Light at Light Speed

[Emil]

This Newtonian physics is some wild, wild stuff, just mind boggling, huh?:bugeye:

Not for all

Who is this "we"? Authors of freshman physics problems?


That assumes a uniform density. A better model is that gravitational acceleration due to the Earth increases linearly from zero at the center of the Earth to just below the core-mantle boundary and then remains constant to the surface of the Earth. An even better model is that gravitational acceleration increases non-linearly from zero to a maximum of about 10.7 m/s[sup]2[/sup] at the core-mantle boundary, decreases to a minimum of about 9.9 m/s[sup]2[/sup] somewhere in the lower mantle, increases again to just over 10 m/s[sup]2[/sup] at the upper/lower mantle boundary, and then decreases to about 9.8 m/s[sup]2[/sup] at the surface of the Earth.

Above the Earth's surface you can model the Earth as a point mass as a zeroth order approximation. If you want to predict the behavior of satellites with any kind of precision at all it is necessary to model the non-spherical nature of the Earth. At the very minimum you need to model the fact that the Earth has an equatorial bulge. For greater accuracy, the standard approach is to use a spherical harmonics model of the Earth's potential field. Here are descriptions of three such models:

• 1996 Earth Gravity Model, EGM96 http://cddis.nasa.gov/926/egm96/egm96.html
• Grace Gravity Model, GGM02C http://www.csr.utexas.edu/grace/gravity/ggm02/GGM02_JGeod_190_0480.pdf
• 2008 Earth Gravity Model, EGM2008 http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm2008/first_release.html

 

[origin]

Well, it is a great pleasure as you agree that the speed of light between an object (that has the speed 150.000km / s) and the light is 150.000km / s.

This is of course true from the reference frame of earth as you stipulated. I can only assume that you have had a major misunderstanding of relativity and are pleased that you have cleared it up in your mind. So congradulations!

 

[origin]

Originally Posted by origin
This Newtonian physics is some wild, wild stuff, just mind boggling, huh?

Not for all

Agreed - I was just trying to be empathetic to Motor Daddy and his confusion.

 

[Motor Daddy]

Agreed - I was just trying to be empathetic to Motor Daddy and his confusion.

No confusion on my part. You seem to be the one confused, thinking the beach ball's velocity is increasing at the rate of 9.81 m/s^2 while it is on the beach.

So you are telling me that if the velocity on the beach is 0 m/s, that after 1 second the ball has a velocity of 9.81 m/s towards the center of the earth, and after two seconds the ball has a velocity of 19.62 m/s towards the center of the earth?

So tell me, then, how much time elapses before the ball actually reaches the center of the Earth??

You are a very confused individual..

 

[Emil]

This is of course true from the reference frame of earth as you stipulated. I can only assume that you have had a major misunderstanding of relativity and are pleased that you have cleared it up in your mind. So congradulations!

Not so, but you don't realize.
Before you have relativity you made certain assumptions. (Do not forget, we can not talk yet about relativity.)
And I've shown you that yours assumptions are wrong.

 

[origin]

No confusion on my part. You seem to be the one confused, thinking the beach ball's velocity is increasing at the rate of 9.81 m/s^2 while it is on the beach.

So you are telling me that if the velocity on the beach is 0 m/s, that after 1 second the ball has a velocity of 9.81 m/s towards the center of the earth, and after two seconds the ball has a velocity of 19.62 m/s towards the center of the earth?

So tell me, then, how much time elapses before the ball actually reaches the center of the Earth??

You are a very confused individual..

When did I talk about a beach ball? I certainly do not think that a ball sitting on the ground has a velocity relative to the ground. You really come up with some interesting stuff.

 

[origin]

Before you have relativity you made certain assumptions. (Do not forget, we can not talk yet about relativity.)

Did I? (Oh - shhhhhh)

And I've shown you that yours assumptions are wrong.

You have? What assumptions? Are we allowed to talk about them?

 

[Emil]

Did I? (Oh - shhhhhh)

You have? What assumptions? Are we allowed to talk about them?

As I said, you just don't realize.

 

[Motor Daddy]

When did I talk about a beach ball? I certainly do not think that a ball sitting on the ground has a velocity relative to the ground. You really come up with some interesting stuff.

You said:

Why do you keep saying that? It sounds really strange. It is like saying, "I am so pleased that you agree with me that the acceleration of gravity at the earths surface is 9.81 m/s^2."

It is not that we agree with you, it is simply that you agree with reality, which is rather trivial.:shrug:

The reality is, an object such as a beach ball at the earth's surface (the beach) is not accelerating. What makes you think that the ball's velocity is increasing at the rate of 9.81 m/s^2 while on the beach? At the surface of the earth, while the ball is at the surface of the earth the velocity towards the center of the earth is 0 m/s, and since I can sit there and watch the ball for two and a half hours, and the ball doesn't get any closer to the center of the earth, then I can correctly say that the ball did not accelerate, which means its velocity did not change in 2 1/2 hours, its velocity is still 0 m/s, and therefore it's acceleration is 0 m/s^2!

