lots of energy needed for accelerating the earth

a = dv/dt (classic and relativistic)

Dv and dt change at relativistic speeds.

t = (c/a) sh(aT/c) =  sqrt[(d/c)2 + 2d/a]
d = (c2/a) [ch(aT/c) - 1] =  (c2/a) (sqrt[1 + (at/c)2] - 1)
v = c th(aT/c) = at / sqrt[1 + (at/c)2]
T = (c/a) sh-1(at/c) = (c/a) ch-1 [ad/c2 + 1]
γ = ch(aT/c) = sqrt[1 + (at/c)2] = ad/c2 + 1

These equations are valid in any consistent system of units such as seconds for time, metres for distance, metres per second for speeds and metres per second squared for accelerations.  In these units c = 3 ? 108 m/s (approx).  To do some example calculations it is easier to use units of years for time and light years for distance.  Then c = 1 lyr/yr and g = 1.03 lyr/yr2.  Here are some typical answers for a = 1g.

    T              t            d              v                       γ
    1 year    1.19 yrs  0.56 lyrs   0.77c                 1.58 
    2            3.75      2.90          0.97                  3.99
    5            83.7      82.7          0.99993              86.2
    8           1,840     1,839         0.9999998         1,895
   12       113,243   113,242       0.99999999996   116,641

Quote from: GeertD on January 19, 2009, 08:46:16 AM

a = dv/dt (classic and relativistic)

Dv and dt change at relativistic speeds.

t = (c/a) sh(aT/c) =  sqrt[(d/c)2 + 2d/a]
d = (c2/a) [ch(aT/c) - 1] =  (c2/a) (sqrt[1 + (at/c)2] - 1)
v = c th(aT/c) = at / sqrt[1 + (at/c)2]
T = (c/a) sh-1(at/c) = (c/a) ch-1 [ad/c2 + 1]
γ = ch(aT/c) = sqrt[1 + (at/c)2] = ad/c2 + 1

These equations are valid in any consistent system of units such as seconds for time, metres for distance, metres per second for speeds and metres per second squared for accelerations.  In these units c = 3 ? 108 m/s (approx).  To do some example calculations it is easier to use units of years for time and light years for distance.  Then c = 1 lyr/yr and g = 1.03 lyr/yr2.  Here are some typical answers for a = 1g.

    T              t            d              v                       γ
    1 year    1.19 yrs  0.56 lyrs   0.77c                 1.58 
    2            3.75      2.90          0.97                  3.99
    5            83.7      82.7          0.99993              86.2
    8           1,840     1,839         0.9999998         1,895
   12       113,243   113,242       0.99999999996   116,641

FE-theory says gravity doesn't exists but you just used it

Quote from: Robbyj on January 19, 2009, 02:25:13 PM

Quote from: GeertD on January 19, 2009, 08:46:16 AM

a = dv/dt (classic and relativistic)

Dv and dt change at relativistic speeds.

t = (c/a) sh(aT/c) =  sqrt[(d/c)2 + 2d/a]
d = (c2/a) [ch(aT/c) - 1] =  (c2/a) (sqrt[1 + (at/c)2] - 1)
v = c th(aT/c) = at / sqrt[1 + (at/c)2]
T = (c/a) sh-1(at/c) = (c/a) ch-1 [ad/c2 + 1]
γ = ch(aT/c) = sqrt[1 + (at/c)2] = ad/c2 + 1

These equations are valid in any consistent system of units such as seconds for time, metres for distance, metres per second for speeds and metres per second squared for accelerations.  In these units c = 3 ? 108 m/s (approx).  To do some example calculations it is easier to use units of years for time and light years for distance.  Then c = 1 lyr/yr and g = 1.03 lyr/yr2.  Here are some typical answers for a = 1g.

