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36

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 ...


33

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 ...


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 ...


12

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 ...


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

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

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

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

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

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

The complete relevant text in the book is The de Broglie wave equation relates the velocity of the electron with its wavelength, λ = 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 ...


8

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 ...


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

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

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, ...


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

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

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

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 ...


7

Terminal velocity is the maximum velocity that you can reach during free-fall. If you imagine yourself falling in gravity, and ignore air resistance, you would fall with acceleration $g$, and your velocity would grow unbounded (well, until special relativity takes over). This effect is independent of your mass, since $F = ma = mg \Rightarrow a = g$ Where ...


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

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 instaneous velocity was zero. Obviously this was at the top of it's travel. When you throw the ball ...


7

Short, short version: It's complicated. Slightly longer version: Internal combustion engines have at least two relevant performance characteristics: power and torque. Furthermore the maximum attainable values for both are functions of the current engine speed (RPM). Acceleration will cease if the current requirement for either power or torque equal the ...


7

Here's my guess: As you know, internal combustion engines burn fuel. The power output of the engine is a function of both the current RPM and the amount of fuel you inject. But there's a catch: the engine can burn a limited amount of fuel per cycle, and therefore the higher the RPM, the more fuel can be combusted. So (as @dmckee stated), at a given gear the ...


7

At low velocities like this you can ignore special relativity and simply add the two velocities. This is really easy to see if you imagine yourself standing still and the Earth moving under you. Relative to you the gun should fire just like you were standing still. This is called an inertial frame of reference. You see the bullet leave at $400\: ...



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