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I just saw the most annoying video about an interesting physics question. Apart from the over the top stereotypical German (supposed to be funny or is he for real?) and arrogant " look how clever we physicists are " attitude, the most annoying is that the really interesting stuff about the question isn't even mentioned. Arrrggg - how can one be so ignorant?
As he mentions toward the end of the video – if you make it that far without smashing the screen – falling cheap hotels new york through a tunnel drilled all the way through the center of the earth to the other side results in a so called harmonic motion (neglecting air resistance of course). This basically means that you would move precisely as if you were attached cheap hotels new york to a spring cheap hotels new york that obeys Hooke's law .
Why? Well, the mass that still attracts you is proportional to the volume below you, which is proportional to the radius cubed . Gravitational attraction goes proportional to one divided by the radius squared . Thus, the force while falling is all the time simply proportional to the radius that you are at, which is precisely like Hooke's spring law. But this is only half of the interesting story that should be told.
Why is it that the time you take to fall through to the other side is precisely the same as a satellite needs to circle the earth at just above the surface of the earth (again, all neglecting that there is air and mountains and so on)?
It would be the perfect physics cheap hotels new york instructor suicide: Shooting a bullet horizontally, giving it the velocity to just circle earth at a constant height, shooting simultaneously with jumping off into the tunnel, the instructor cheap hotels new york falls and is hit by the bullet just when he reappears at the antipode at the opposite side of the tunnel.
Why does this perfect cheap hotels new york physics instructor suicide work? And please, cheap hotels new york NO MATHS ! I want a "philosophical" answer that shows deeper insight into physics – math trickery is for those who can't do any better. The answer is almost already given above.
This problem has a good use by the way. The velocity v of a satellite is simply gotten from the centrifugal force v 2 m / R ,and the way half around the earth is simply Pi times the radius cheap hotels new york R . Thus, the time t for falling through is also simply t = Pi* R / v , the same time as that for the satellite to go around. This way can be easier than getting the time from the harmonic motion, especially cheap hotels new york for those who do not like infinitesimal calculus but do for whatever reason find the formulas for circular motion and the accompanying forces easy to understand.
UPDATE: Sorry, why not at least give the time - also proves how easy it is this way: With the well known F = m g , it follows v 2 = g R , thus t = Pi ( R / g ) 1/2 . The radius is about 6370 km and g about 9.8 m/s 2 , thus the time to fall through is about 42 minutes .
Notice that the Pi is coming from earth's circumference. If going through Hooke's law instead (which is done here ), it takes longer and either you remember less memorable equations or you know how to derive cheap hotels new york them via integral calculus, and Pi will emerge from the "magic maths". I like it this way much better, because the the centrifugal (centripetal in the land where no hair is left undivided) is easy to remember as it just expresses the obvious proportionalities involved (OK, now punters will attack me on this point. Whatever.).
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Okay, I'm a first year physics student, so I may be way off here, but here's my attempt. SHM can be thought of as the 2d projection of a body rotating in 3d space. (Imagine cheap hotels new york a huge screen behind the Earth onto which you project the shadow of the bullet as it orbits the Earth.) In this case, both the instructor and the bullet cheap hotels new york have the same amplitude: approx. the radius of the Earth. Both are accelerating due to the force of gravity, but the instructor has a varying acceleration (as he travels towards the centre of the Earth, there is more mass "above" him than "below" him), whereas the bullet experiences a constant accelerating force. So even if the displacement of the bullet is larger, it travels that distance faster. (Though if you could view it on the screen, it would appear cheap hotels new york to travel at the exact same speed as the instructor.)
Imagine a huge screen behind the Earth onto which you project the shadow of the bullet as it orbits the Earth. Yes - this is the way! Now add independence of orthogonal forces and the fact that the forces at the beginning are the same. Then once more think about the shadow and the nature of harmonic motion as you already did (you did the most important part as far as I can tell), and it becomes totally obvious that the shadow of the bullet and the shadow of the instructor cheap hotels new york will always stay together. No derivatives, no wannabe scientific least action principles. Thank you for giving me back my hope in humanity today.