    T              t            d              v                       γ
    1 year    1.19 yrs  0.56 lyrs   0.77c                 1.58 
    2            3.75      2.90          0.97                  3.99
    5            83.7      82.7          0.99993              86.2
    8           1,840     1,839         0.9999998         1,895
   12       113,243   113,242       0.99999999996   116,641

FE-theory says gravity doesn't exists but you just used it

So far every equation is wrong here, you can not use a non relativistic equation to explain a relativistic situation, and you cna not use an equation the uses a new point of origin for every calculation.

FE-theory says gravity doesn't exists but you just used it

No, it sets acceleration equal to g which is what your question concerned.  Set acceleration to whatever you please.  The outcome will be the same.

and you cna not use an equation the uses a new point of origin for every calculation.

Then use the hyperbolic function if you prefer.

Quote from: GeertD on January 19, 2009, 02:29:38 PM

FE-theory says gravity doesn't exists but you just used it

No, it sets acceleration equal to g which is what your question concerned.  Set acceleration to whatever you please.  The outcome will be the same.

g is the symbol for the standard gravity
http://en.wikipedia.org/wiki/Standard_gravity

Quote from: Robbyj on January 19, 2009, 02:33:38 PM

Quote from: GeertD on January 19, 2009, 02:29:38 PM

FE-theory says gravity doesn't exists but you just used it

No, it sets acceleration equal to g which is what your question concerned.  Set acceleration to whatever you please.  The outcome will be the same.

g is the symbol for the standard gravity
http://en.wikipedia.org/wiki/Standard_gravity

g and a are the same thing.

Quote from: Robbyj on January 19, 2009, 03:56:55 AM

Quote from: GeertD on January 18, 2009, 11:53:53 PM

What you are saying is:
u = 0 , u is the speed of the reference in space (earth in this case) => earth stands still
v is speed of object on earth
w is speed of object in space
if we calculate this we have the wonderful conclusion that v=w
if u and v are little compared to the speed light this is reduced to w = u + v
This is a complete wrong formula to prove this

Incorrect.

Don't just say "incorrect" but corect it if it's wrong

Don't let one mistake cloud the soundness of the other argument. First the wrong argument: no matter what speed you have reached with respect to a given frame of reference, you can accelerate at any rate (9.8 m/s/s, for example) with respect to the frame of reference you are in right now. An observer placed on wherever the Earth started accelerating would see an Earth that is infinitesimally under the speed of light, and accelerating infinitesimally towards the speed of light. We would see an Earth that accelerates at 9.8 m/s/s forever.

Now the right argument: the amount of energy required to accelerate Earth at 9.8 m/s/s forever is far beyond what anyone can explain. Somewhere I calculated that it would be something like having Hiroshima-like atom bombs carpeting the underbelly of Earth, side by side all under Earth (one per square meter), exploding every few hours. Your calculations in this respect are sound. I have not seen the details, but at least the order of magnitude is most probably right. You can also see it like this: a Saturn V rocket is required to accelerate a few tons of command module at an acceleration of some 13 m/s/s (if it were in outer space) for a few minutes. We are talking an acceleration of 9.8 m/s/s for the last 4.5 billion years or more.

Quote from: GeertD on January 19, 2009, 08:46:16 AM

a = dv/dt (classic and relativistic)

Dv and dt change at relativistic speeds.

t = (c/a) sh(aT/c) =  sqrt[(d/c)2 + 2d/a]
d = (c2/a) [ch(aT/c) - 1] =  (c2/a) (sqrt[1 + (at/c)2] - 1)
v = c th(aT/c) = at / sqrt[1 + (at/c)2]
T = (c/a) sh-1(at/c) = (c/a) ch-1 [ad/c2 + 1]
γ = ch(aT/c) = sqrt[1 + (at/c)2] = ad/c2 + 1

These equations are valid in any consistent system of units such as seconds for time, metres for distance, metres per second for speeds and metres per second squared for accelerations.  In these units c = 3 ? 108 m/s (approx).  To do some example calculations it is easier to use units of years for time and light years for distance.  Then c = 1 lyr/yr and g = 1.03 lyr/yr2.  Here are some typical answers for a = 1g.

