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43

One of the results of special relativity is that a particle moving at the speed of light does not experience time, and thus is unable to make any measurements. In particular, it cannot measure the velocity of another particle passing it. So, strictly speaking, your question is undefined. Particle #1 does not have a "point of view," so to speak. (More ...


34

The mass (the true mass which physicists actually deal with when they calculate something concerning relativistic particles) does not change with velocity. The mass (the true mass!) is an intrinsic property of a body, and it does not depends on the observer's frame of reference. I strongly suggest to read this popular article by Lev Okun, where he calls the ...


21

The complete answer to that question is an open problem in fluid mechanics, as exact solutions to the irrotational water wave equations are unknown. However, under certain asymptotic approximations, we can estimate the speed of these waves. Irrotational inviscid surface waves are governed by Laplace's equation, i.e. $$\nabla^2 \phi = 0$$ where $\phi$ is ...


14

No, the escape velocity doesn't need to be maintained for any length of time. Escape velocity is the minimum speed you need to have at the Earth's surface to be able to escape the gravitational pull, without using a rocket or other continuous propulsion. In other words, ignoring all sources of gravity other than the Earth, if you launch a projectile ...


14

One's naive expectation would be that as the object moves through the medium, it collides with molecules at a rate proportional to $v$. The volume swept out in time $t$ is $A v t$, where $A$ is the cross-sectional area, so the mass with which it collides is $\rho A v t$. The impulse in each collision is proportional to $v$, and therefore the drag force ...


13

This is what special relativity is all about.. In special relativity you cannot simply state that particle 2 is moving at c+c=2c in a reference frame where particle 1 is at rest. Speeds add like this (easily found in wikipedia): $$v_2^{'} = \frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$$ i.e. the speed of particle 2 $v_2'$ in a reference frame where particle 1 is ...


13

Photons always travel at $c$ (not completely true, but a good simplification for this question's purposes). Common sense tells us that if person A running at velocity $v$ is chasing person B with velocity $u$, the velocity of person B with respect to person A ($w$) is: $$w=u-v$$ But our common sense is misleading, and this equation is only an approximation ...


11

No. Any circular orbital velocity is about 70% ($1/\sqrt{2}$) of the required escape velocity. To find circular orbital velocity, equate the centripetal force to the force of gravity: $$ \frac{m v^2}{r} = G \frac{ M m}{r^2} \rightarrow \boxed{ v_\textrm{circ} = \sqrt{\frac{GM}{r}}}$$ To find escape velocity, equate the magnitude of the potential energy to ...


10

Terminal velocity is reached when the drag force due to moving through air is equal (but opposite) to the gravitational force. Now, the gravitational force is proportional to the mass, while the drag force has nothing to do with mass, but everything to do with how large and "streamlined" the object is. Suppose object A is twice as heavy as object B. If ...


9

I endorse Ron's answer – it's the systematic way to proceed. The velocity $v/c$ may be written as $\tanh \eta$ where $\eta$, the rapidity or whatever, is the hyperbolic (Minkowski) counterpart of the (Euclidean) angle. The addition of velocities then boils down to an addition formula for $\tanh(\eta_1+\eta_2)$ because the rapidities just add additively. Let ...


9

Rocket fuels initially, followed by a series of gravitational assists (slingshots): http://en.wikipedia.org/wiki/Gravity_assist The linked article mentions Voyager 1 mission as an example.


9

Instantaneous velocity can never be measured since there is no way in the real world to do anything instantaneously. All measurements take some amount of time to peform. For example the comment to the question mentioned using the Doppler effect to measure instantaneous velocity. That is not possible since to measure the frequency of a wave you have to ...


9

The complete relevant text in the book is The de Broglie wave equation relates the velocity of the electron with its wavelength, $\lambda = h/mv$ ... However, the equation breaks down when the electron velocity approaches the speed of light as mass increases. ... Actually, the de Broglie wavelength should be $$ \lambda = \frac hp, $$ where $p$ is the ...


9

He "only" flew at the maximum speed of 370 m/s or so which is much less than the speed of the meteoroids – the latter hit the Earth by speeds between 11,000 and 70,000 m/s. So he was about 2 orders of magnitude slower. The friction is correspondingly lower for Baumgartner. Note that even if he jumped from "infinity", he would only reach the escape velocity ...