You want to say the beach ball is accelerating at the rate of 9.81 m/s^2 at the surface of the earth??? Prove it!!!

 

[origin]

As I said, you just don't realize.

Oooooo, you have such deep and mysterious knowledge of the universe. Mere mortals cannot hope to comprehend the depths to which you understand high school algebra.

We are not worthy.

 

[origin]

You want to say the beach ball is accelerating at the rate of 9.81 m/s^2 at the surface of the earth??? Prove it!!!

Why don't we let you prove it.
Follow these steps.

1. Jump into the air.
2. If you post again I will assume that you are not in outerspace and you have proved it.

Like I said these Newtonian physics are just, so, like you know, wicked hard to understand.... it's mind boogling dude.

 

[Emil]

Oooooo, you have such deep and mysterious knowledge of the universe. Mere mortals cannot hope to comprehend the depths to which you understand high school algebra.

We are not worthy.

LOL...Your scientific arguments are delectable.

 

[origin]

LOL...Your scientific arguments are delectable.

If you have more continue or does your explanation of your idea end with "you do not realize"?

 

[Motor Daddy]

Why don't we let you prove it.
Follow these steps.

1. Jump into the air.
2. If you post again I will assume that you are not in outerspace and you have proved it.

Like I said these Newtonian physics are just, so, like you know, wicked hard to understand.... it's mind boogling dude.

You said the acceleration at the surface of the earth is 9.81 m/s^2. Well, I'm at the surface of the earth, and I am NOT accelerating towards or away from the center of the earth. My distance from the center of the earth is remaining unchanging. Therefore I have a 0 m/s velocity towards the center of the earth, and the rate at which my velocity towards the center of the earth is increasing or decreasing is 0 m/s^2.

So you say when I am laying on the beach my acceleration is 9.81 m/s^2, and I say my acceleration is 0 m/s^2.

The distance between me and the center of the earth remains unchanged, so you are wrong and I am correct. You see how that works?

Are you living in your imaginary world again where you like to pretend beach balls on the beach are accelerating at the rate of 9.81 m/s^2, when really the rate of change of velocity is 0 m/s^2? You really need to start living in reality and stop pretending.

 

[origin]

You said the acceleration at the surface of the earth is 9.81 m/s^2. Well, I'm at the surface of the earth, and I am NOT accelerating towards or away from the center of the earth. My distance from the center of the earth is remaining unchanging. Therefore I have a 0 m/s velocity towards the center of the earth, and the rate at which my velocity towards the center of the earth is increasing or decreasing is 0 m/s^2.

So you say when I am laying on the beach my acceleration is 9.81 m/s^2, and I say my acceleration is 0 m/s^2.

The distance between me and the center of the earth remains unchanged, so you are wrong and I am correct. You see how that works?

Are you living in your imaginary world again where you like to pretend beach balls on the beach are accelerating at the rate of 9.81 m/s^2, when really the rate of change of velocity is 0 m/s^2? You really need to start living in reality and stop pretending.

This is very basic stuff here. I am not going to try and teach you elementary physics. Suffice to say that accleration can be calculated by the change in velocity or it can be calculated by the dividing the force of an object by the mass of the object. In the case of gravity you can derive the acceleration by measuring the change in the velocity of a falling object or by weighing an object and dividing the weight (force) by the mass. Either way you are calculating the acceleration due to gravity. The fact that this is something that you think is imaginary is really not surprising. I recomend that you do not test your theory by jumping off a bridge.

You really need to take an introductory course in physics, especially since you have shown (in your mind) that you are smarter than Einstein, and the millions of people that have actually studied physics.

 

[Motor Daddy]

I recomend that you do not test your theory by jumping off a bridge.

How do you recommend I jump off a bridge at the surface of the earth? Are saying I need to get above the surface of the earth, and free fall until I reach the surface of the earth??

Now you are changing your story. Before you said I would be accelerating at 9.81 m/s^2 AT the surface of the earth, not while I'm above the surface free falling towards the surface. You really are confused, aren't you??

 

[rpenner]

As I said, you just don't realize.

Mere mortals cannot hope to comprehend the depths to which you understand high school algebra.

Sarcasm noted.

So let's recap:
Considering the approximations made​​.
Time = 8min (480sec) according to Earth, distance = L / 2 (72.000.000km) according to Earth, speed V =150.000km / s according to Earth.
Is this correct?

But, origin, by whose curriculum standards is this "high school algebra"? I think you have overestimated Emil's contributions to this thread. It looks like literally elementary school arithmetic, circa age 11, to me.
By the standards, I thought were in place, I thought this was high school algebra:

Another alternative is that the family of functions $$f_K (u,v) = \frac{u - v}{1 - K u v}$$ will work. And since the consistency of the speed of light means $$c = f_K (c,v) = \frac{c - v}{1 - K c v} = c \frac{1 - \frac{v}{c}}{1 - K c v}$$ which means $$1 - \frac{v}{c} = 1 - K c v$$ which means $$K = \frac{1}{c^2}$$.