    T              t            d              v                       γ
    1 year    1.19 yrs  0.56 lyrs   0.77c                 1.58 
    2            3.75      2.90          0.97                  3.99
    5            83.7      82.7          0.99993              86.2
    8           1,840     1,839         0.9999998         1,895
   12       113,243   113,242       0.99999999996   116,641

Please look up the term 'proper frame of reference' and 'proper acceleration' and learn some calculus.

Quote from: Robbyj on January 19, 2009, 02:25:13 PM

Quote from: GeertD on January 19, 2009, 08:46:16 AM

a = dv/dt (classic and relativistic)

Dv and dt change at relativistic speeds.

t = (c/a) sh(aT/c) =  sqrt[(d/c)2 + 2d/a]
d = (c2/a) [ch(aT/c) - 1] =  (c2/a) (sqrt[1 + (at/c)2] - 1)
v = c th(aT/c) = at / sqrt[1 + (at/c)2]
T = (c/a) sh-1(at/c) = (c/a) ch-1 [ad/c2 + 1]
γ = ch(aT/c) = sqrt[1 + (at/c)2] = ad/c2 + 1

These equations are valid in any consistent system of units such as seconds for time, metres for distance, metres per second for speeds and metres per second squared for accelerations.  In these units c = 3 ? 108 m/s (approx).  To do some example calculations it is easier to use units of years for time and light years for distance.  Then c = 1 lyr/yr and g = 1.03 lyr/yr2.  Here are some typical answers for a = 1g.

    T              t            d              v                       γ
    1 year    1.19 yrs  0.56 lyrs   0.77c                 1.58 
    2            3.75      2.90          0.97                  3.99
    5            83.7      82.7          0.99993              86.2
    8           1,840     1,839         0.9999998         1,895
   12       113,243   113,242       0.99999999996   116,641

Please look up the term 'proper frame of reference' and 'proper acceleration' and learn some calculus.

This is incorrect: the frame of reference should be the same but here it's always changing

Quote from: GeertD on January 18, 2009, 11:58:48 PM

@ "TheEngineer": Do you want an explanation for the magic in your sign?

Please. 

And check your math.  Your equations are wrong.

I'm waiting.

Quote from: John Jackson on January 19, 2009, 03:34:50 PM

Quote from: Robbyj on January 19, 2009, 02:25:13 PM

Quote from: GeertD on January 19, 2009, 08:46:16 AM

a = dv/dt (classic and relativistic)

Dv and dt change at relativistic speeds.

t = (c/a) sh(aT/c) =  sqrt[(d/c)2 + 2d/a]
d = (c2/a) [ch(aT/c) - 1] =  (c2/a) (sqrt[1 + (at/c)2] - 1)
v = c th(aT/c) = at / sqrt[1 + (at/c)2]
T = (c/a) sh-1(at/c) = (c/a) ch-1 [ad/c2 + 1]
γ = ch(aT/c) = sqrt[1 + (at/c)2] = ad/c2 + 1

These equations are valid in any consistent system of units such as seconds for time, metres for distance, metres per second for speeds and metres per second squared for accelerations.  In these units c = 3 ? 108 m/s (approx).  To do some example calculations it is easier to use units of years for time and light years for distance.  Then c = 1 lyr/yr and g = 1.03 lyr/yr2.  Here are some typical answers for a = 1g.

    T              t            d              v                       γ
    1 year    1.19 yrs  0.56 lyrs   0.77c                 1.58 
    2            3.75      2.90          0.97                  3.99
    5            83.7      82.7          0.99993              86.2
    8           1,840     1,839         0.9999998         1,895
   12       113,243   113,242       0.99999999996   116,641

Please look up the term 'proper frame of reference' and 'proper acceleration' and learn some calculus.

