[McQueen (OP)]

I have just been reading Brian Greene’s “The Elegant Universe” .  What struck me while going through the book was the equivalence principle quoted by Einstein in support of General Relativity. . Einstein stated it thus:
“we ... assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system”.  — Einstein, 1907
What it means I suppose is that the acceleration of bodies towards the centre of the Earth at a rate of 1g (g= 9.81 m/s² being a standard reference of gravitational acceleration at the Earth's surface) is equivalent to the acceleration of an inertially moving body that would be observed on a rocket in free space being accelerated at a rate of 1g. The key insight that Einstein had was that gravity and acceleration are indistinguishable from each other.  For instance, if your compartment is being accelerated upward you will feel the force of the floor on your feet. Einstein's realization was that within the confines of your tiny compartment, you will not be able to distinguish these accelerated situations from ones without acceleration but with gravity: When their magnitudes are judiciously adjusted, the force you feel from a gravitational field or from accelerated motion are indistinguishable.
   But think about this is it really true that gravity and accelerated motion are interwoven or the same? Take the example given of a space craft somewhere deep in interstellar space, where gravitational forces are at an absolute minimum. In these conditions the only gravity available is the gravitation due to the mass of the space craft. Normally gravity is calculated by simply multiplying the mass of the object by the acceleration due to gravity, which on earth is 9.8m/s². In interstellar space there is no earth and hence no acceleration due to gravity. Let’s just suppose (as a special case) that in interstellar space there is a planet exactly like earth about 50 billion miles away from the space ship, which weighs 2 tons or 2000 Kg. Then the acceleration due to that  planet would be 9.8m/s² and the gravitational  attraction between the planet and the space ship would be  G x (5.972 x 10²⁴ x 2x 10³)/((5X10¹⁰)2) = 0.000318860846 m³ kg^-1 s^-2 . It is now possible to see that the gravity exerted on the space ship is very small, therefore if the mass of the spaceship were to be multiplied by the force of gravity a total force of 0.00031 N which is an extremely small force would be exerted by gravity. If the ship were being accelerated at 9.8m/s² then the force exerted on the space ship would be equal to F = 2000 Kg x 0.00003 m/s¹ = 0.62 N which is a ridiculously small force and certainly would not have the effect of pushing anyone in spaceship back against their seats. Where am I going with this? I am trying to show that the equivalence principle of Einstein’s has absolutely no real meaning, it is just using gravitation as an explanation of itself. Here we have an object in space being accelerated to an equivalent acceleration of the acceleration due to the gravity on earth (i.e., 9.8 m/s² ) yet it has no effect, some gravitational mass is needed close by to make it work.

[Bored chemist]

Then the acceleration due to that  planet would be 9.8m/s2

No, it wouldn't.
Unless you were on the surface of the planet.

the gravitational  attraction between the planet and the space ship would be  G x (5.972 x 1024 x 2x 103)/((5X1010)2) = 0.000318860846 m3 kg-1 s-2 .

The attraction would be a force which does not have units of "m3 kg-1 s-2".

If the ship were being accelerated at 9.8m/s2 then the force exerted on the space ship would be equal to F = 2000 Kg x 0.00003 m/s1 = 0.62 N

No, it's not.
If a ship with a mass of 2000 Kg is accelerated (for example, by the engines) at 9.8 m/s/s  then the force on it can be calculated for F=MA
So it's 2000*9.8
That's 19600 Newtons.

The equivalence principle has been experimentally validated to very high precision.
Basically, you seem to be saying relativity is wrong, but you have yet to understand Newtonian physics.

[Janus]

Let’s just suppose (as a special case) that in interstellar space there is a planet exactly like earth about 50 billion miles away from the space ship, which weighs 2 tons or 2000 Kg. Then the acceleration due to that  planet would be 9.8m/s2 and the gravitational  attraction between the planet and the space ship would be  G x (5.972 x 1024 x 2x 103)/((5X1010)2) = 0.000318860846 m3 kg-1 s-2 .

