# Tag Info

## Hot answers tagged energy-conservation

88

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 ...

48

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 ...

48

[5/3 - Extended the answer, made some corrections, and responded to John Duffield's comment] This is actually the paradox that led Einstein 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 ...

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 ...

19

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 ...

18

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 ...

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

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

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

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 ...

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

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 ...

12

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 ...

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

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 ...

11

Noether's theorem states that to every continuous symmetry of a physical system there is an associated, conserved quantity. The conserved quantity associated with time translation invariance (i.e. it doesn't matter if you perform an experiment now or tomorrow, provided you set it up the same way) is what we call energy. Therefore, somewhat tautologically, ...

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

SECTION A : The example in Feynman's Lectures Let a body P (Planet or Particle or whatever) moving in orbit around a center of attraction called $\:\rm{SUN}$, as in above Figure. Suppose that the attractive force $\:\mathbf{f}\left(r\right)\:$ depends continuously only on the distance $\:r\:$ of the body P from the center $\:\rm{SUN}$. Here it's not ...

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 ...

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

The mass of a free neutron is 939.566 MeV/c$^2$ (almost 1 GeV/c$^2$, so that's probably where your instructor got the "1" value), and the mass of a free proton is 938.272 MeV/c$^2$. A free neutron will decay into a free proton, free electron ($\beta^-$), and an anti-neutrino, $\bar{\nu}$. The mass of the electron is 0.511 MeV/c$^2$, and of the ...

7

Ever since Newton and the use of mathematics in physics, physics can be defined as a discipline where nature is modeled by mathematics. One should have clear in mind what nature means and what mathematics is. Nature we know by measurements and observations. Mathematics is a self consistent discipline with axioms, theorems and statements having absolute ...

7

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 ...

7

Gravitational potential energy is usually measured as a negative value. We do this because an object that is so far away from a gravity well that it practically is unaware of it shouldn't be considered as having any potential energy. So as $r\to\infty$, $PE\to0$. As an object falls into a gravity well, it loses potential energy, so gravitational PE is a ...

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