This is incorrect: the frame of reference should be the same but here it's always changing

If that is wrong and you know the "correct" way to calculate it, you should stop wasting time on this forum and start preparing your Nobel Prize acceptance speech.

Quote from: GeertD on January 19, 2009, 03:37:52 PM

Quote from: John Jackson on January 19, 2009, 03:34:50 PM

Quote from: Robbyj on January 19, 2009, 02:25:13 PM

Quote from: GeertD on January 19, 2009, 08:46:16 AM

a = dv/dt (classic and relativistic)

Dv and dt change at relativistic speeds.

t = (c/a) sh(aT/c) =  sqrt[(d/c)2 + 2d/a]
d = (c2/a) [ch(aT/c) - 1] =  (c2/a) (sqrt[1 + (at/c)2] - 1)
v = c th(aT/c) = at / sqrt[1 + (at/c)2]
T = (c/a) sh-1(at/c) = (c/a) ch-1 [ad/c2 + 1]
γ = ch(aT/c) = sqrt[1 + (at/c)2] = ad/c2 + 1

These equations are valid in any consistent system of units such as seconds for time, metres for distance, metres per second for speeds and metres per second squared for accelerations.  In these units c = 3 ? 108 m/s (approx).  To do some example calculations it is easier to use units of years for time and light years for distance.  Then c = 1 lyr/yr and g = 1.03 lyr/yr2.  Here are some typical answers for a = 1g.

    T              t            d              v                       γ
    1 year    1.19 yrs  0.56 lyrs   0.77c                 1.58 
    2            3.75      2.90          0.97                  3.99
    5            83.7      82.7          0.99993              86.2
    8           1,840     1,839         0.9999998         1,895
   12       113,243   113,242       0.99999999996   116,641

Please look up the term 'proper frame of reference' and 'proper acceleration' and learn some calculus.

This is incorrect: the frame of reference should be the same but here it's always changing

If that is wrong and you know the "correct" way to calculate it, you should stop wasting time on this forum and start preparing your Nobel Prize acceptance speech.

I know if one of us is going to win the nobel prize it's me, you are transforming the speed from relative to absolute then you take the absolute speed as the speed of your frame of reference and transform the speed again in this new frame of reference.
This is what you do:

• An object on earth has a speed and the earth don't move. The absolute speed of the object is the same as the relative speed of the object.
• The speed of your frame of reference is now the absolute speed of the object. Now your frame of reference (now it's moving). Again you take an object with a relative speed to the earth.
• You transform it's relative speed to the absolute speed.
• And now you do step 2 and 3 again and again.

You don't calculate the speed of the earth with this but the speed of an object that's moving to an object that's moving to an object ... that's moving to an object that's moving to the earth wich don't move in your calculations.
The earth isn't moving at all. It's like you have an object that's moving to the earth with a certain speed. . And

Quote from: TheEngineer on January 19, 2009, 04:42:23 AM

Quote from: GeertD on January 18, 2009, 11:58:48 PM

@ "TheEngineer": Do you want an explanation for the magic in your sign?

Please. 

And check your math.  Your equations are wrong.

I'm waiting.

Probably you haven't read everything what I wrote:

There is a power between to objects:
F = c* m*M/r²

where c is a constant factor this can be experimental verified
m is mass of object, M is mass of other object
r the distance between the objects

here: M is mass earth
m is the mass of the object (eg a person)
r the distance to the center of earth

I shall give some extra explanation:
You can verify the existence of this force with the Cavendish experiment: http://en.wikipedia.org/wiki/Cavendish_experiment
With this experiment you can also calculate the constant.

http://en.wikipedia.org/wiki/Gravitational_constant

The constant that I've called c is here called G

you are transforming the speed from relative to absolute

No, that is calculating the velocity relative to an initial inertial observer.  There is no absolute reference frame.

The simple answer is that it is actually traveling the opposite direction and slowing down at a rate of 9.8m/s/s due to the properties of the space medium.

That also works.

