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1

The velocity addition formula you cite does not quantify what you think it quantifies. I'm going to say see and when I do, I mean "computes relative to its own frame". The velocity addition formula describes the following setup. Frame 0 sees a particle (particle 1) moving in a direction at speed $v_1$. Frame 1 sees that particle 1 at rest and frame 1 ...


1

Let's say you have 3 systems. $B$ moving relative to $C$ with velocity $u$ and $A$ moving relative to $C$ with velocity $v$, all along one axis. $A$ will "measure" for the velocity of $B$: $$ u' = \frac{u-v}{1-\frac{uv}{c^2}} $$ While $B$ will "measure" for the velocity of $A$: $$ v' = \frac{v-u}{1-\frac{uv}{c^2}} = -u' $$ It holds true that the velocity ...


0

When two object orbit each other, the shape of their orbit is independent of their relative mass. In fact they will each have the same shape of orbit, but scaled by the mass of the other. So if you have two objects of mass $m$ and $2m$ respectively, then the former will have an orbit (circle or ellipse) that is exactly twice as big as that of the other. ...


3

To say that the orbit becomes more circular the greater the Sun's mass is not true. Instead, the eccentricity (i.e. how much the shape of an orbit varies from being circular) is governed by a couple of factors. If you have a planet orbiting about the Sun with a mass much less than that of the Sun, and you know the following for an instantaneous point in the ...


4

No. The shape of the orbit, i.e. how elliptical it is, does not depend on the relative masses of the two bodies. All objects in the solar system orbit around the centre of mass of the solar system. For obvious reasons, namely that the Sun contain far and away most of the mass of the solar system, the centre of mass of the solar system is quite close to the ...


2

Yes, objects with mass all attract to each other and move each other of course, except that the star doesnt change it's theoretical orbital shape depending on it's mass, it probably just experiences small tidal forces that aren't much bigger than it's own centrifugal forces. The oscillation of the star position is a complex dynamic based on it's surrounding ...


3

One has to realize that Kepler's laws are a mere approximation. The motion of planets around the sun is a two-body problem. In case of such two-body problems, both the bodies revolve around the center of mass. But it turns out that Sun is much much heavier than the planets. So the center of mass of the system is very close to the Sun and Hence it is a good ...


2

For two objects to remain in a stable circular orbit, the force acting on them must be equal to the centripetal force corresponding to their rotation. $$F=\frac{mv^2}{r}$$ or in terms of angular velocity $$F=m\omega^2r$$ where $r$ is the radius of orbit in this case. As the gravitational force acting on the two stars is the same. $F$ is equal in both ...


3

I will be assuming that the system has some angular momentum about the center of mass initially. If the system has no angular momentum then both the stars would accelerate towards each other and end up colliding. The problem is a two body problem. In such cases, both the stars would revolve around the center of mass. If the distance between two stars is ...


0

Galilean relativity does not automatically hold in Newtonian mechanics. You are correct that Newton's first two laws appear at first sight to be invariant under galilean transformations, but a galilean transformation only transforms spatial coordinates,they don't actually tell you how forces transform. If your force laws just has forces that depend only on ...


2

Let's consider a hover craft or a rocket that will accelerate up to a certain elevation and keep hovering at it, as a better vehicle for our purposes than a plane. Let's forget about air momentarily. What will happen is that the moment our air craft leaves contact with the ground, its initial tangential velocity will be the same tangential velocity of ...


1

Where is the flaw in my thinking? In your concept of time. It's little more than a cumulative measure of local motion, see A World without Time: The Forgotten Legacy of Godel and Einstein]. Your macroscopic motion relative to some other guy results in you measuring his local motion to be slow, whilst he measures your local motion to be slow. This sounds ...


3

To measure time, a duration, you need two moments – when you press "start" and "stop" button on the stopwatch, respectively. But because the two objects are moving relatively to each other, it isn't possible for them to "meet" at both moments. If their locations coincide at the "start" moment, for example, so that their clocks may be compared at this "start" ...


0

To answer my question #1: As Demosthene said, $Λ^μ_ν$ depends only on the relative velocity between frames $S$ and $S'$, but not the relative velocity between particles. This allows me to make the assumption, which I numbered 6 in my initial question, thus solving my problem.


0

EM waves are mass-less. The time dilation effect you mention only applies when a travelling body has mass. Note that for a body that has mass it would take an infinite amount of energy to travel at c.


0

you are correct that the EM field propagates at c, but as the comment from Acuriousmind mentioned, there is no frame of reference traveling at that speed. What that means in simpler words, is that the equations break or diverge for anything traveling at the speed of light. Light is not bound to time dilation or length contraction for that same reason. Only ...


0

Electromagnetic waves rely upon oscillations to propagate Electromagnetic waves are propagating disturbances (or oscillations) in the electromagnetic field and the electromagnetic field exists everywhere and everywhen. But the electromagnetic field itself does not 'travel' at $c$ so your reasoning isn't clear to me.



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