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87

Cory, here's a different way of thinking about gravity assists that may help: First is my short answer for readers in a hurry: What is really going on is a giant game of pool, with fast-moving planets acting as massive cue balls that impart some of their energy when they whack into tiny spacecraft. Since you can't bounce a spacecraft directly off the ...


61

Can photons push the source which is emitting them? Yes. If yes, will a more intense flashlight accelerate me more? Yes Does the wavelength of the light matter? No Is this practical for space propulsion? Probably not Doesn't it defy the law of momentum conservation? No In fact that last question is the key one, because photons ...


46

When light is propagating in glass or other medium, it isn't really true, pure light. It is what (you'll learn about this later) we call a quantum superposition of excited matter states and pure photons, and the latter always move at the speed of light $c$. You can think, for a rough mind picture, of light propagating through a medium as somewhat like a ...


43

This is actually the paradox that led Einstein to the equivalence of mass and energy, and to General Relativity. Consider a special case: An electron and positron are at the Earth's surface. Bring them together and they annihilate, creating gamma rays (which is very energetic light). The gamma rays travel up to the Space Station, where they are converted ...


34

Just for completeness, I'll explain how to obtain the time taken for an arbitrary curve. If $h$ is the initial height of the child and $y$ the height once he has started falling. By energy conservation: $$mgh=mgy+\frac{1}{2}mv^2\implies v=\sqrt{2g(h-y)}\tag{1} $$ We know know the speed at any time. Let us denote the horizontal position as $x$. The ...


34

Energy is in fact conserved, even in gravitational slingshots. After the slingshot, the velocity of the spacecraft may indeed change, which means its kinetic energy will also change. If this happens, the energy increase (or decrease) will be made up by a commensurate decrease (or increase) in the kinetic energy of the planet. In plain English: The planet ...


25

The shape of the slide definitely determines how long it takes to go down it. Consider if the slide was completely vertical. Now, a certain famous [recently deceased :( ] comedian had the astute observational powers to point out that this would, in fact, be a drop, not a slide. Nevertheless, you would quickly reach the bottom. Now imagine if the slide was ...


23

Why does the author say that we would need to know the shape of the slide to find the time taken for the child to reach bottom of the slide? As you've discovered, the speed going down a frictionless slide only depends on the vertical distance. This speed is not the vertical component of velocity. It is the magnitude of the velocity. The vertical ...


18

MSalters already said "yes". I would like to expand on that by computing the change. Let's take a 10 kg cannon ball, made of lead. Heat capacity of 0.16 J/g/K means that in dropping from 1000 K to 100 K it has lost $10000\cdot 900 \cdot 0.16 \approx 1.4 MJ$. This corresponds (by $E=mc^2$) to a mass of $1.6 \cdot 10^{-11} kg$ or one part in $6\cdot 10^{11}$. ...


16

Work is calculated as force times distance. $$W = Fd$$ The purpose of a simple machine like a screw jack is to lessen the force required. However, the work needed is still the same, so the distance over which you exert the force has to increase. Halving the force requires doubling the distance. In this problem, you want to lift 2000 lbs a distance of 1 ...


16

1. If the universe is expanding, what is it expanding into? The universe isn't expanding into anything. Space-time isn't curving into a higher-dimensional space. So what do we mean by "curved" and "expanding", words usually having a meaning only for objects in space? The answer is it is just an analogy. Mathematicians have found properties of space an ant ...


16

Of course, it does, since: $$\frac{\partial E}{\partial t} = \frac{\partial }{\partial t} \left(m \cdot c^2 \right) $$ Very little, though


16

Can photons push the source which is emitting them? Yes, photons have momentum and momentum must be conserved. The source is pushed in the opposite direction of the photons. If yes, will a more intense flashlight accelerate me more? Yes, more photons means greater momentum. Does the wavelength of the light matter? Yes, shorter wavelength ...


16

Your guess at the solution to this paradox is correct. "Pumping energy up" to the space station, regardless of the method you choose, would require an input of at least the amount of energy you would gain in kinetic energy on the way down. This is just a variation on the impossible perpetual motion machine concept. In practice, you would not only not gain ...


15

If you could take from orbital energy, then it would decrease, until at some point in the future it would zero. Hence, it can't be perpetual.