Yes, but it's the orthogonality that begs the question. You may as well do the whole thing back to front and insist on two half-bullets performing HM at right angles through the earth and in quadrature and adding the two together to get one bullet in orbit. Matter of fact I thought that might be what you were driving at but as I could see it turning into an argument about superposition I decided not to comment :)
The energy cheap hotels new york of the man is being swapped between kinetic in Y and potential, and the energy of the bullet is being swapped between kinetic in Y and kinetic in X. Perfect circular orbit for bullet and falling man SHM means energy is moving to and from Y with exactly the same pattern. Ie. falling man Y and bullet Y are in lockstep.
Does this mean that if the bullet had greater initial X velocity, then falling man Y and bullet Y would still be in lockstep? (I feel in my water that it does, but it's beyond me to prove it. Something like: X is orthogonal to Y so the scale of X is irrelevant. The timing of energy flow peaks remains the same.)
.. except, that doesn't make sense, does it .. what if bullet initial velocity X is escape velocity for that angle. That would mean the bullet would never return. = no lockstep between falling man Y and bullet Y. So, i guess, if the bullet has a greater initial velocity, and it has elliptic orbit, then it will take longer to arrive at the opposite side of the planet than falling man.
[red faced with embarrassment] .. lol .. it's even worse! .. The nadir of the bullet's Y (for elliptic orbit) cheap hotels new york may be positioned above the "south pole", but it would be way above it. Bullet Y and falling man Y not in lockstep at all. [/red faced with embarrassment]
The general elliptic case is where energy is swapping cheap hotels new york between kinetic X, kinetic Y, and potential energy. (I assumed scaling in X would make no difference, but force of gravity cheap hotels new york (potential energy) doesn't scale with geometry.) The two special cases are where kinetic X is constant (falling through the earth), and where potential is constant (circular orbit). So these 2 can be directly compared. Which means my assumption of simply comparing falling through the earth to the general case is rubbish.
You got there. The assumed rule that the acceleration depends only on the mass below you and the radius, only works with a solid sphere. If you shoot a bullet horizontally with enough velocity it will simply escape. With a bit less, it may take up an elliptical orbit but the orbital frequency will be less than that of the direct route through the earth or a circular orbit. So it only works if the bullet makes a circular orbit. Semicircular if an instructor pops out of the hole.
Gravity within and up to the boundary of the sphere of constant density is like a linear spring (with stiffness cheap hotels new york mg/R). So a mass in the field is a 2nd order undamped linear system (with (2.pi.f)^2 = stiffness/mass = g/R). But outside the sphere, gravity is no longer linear. So no more 2nd order linear system. (wow .. it took some time to clear enough barnacles off this bit of my brain)
And, because gravity upto the boundary cheap hotels new york of the globe is linear we can just examine the vertical aspect of the bullet because the horizontal aspect is independent. Which means, vertically, we're comparing a bullet falling from rest through the globe with a man falling from rest through the globe. Movement (frequency) is independent of mass, so they have exactly the same pattern of movement vertically.
You don't need the psuedo-Hook's law: any law that keeps the objects bouncing in the hole or circulating in orbit will do. You don't need to solve the differential equation. All you need is the two facts: that the x and y systems are independent and are resolved by cosine projection. edited: Note, I am not saying that other solutions exist, I am saying that in order to show that the arrival times are the same it is not necessary even to use the implicit maths that Sascha introduced. cheap hotels new york Principles like resolution, yes, actual solutions, be they Hook's law or harmonic motion, no :)
This is my dredged up understanding: .. it has to be a linear system. The falling man needs to be a linear second order system within the body of the planet, else he won't have sinusoidal falling. And that means he has constant mass, and the combination man/gravitational field has constant "stiffness". (We already knew he had sinus
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