For one, you screwed up your equation.  You put the (2 x 10^3), which I assume was a rough attempt to convert miles to meters*,  in the wrong half of the equation   It needed to be multiplied by the 5 x 10 so that the you had
 G x (5.972 x 1024 )/((5X1010x 2x 103)2) . In other words, the numerical value of your number was off by a factor of at ~ 8e10. in addition to being expressed in the wrong units.  Correctly used, the equation gives an answer in m/s^2 or acceleration.  For example, the equation used for the surface of the Earth (a distance of ~6378000 meters from the Earth center gives
G(5.972e24/6378000^2 = 9.8 m/s^2 or the acceleration due to gravity at the surface.)
A more accurate value for your scenario is 6.23e-14 m/s^2.    Which is the acceleration due to gravity at this distance from the Earth. 

It is now possible to see that the gravity exerted on the space ship is very small, therefore if the mass of the spaceship were to be multiplied by the force of gravity a total force of 0.00031 N which is an extremely small force would be exerted by gravity. If the ship were being accelerated at 9.8m/s2 then the force exerted on the space ship would be equal to F = 2000 Kg x 0.00003 m/s1 = 0.62 N which is a ridiculously small force and certainly would not have the effect of pushing anyone in spaceship back against their seats.

The above arrived at value 6.23e-14 m/s^2 multiplied by the 2000 kg of the rocket gives the amount of thrust or force the rocket would have to generate in order to "hold station" at this distance and not slowly begin to drift towards the planet (1.25e-10 N).
A 80 kg man standing on the floor of said rocket would feel a force of ~5e-12 N Small, yes, but no more or less than what he would feel if standing on the surface of a planet with a mass equal to that of the Earth and 50 billion miles in radius.  It is also exactly the same force he would feel pressing him against the floor if the rocket were accelerating at 6.23e10-14 m/s^2 in empty space with no gravitational forces present.  The point of the equivalence principle is that our passenger, would not be able to distinguish between "hovering" 50 billion miles from an Earth massed planet and the rocket accelerating at 6.23e10-14 m/s^2 through free space with there being no gravity.  The fact that the force he feels is in all practical effect too small to be measured doesn't mean that it is the same in both cases.

Where am I going with this? I am trying to show that the equivalence principle of Einstein’s has absolutely no real meaning, it is just using gravitation as an explanation of itself. Here we have an object in space being accelerated to an equivalent acceleration of the acceleration due to the gravity on earth (i.e., 9.8 m/s2 ) yet it has no effect, some gravitational mass is needed close by to make it work.

No. What you have shown here is that you don't actually grasp the meaning behind the equivalence principle. The flaw is in your understanding, not the principle itself.

*more accurate would have been 8/5 x 10^3. And if you are going to rough out this by this degree (2 instead of 1.6, or only to to 1 significant digit), it really doesn't make sense to carry out the Earth's mass to 5 significant digits.

[McQueen (OP)]

Bored Chemist: As usual you are so wrapped up in being bored that you are missing the wood for the trees. Wake up!  Your reply seems to imply that in your opinion there is no difference for a passenger in a space ship accelerating away from the vicinity of the earth, say at a distance of 1000 m and a passenger in a space ship 50 billion kilometres away from an earth like planet. The gravitational force of a huge mass of 5.97 x 1024 Kg is affecting the passenger of the space ship in the vicinity of the earth. ( 6.67 x 10¹¹) x( (5.97 x 10²⁴) x(2 x 10³)/((10³)²)) =8.9 x 10²⁰N.  The space ship that is one kilometre away from the earth is experiencing a force of  8.9 x 10²⁰ kgm/s² . Whereas on the spaceship that is 50 billion kilometres from the earth like planet the force exerted by gravity is only  0.00003kgm/s² , surely this makes some difference as to what the passenger on the spaceship 50 billion kilometres away from the earth like planet is experiencing?