13

In your working you have assumed that $a=g$ - this is true if the slide is vertical. Slides will have some angle, $\theta$ (e.g. $45^\circ$), which will mean that the acceleration, $a$ is given by $$a = g ~sin \theta$$ Note that $a$ will be less than $g$ because the value of the $sin$ term will be between $0$ and $1$. (except in the case of a vertical ...


13

In step 1 you lower the mass and this generates energy. Let's say you store this energy is a spring, and for the sake of argument let's say the energy stored is 1J. The energy has to come from somewhere, and of course it comes from the rotational energy of the torus so the rotational energy of the torus is now 99J. In step 2 you slow the torus, perhaps by ...


13

I believe, the answer is a small but quantifiable, yes, there is a non flat road configuration that would lead to better gas mileage between any two points at the same height. I have numerically solved for such an optimal path. I believe I can give a nice explanation of why that is, but it will take some work, so bear with me. Granted, you can only expect ...


11

A classical explanation to supplement Rod's excellent quantum mechanical one: If you make a Huygens construction of wave propagation (I assume you know how to do that) then every point on the wave front is treated as the source of a new wave of the same frequency and phase. How that wave propagates depends on the medium it encounters. So the Huygens ...


11

The idea of partitioning energy into different forms like "mechanical energy" or "chemical energy" and such is actually arbitrary. More or less by definition, energy is that which is conserved unter time translations by Noether's theorem. If what you call "mechanical energy" has changed, then there is another term in the Noetherian energy that has changed ...


11

We already harvest energy from the Moon. It causes the tides and stress and strain and motion throughout the Earth. As a result, the Moon keeps getting farther away. (And it causes some heating in the Earth). The Moon at one time had a spin that was not locked to the Earth, and the tidal bulges in the Moon's shape caused by the Earth generated heat in the ...


10

First, the energy expectation value of the superposition state you have written down is $$ \left(\frac{n_1 + n_2}{2} + \frac{1}{2}\right)\hbar\omega $$ and one might naively conclude that therefore the energy of the state lies in between the energy of its constituents. This naive concept doesn't work, though - the "energy" of a state that is not an energy ...


10

A simple counterexample: Imagine two particles with opposite direction and equal speed. The center of mass does not move, yet the kinetic energy of the system is non-zero. Now let both particles come to rest (by friction, hitting a wall, whatever). The kinetic energy is now zero, and total momentum has been conserved, while energy is not. The crucial ...


9

Yes, when you fire a pistol the hammer hits the bullet with a relatively small initial kinetic energy but the kinetic energy of the hammer and bullet after the collision is considerably higher. This may seem a silly example, but I think it actually highlights the important principle involved. In general when two bodies undergo an inelastic collision part of ...


9

You say: Imagine a book that we lift it with a force that is exactly equal to the force of gravity so the forces cancel out and the book moves with a constant velocity. so I'm guessing your reasoning is that the net force on the book is zero so the amount of work done on the book is zero. And you are absolutely correct - no work is done on the book and ...


9

...two roads of the same length. One road is flat, the other road goes up and down some hills. Will an automobile always get the best mileage driving between the two towns on the flat road versus the hilly one..? The question two roads (between two towns A, B) with a different profile cannot be of same length The flat road (1) is shorter, ...


8

"The end state in A and B is the same." This is the fallacy. The total state has to include the potential field of the gravitational field. If you do things via A, the potential field disturbance has not yet propagated to the right-most body, so the potential energy field has not yet settled down. After a few more seconds, the disturbance will reach the ...


8

The answer is yes. The de Broglie wavelengths of freely propagating particles (i.e. forget the influence of interactions and gravity perturbations, just consider the Universe as a whole) are redshifted by the expansion of the universe. Another way of saying this is that their peculiar momenta with respect to a co-moving local volume decrease as the inverse ...


8

Gravity assists don't change speed in the two body problem. An object approaching a lone gravitating body will enter and leave the vicinity of that body with exactly the same speed. All that a lone gravitating body can do is change the direction in which the object is heading. The body that provides the assist needs to be moving with respect to the target ...


8

Even if the laser had perfectly reflecting, i.e. lossless, mirrors at either end of the cavity, and both ends were sealed so no light could escape it would still require a continual power input. That's because excited atoms/molecules can decay by mechanisms that don't involve a photon e.g. collisional de-excitation. The lost energy goes into heating up the ...



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