Likewise, high school algebraic geometry:

For Earth rulers and clocks, C, B, and A happen at the same time.

canvas_sigma by PhysForum Photos, on Flickr

For the outbound plane's rulers and clock, E, D, and A happen at the same time.

canvas_sigma_prime by PhysForum Photos, on Flickr

If Emil was working on the subject of Physics, Emil would be meaningfully addressing the claims and arguments here:

Speed is just someones description of something moving a certain distance in a certain amount of time.

For physics to be consistent, the same formula for "If I see it moving at speed u and I see someone moving at speed v, what speed would that someone see it moving at?" has to apply to light, sound and bullets.

w = f(u,v)

Clearly f(v,v) = 0 since a bullet fired at the same speed as the train would seem to stand still, and f(u,0) = u since a bullet fired at speed u should still look like speed u if seen from the train when the train is stopped. Moreover, f(0,v) = -v means that if I see the train moving at speed v, the train sees me moving at speed -v.

Clearly f(u,v) = u - v will satisfy these conditions, but will not account for the consistency of light speed by various observers.

Another alternative is that the family of functions $$f_K (u,v) = \frac{u - v}{1 - K u v}$$ will work. And since the consistency of the speed of light means $$c = f_K (c,v) = \frac{c - v}{1 - K c v} = c \frac{1 - \frac{v}{c}}{1 - K c v}$$ which means $$1 - \frac{v}{c} = 1 - K c v$$ which means $$K = \frac{1}{c^2}$$.

 

[origin]

How do you recommend I jump off a bridge at the surface of the earth?

Damn, you can seem to understand anything at all. I said, "I DO NOT recommend that you jump off a bridge"!

Are saying I need to get above the surface of the earth, and free fall until I reach the surface of the earth??

Again I personnaly do not recommend that, but you are free to do what you want.

Now you are changing your story.

No, I do not believe that is an accurate statement.

Before you said I would be accelerating at 9.81 m/s^2 AT the surface of the earth, not while I'm above the surface free falling towards the surface. You really are confused, aren't you??

No, I am not confused in the slightest. I gave you 2 ways to calculate acceleration, a change in velocity and F=ma. Did you not read the post or did you not understand the post?

You can choose to have your own definitions of generally accepted terms but it is a bit presumptous to assume that academia will adopt your terminology. Especially since you appear to be completely ignorant of any physics advances since Galileo...

 

[Motor Daddy]

Damn, you can seem to understand anything at all. I said, "I DO NOT recommend that you jump off a bridge"!

You said you don't recommend I jump off a bridge to test my theory. Do you think my theory is that when I do jump off a bridge above the surface of the earth that while in free fall I wouldn't be accelerating? I never said or implied that. You said I would be accelerating at 9.81 m/s^2 at the surface of the earth. I am on the surface of the earth, and I am not accelerating, so please provide some evidence that my velocity is increasing or decreasing at the rate of 9.81 m/s^2 in order to back up your wild imagination with some type of fact.

Again I personnaly do not recommend that, but you are free to do what you want.

Why would you recommend that, that would be silly, because that would require me to be above the surface of the earth, not AT the surface of the earth, of which you made statements about. While we're on the subject, do you acknowledge that if I were to be free falling towards the earth, that my acceleration rate would be increasing the closer I got to the earth? How do you reconcile that fact in your equations of acceleration while falling towards the earth? It is not just a particular acceleration rate, because if you are accelerating, you are getting closer to the earth while falling towards the earth, and the distance is decreasing, increasing the acceleration rate while accelerating. How do you account for that?

No, I am not confused in the slightest. I gave you 2 ways to calculate acceleration, a change in velocity and F=ma. Did you not read the post or did you not understand the post?

Well a change in velocity towards the center of the earth is not happening while I'm standing on the beach, so that rules that out. Now you mention f=ma. Isn't that a net force? In order to have acceleration, there must be a net force, correct?

You can choose to have your own definitions of generally accepted terms but it is a bit presumptous to assume that academia will adopt your terminology. Especially since you appear to be completely ignorant of any physics advances since Galileo...

I use the standard definition of acceleration, which is the rate of change of velocity. Is that not an acceptable definition of acceleration in your physics world?

 

[arfa brane]

I think what Motor Daddy is overlooking completely, probably because he hasn't studied physics at high school, is that you can be accelerating in time.

Although you have a speed of 0 m/s at the surface of the earth, you experience a force. To explain the force, you say gravity is accelerating your mass towards the centre of the earth. Since the universe is 4-dimensional, you can accelerate in one dimension; that leaves three in which you aren't accelerating.

On the surface of the earth the three remaining dimensions are spatial; you accelerate in the time dimension. This is the 'force' of gravity with acceleration g = F/m.