9

Suppose you throw a ball upwards at some velocity $v$. When you catch it again it's traveling downwards at (ignoring air resistance) a velocity of $-v$. So somewhere in between throwing and catching the ball it must have been stationary for a moment i.e. it's instantaneous velocity was zero. Obviously this was at the top of its travel. When you throw the ...


9

It's simpler than you (probably) think. In your example of defining speed: this is a change of position $s$ in a time $t$. The units of distance are metres and the units of time are seconds, so the units of velocity are metres per second. So far so good. Now consider acceleration: this is a change of velocity $v$ in a time $t$, so the units of acceleration ...


9

How can kinetic energy be proportional to the square of velocity, when velocity is relative? Without reading the rest of your question, I must first reply that one has nothing to do with the other. Kinetic energy is frame dependent, just as velocity is. Momentum is proportional to velocity and is frame dependent too, just as velocity is. Now, ...


9

Technically, yes, the plank helps. However, in practise in a real situation the effect isn't really significant. The first advantage to falling with a plank under you as you show in your bottom left picture is that it adds wind resistance. If the plank can stay oriented flat to the upward rushing air (from your perspective), then it will push upwards on ...


9

The velocity is constantly increasing due to a constant acceleration. Exactly at 1 s the velocity is 10 m/s, but this does not mean that velocity was at 10 m/s in preceding second. In fact, given the distance 5 m moved in this second, the average velocity in this second was 5 m/s. And this should make sense to you, because in this first second the velocity ...


8

If you are not interested in relativistic effects, the answer to your question is easy to workout. According to Wikipedia, Alpha Centauri is 4.24 ly away (4.0114x$10^{16}\mathrm{m}$). So to get there in 60 years ($1892160000\mathrm{s}$). So your non-relativistic answer is $v = \frac{d}{t} = \frac{4.0114 \times 10^{16}}{1892160000} = 21200000 ...


8

The distance between Earth and Alpha Centauri is $4.4\,\text{ly}$. Dividing by $60\,\text{years}$ it's approximately $22000\,\text{km/s}$. The relativistic factor (I mean $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$ for this is almost $1$. If we take a constant acceleration of $2g$ (it's possible) it would take only $320\,\text{hours}$ to reach this speed (and, ...


8

The problem here is that the statement "The current time is 'x'" CANNOT be globally true. It is true only with respect to a given reference frame, so what you've implied with the statements involved in your question (i.e. that it is either 2014 or not 2014 everywhere) is simply invalid in relativity. "At earth" doesn't cut it, since the earth is an extended ...


8

Yes, the object would slow down to its terminal velocity. To see why, notice that the net force on a falling object of mass $m$ near the surface of the earth is \begin{align} F = F_\mathrm{drag} - mg \end{align} where $F_\mathrm{drag}$ is the force due to air resistance, and here I have assigned "up" to be the positive direction. On the other hand, the ...


8

Like all paradoxes, there is no contradiction here, just misuse of logic. How do you define velocity? If you say the distance traveled in an extended period of time, divided by that time well then of course there's no such thing as instantaneous velocity. Asking what something's instantaneous velocity is under this definition is logically equivalent ...


8

There's a simple way to look at this that doesn't involve any maths. Suppose the two cars are parked and are stationary, and you accelerate past them in your car. If you are accelerating forwards then from your perspective it looks as if the two cars are accelerating backwards (at the same rate). But the cars are at rest, so the distance between them can't ...


7

Here is a very basic estimation: The kinetic energy of a 1000 kg car moving at 60 km/h is $$E=\frac{mv^{2}}{2}=\frac{1000kg(16.7m/s)^{2}}{2}=138.9 kJ$$ The heat of gasoline combustion is 47 MJ/kg = 35000 kJ/litre. Assuming 10% efficiency of the car's engine, you would need to burn $$\frac{138.9 kJ}{0.1\cdot35000kJ/l}=0.04 litre$$ of gasoline to accelerate ...


7

There is a point of view, that under the term "the mass" one must mean "the rest mass". From that point of view there is obviously no dependence of the (rest) mass on the speed of an object. And, therefore, the mass of an object does not increase when its speed increases. The correct (from that point of view) way to talk about the phenomenon is to say that ...



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