level would be about 39 hours ahead of a clock set at sea level.
Let me translate it into easy physics: a person who lives in the Himalayas becomes older than his twin brother who lives in New Delhi does.
Is this correct? Or younger?
The following conversation took place between two people travelling in space:
"Did that object suddenly turn blue?"
"Yes - that's an example of 'blueshift.' As you approach an object, its color shifts towards the blue end of the spectrum."
"Except...we're moving away from it."
Explain what is actually happening.
Quote: ThatDonGuyOkay, time for an actually easy physics puzzle - probably a little too easy
The following conversation took place between two people travelling in space:
"Did that object suddenly turn blue?"
"Yes - that's an example of 'blueshift.' As you approach an object, its color shifts towards the blue end of the spectrum."
"Except...we're moving away from it."
Explain what is actually happening.That definition of blueshift is not entirely accurate
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