[McQueen (OP)]

Janus
 I really do appreciate the trouble you have taken. I must say that I also see some humour in the post. Literally rolling on the floor laughing!

[Kryptid]

Your reply seems to imply that in your opinion there is no difference for a passenger in a space ship accelerating away from the vicinity of the earth, say at a distance of 1000 m and a passenger in a space ship 50 billion kilometres away from an earth like planet.

His point is that the force due to acceleration is in no way dependent upon the force due to gravity from a distant planet. If you want to know the total force, of course you would need to add these two forces together. However, the thought experiment assumes a world where the only force acting on the observer is that of the accelerating spacecraft. If it is accelerating upwards at 1 G, then the observer experiences a downward force of 1 G. No planet needed.

[Bored chemist]

  The space ship that is one kilometre away from the earth is experiencing a force of  8.9 x 10²⁰ kgm/s2 .

Where did you get that number?

A " space ship, which weighs 2 tons or 2000 Kg." near the surface of the Earth experiences a force- generally referred to as its weight- which is  2000 g
19600 Kg

Whereas on the spaceship that is 50 billion kilometres from the earth like planet the force exerted by gravity is only  0.00003kgm/s2 , surely this makes some difference as to what the passenger on the spaceship 50 billion kilometres away from the earth like planet is experiencing?

Well, while the ship is parked on the surface of the Earth, the occupants are pulled down by their weight,  a force equal to the product of g and their mass.
And, if the ship falls over, the people accelerate towards the centre of the Earth at 9.8 m/s.

Out in deep space 5 billion km away then they weigh less.
The factor by which they weigh less is the square of the ratio of the distances so it's (50 bn / the radius of the earth (in km) ) squared
Of course it has less effect- it's further away,
And, as a consequence of the much smaller force, they have a much smaller acceleration.

This has little or nothing to do with the principle of equivalence.

What do you think that principle actually means?

[McQueen (OP)]

Kryptid: The point I was trying to make is that an acceleration of 9.8m/s², 50 billion kilometres from earth or other gravitational influences,is in no way representative of the gravitational force on earth. Instead as Janus points out the force that would be experienced by the spaceship would be the same force that it would experience if it were on an earth that possessed a radius of 50 billion kilometres ! This reinforces my original conjecture that in the absence of a large mass close by, Einstein's thought experiment makes no sense. He is using gravity as an explanation of itself.

[McQueen (OP)]

Bored Chemist: What it means I suppose is that the acceleration of bodies towards the centre of the Earth at a rate of 1g (g= 9.81 m/s2 being a standard reference of gravitational acceleration at the Earth's surface) is equivalent to the acceleration of an inertially moving body that would be observed on a rocket in free space being accelerated at a rate of 1g. If the equivalence principle does not mean this but that an acceleration that results in a force of 1 g has to be applied then the experiment makes no sense.

[Janus]

Kryptid: The point I was trying to make is that an acceleration of 9.8m/s², 50 billion kilometres from earth or other gravitational influences,is in no way representative of the gravitational force on earth. Instead as Janus points out the force that would be experienced by the spaceship would be the same force that it would experience if it were on an earth that possessed a radius of 50 billion kilometres ! This reinforces my original conjecture that in the absence of a large mass close by, Einstein's thought experiment makes no sense. He is using gravity as an explanation of itself.