I know if one of us is going to win the nobel prize it's me, you are transforming the speed from relative to absolute then you take the absolute speed as the speed of your frame of reference and transform the speed again in this new frame of reference.
This is what you do:

• An object on earth has a speed and the earth don't move. The absolute speed of the object is the same as the relative speed of the object.
• The speed of your frame of reference is now the absolute speed of the object. Now your frame of reference (now it's moving). Again you take an object with a relative speed to the earth.
• You transform it's relative speed to the absolute speed.
• And now you do step 2 and 3 again and again.

You don't calculate the speed of the earth with this but the speed of an object that's moving to an object that's moving to an object ... that's moving to an object that's moving to the earth wich don't move in your calculations.
The earth isn't moving at all. It's like you have an object that's moving to the earth with a certain speed. . And

Those numbers are for an inertial observer calculating the speed of the FE at any given time.  There is no second object.

Quote from: TheEngineer on January 19, 2009, 05:08:24 PM

Quote from: TheEngineer on January 19, 2009, 04:42:23 AM

Quote from: GeertD on January 18, 2009, 11:58:48 PM

@ "TheEngineer": Do you want an explanation for the magic in your sign?

Please. 

And check your math.  Your equations are wrong.

I'm waiting.

Probably you haven't read everything what I wrote:

Quote from: GeertD on January 19, 2009, 08:50:12 AM

There is a power between to objects:
F = c* m*M/r²

where c is a constant factor this can be experimental verified
m is mass of object, M is mass of other object
r the distance between the objects

here: M is mass earth
m is the mass of the object (eg a person)
r the distance to the center of earth

I shall give some extra explanation:
You can verify the existence of this force with the Cavendish experiment: http://en.wikipedia.org/wiki/Cavendish_experiment
With this experiment you can also calculate the constant.

http://en.wikipedia.org/wiki/Gravitational_constant

The constant that I've called c is here called G

So, where is that explanation?  All you have given me is a formula that has been 'adjusted' to fit the observed data.

Quote from: GeertD on January 20, 2009, 02:07:24 AM

Quote from: TheEngineer on January 19, 2009, 05:08:24 PM

Quote from: TheEngineer on January 19, 2009, 04:42:23 AM

Quote from: GeertD on January 18, 2009, 11:58:48 PM

@ "TheEngineer": Do you want an explanation for the magic in your sign?

Please. 

And check your math.  Your equations are wrong.

I'm waiting.

Probably you haven't read everything what I wrote:

Quote from: GeertD on January 19, 2009, 08:50:12 AM

There is a power between to objects:
F = c* m*M/r²

where c is a constant factor this can be experimental verified
m is mass of object, M is mass of other object
r the distance between the objects

here: M is mass earth
m is the mass of the object (eg a person)
r the distance to the center of earth

I shall give some extra explanation:
You can verify the existence of this force with the Cavendish experiment: http://en.wikipedia.org/wiki/Cavendish_experiment
With this experiment you can also calculate the constant.

http://en.wikipedia.org/wiki/Gravitational_constant

The constant that I've called c is here called G

So, where is that explanation?  All you have given me is a formula that has been 'adjusted' to fit the observed data.

It's equivalent with the cavendish experiment. There's a force between tow objects, this can be the earth and a person or 2 masses like in the cavendish experiment. How can you explain it?
Gravitation is the force between the earth and an object on the earth, like a person.

It's equivalent with the cavendish experiment. There's a force between tow objects, this can be the earth and a person or 2 masses like in the cavendish experiment. How can you explain it?
Gravitation is the force between the earth and an object on the earth, like a person.

So like I said, where is the explanation?  All you have given me is an observation and an equation that has to be adjusted so that it will work with that observation.  You still have yet to explain the 'magic' part.

Quote from: GeertD on January 21, 2009, 12:59:19 AM

It's equivalent with the cavendish experiment. There's a force between tow objects, this can be the earth and a person or 2 masses like in the cavendish experiment. How can you explain it?
Gravitation is the force between the earth and an object on the earth, like a person.