This has no bearing on the equivalence principle. 
An example illustrating the principle is this:
Take your ship, put it just above the surface of the Earth firing its engines with just enough thrust to hold it hovering above the Ground at a constant height.  An occupant in the rocket standing on a scale would weight the same as he would standing of the surface, if he drops an object, it falls to the floor, etc.
Now take that rocket and move out to 50 billion miles from the Earth.  But still firing its engines at the sane thrust as before.  Now this of course will be more thrust than the pull of gravity from the Earth and the rocket will accelerate away from the Earth.  An occupant inside the ship however will still weigh the same on a scale in the ship and objects will still fall towards the floor of the ship just like they did when the ship was hovering just at the surface of the Earth.  Relative to the rocket, objects in this accelerating rocket behave just like they would in a gravitational field.
This is the gist of the equivalence principle.  The importance of this is that you can imagine how something like light would behave in the rocket in the second scenario For example, light emitted from the tail would appear red-shifted to an observer in the nose, or if fired sideways to the acceleration would trace a curved path relative to the ship, and then use this to conclude conclude that it would exhibit the same behavior in a gravity field (be red shifted climbing against it or have it path curved by it.
You would also conclude that an observer in the nose of the accelerating rocket would measure a clock at the tail as running slow, and by extension that someone higher in a gravitational field would measure a clock lower than them as running slow.

[Kryptid]

Kryptid: The point I was trying to make is that an acceleration of 9.8m/s2, 50 billion kilometres from earth or other gravitational influences,is in no way representative of the gravitational force on earth.

If you experience an acceleration of 9.8 m/s² due to the firing of the spacecraft's engines, then add that to the force you experience from the distant planet (assuming it is below you), then the total acceleration you feel is 9.8 + 0.0000000000000623 m/s² = 9.8000000000000623 m/s². The difference is so small you wouldn't be able to tell that any force other than the rocket's engines are acting on you.

Instead as Janus points out the force that would be experienced by the spaceship would be the same force that it would experience if it were on an earth that possessed a radius of 50 billion kilometres !

More specifically, that would be the force experienced due to the gravity of the planet. That completely ignores the force experienced due to the acceleration of the spacecraft, which is much, much larger.

This reinforces my original conjecture that in the absence of a large mass close by, Einstein's thought experiment makes no sense.

No, it doesn't. It only reinforces that you don't understand how acceleration works.

He is using gravity as an explanation of itself.

No, he isn't.

[Bored chemist]

The point I was trying to make is that an acceleration of 9.8m/s2, 50 billion kilometres from earth or other gravitational influences,is in no way representative of the gravitational force on earth.

How would you tell the difference?

[McQueen (OP)]

Bored Chemist: Kryptid:  How would you tell the difference?

The person in the spaceship will know that he is not in a gravitational field because he is floating around. Just as the astronauts in the space station are floating around even though they are in a state of accelerated motion around the earth.  The International Space Station travels in orbit around Earth at a speed of roughly 27,440 kilometres  per hour (that's about 8 kilometres per second!). This means that the Space Station orbits Earth (and sees a sunrise) once every 92 minutes!  According to what you state , moving at this speed the astronauts should be plastered to one side of the walls of the space station. Unable to move.

[Kryptid]

Bored Chemist: The person in the spaceship will know that he is not in a gravitational field because he is floating around. Just as the astronauts in the space station are floating around even though they are in a state of accelerated motion around the earth. 

And yet those astronauts floating around in space are, indeed, in a gravitational field.

[McQueen (OP)]

Kryptid:And yet those astronauts floating around in space are, indeed, in a gravitational field.

You will forgive me if your answer appears somewhat inane. Would you go as far as to say they (the astronauts) are in a diminished gravitational field. The point is even though they are in an accelerated state, there is no sense of up and down, as you suggest there should be?

[Kryptid]

You will forgive me if your answer appears somewhat inane. Would you go as far as to say they (the astronauts) are in a diminished gravitational field.

Diminished only in the sense that they are a little further away from the planet's surface than you and I.

The point is even though they are in an accelerated state, there is no sense of up and down, as you suggest there should be?

Now you are putting words in my mouth. Please quote where I said anything about there being a sense of up and down.

It's the same kind of situation where you are sealed inside of a box and then dropped out of an airplane. Inside the box, you would be floating just like the astronauts in space. You would have no sense of up or down even though you are very much inside of the Earth's gravitational field. If you couldn't see or hear anything outside of the box, you'd have no way of knowing whether you were free-floating in interstellar space or in free fall in the Earth's atmosphere.

[McQueen (OP)]

kryptid : It's the same kind of situation where you are sealed inside of a box and then dropped out of an airplane. Inside the box, you would be floating just like the astronauts in space.

Thanks, it is a good answer. It still leaves a lot to be desired though. According to Newton’s third law, every reaction has an equal and opposite reaction.  The astronauts are subject to acceleration since they are travelling in a circular motion.  According to Newton’s third law the astronauts should be exerting an equal and opposite reaction to that which they are experiencing. They are experiencing a force of acceleration and they should press back, this is what makes things stick to the ground instead of floating away. According to Einstein, if the astronauts in a space ship are experiencing an acceleration they should not be able to tell it apart from gravity. According to the example you have given they do not experience a force because they are weightless (i.e., in free fall). My point is that astronauts in my example (i.e., in a spaceship far from any gravitational influences), would also be weightless and in free fall and hence the force of acceleration exerted on the ship would not affect them. Ergo, Einstein’s equivalence principle is wrong.

[Kryptid]

According to Newton’s third law the astronauts should be exerting an equal and opposite reaction to that which they are experiencing.

They are. They are pulling on the Earth just as strongly as the Earth is pulling on them.

According to the example you have given they do not experience a force because they are weightless (i.e., in free fall).

They do experience a force because they are accelerating. F=ma. But the spacecraft they are in is accelerating at the same rate as they are, so they can't actually tell that they are accelerating if they don't look outside.

My point is that astronauts in my example (i.e., in a spaceship far from any gravitational influences), would also be weightless and in free fall and hence the force of acceleration exerted on the ship would not affect them.

Only if this "force of acceleration" you speak of is the force caused by the gravity of the planet. The acceleration caused by the rocket engines would still very much be felt. Unless you are somehow under the bizarre assumption that G-forces due to acceleration don't exist, which would also mean that pilots shouldn't experience blackouts from turning their aircraft too hard. But they do. So acceleration produces a force whether there is gravity around or not.

[McQueen (OP)]

Only if this "force of acceleration" you speak of is the force caused by the gravity of the planet. The acceleration caused by the rocket engines would still very much be felt…….Unless you are somehow under the bizarre assumption that G-forces due to acceleration don't exist…..

Now we are coming to the nub of the matter. You claim that the acceleration due to a rocket is “different” from the acceleration due to gravity.  How could this be? When an object is experiencing acceleration it doesn’t matter what kind of force is giving rise to that acceleration or what machine is responsible for it. If an acceleration is present it should have some effect on the spaceship and on its passengers also. In the space station the astronauts are under the influence of a constant acceleration of around 8.4m/s²  why should you claim that this does not affect the astronauts?  In effect what you are saying, namely that there is a difference in acceleration due to gravity and that due to a rocket engine, is a complete right about turn to what the equivalence principle states, namely that there is no difference between the force caused by gravitation and that caused by an acceleration.  In my example, in a space craft far from anywhere else, the passengers are weightless, they are free falling around space, so the application of an acceleration would have no effect on them. As long as they are weightless, acceleration should not affect them.  Or can you claim that these astronauts have weight?

[McQueen (OP)]

Now take that rocket and move out to 50 billion miles from the Earth.  But still firing its engines at the sane thrust as before.  Now this of course will be more thrust than the pull of gravity from the Earth and the rocket will accelerate away from the Earth.  An occupant inside the ship however will still weigh the same on a scale in the ship and objects will still fall towards the floor of the ship just like they did when the ship was hovering just at the surface of the Earth.  Relative to the rocket, objects in this accelerating rocket behave just like they would in a gravitational field.
This is the gist of the equivalence principle.

That’s just it they don’t. Astronauts in the space station are spinning around the earth at about 27,500 kmh, they are under a constant state of acceleration, which according to your post, the astronauts “should still weigh the same on a scale in the ship and the objects will still fall towards the floor.” News update for you. They don’t!