So like I said, where is the explanation?  All you have given me is an observation and an equation that has to be adjusted so that it will work with that observation.  You still have yet to explain the 'magic' part.

And you call yourself "TheEngineer"?

Quote from: GeertD on January 21, 2009, 12:59:19 AM

It's equivalent with the cavendish experiment. There's a force between tow objects, this can be the earth and a person or 2 masses like in the cavendish experiment. How can you explain it?
Gravitation is the force between the earth and an object on the earth, like a person.

So like I said, where is the explanation?  All you have given me is an observation and an equation that has to be adjusted so that it will work with that observation.  You still have yet to explain the 'magic' part.

I've a better idea: Can you give an explanation for the Cavendish experiment.

Quote from: TheEngineer on January 21, 2009, 04:43:44 AM

Quote from: GeertD on January 21, 2009, 12:59:19 AM

It's equivalent with the cavendish experiment. There's a force between tow objects, this can be the earth and a person or 2 masses like in the cavendish experiment. How can you explain it?
Gravitation is the force between the earth and an object on the earth, like a person.

So like I said, where is the explanation?  All you have given me is an observation and an equation that has to be adjusted so that it will work with that observation.  You still have yet to explain the 'magic' part.

I've a better idea: Can you give an explanation for the Cavendish experiment.

Can you duplicate that "experiment"?

Quote from: GeertD on January 21, 2009, 05:29:29 AM

Quote from: TheEngineer on January 21, 2009, 04:43:44 AM

Quote from: GeertD on January 21, 2009, 12:59:19 AM

It's equivalent with the cavendish experiment. There's a force between tow objects, this can be the earth and a person or 2 masses like in the cavendish experiment. How can you explain it?
Gravitation is the force between the earth and an object on the earth, like a person.

So like I said, where is the explanation?  All you have given me is an observation and an equation that has to be adjusted so that it will work with that observation.  You still have yet to explain the 'magic' part.

I've a better idea: Can you give an explanation for the Cavendish experiment.

Can you duplicate that "experiment"?

Can you explain it?
I haven't done this experiment but some of my friends did this experiment.
Where I studied physics we must do a lot experiments this was one of the experiments we could do but I've done different experiments.

Quote from: Raist on January 21, 2009, 08:38:35 AM

Quote from: GeertD on January 21, 2009, 05:29:29 AM

Quote from: TheEngineer on January 21, 2009, 04:43:44 AM

Quote from: GeertD on January 21, 2009, 12:59:19 AM

It's equivalent with the cavendish experiment. There's a force between tow objects, this can be the earth and a person or 2 masses like in the cavendish experiment. How can you explain it?
Gravitation is the force between the earth and an object on the earth, like a person.

So like I said, where is the explanation?  All you have given me is an observation and an equation that has to be adjusted so that it will work with that observation.  You still have yet to explain the 'magic' part.

I've a better idea: Can you give an explanation for the Cavendish experiment.

Can you duplicate that "experiment"?

Can you explain it?
I haven't done this experiment but some of my friends did this experiment.
Where I studied physics we must do a lot experiments this was one of the experiments we could do but I've done different experiments.

Oh so when you studied physics... in high school? In university?

Sounds a bit too complex for just an "experiment."

Please go into detail though.

Quote from: GeertD on January 21, 2009, 08:52:14 AM

Quote from: Raist on January 21, 2009, 08:38:35 AM

Quote from: GeertD on January 21, 2009, 05:29:29 AM

Quote from: TheEngineer on January 21, 2009, 04:43:44 AM

Quote from: GeertD on January 21, 2009, 12:59:19 AM

It's equivalent with the cavendish experiment. There's a force between tow objects, this can be the earth and a person or 2 masses like in the cavendish experiment. How can you explain it?
Gravitation is the force between the earth and an object on the earth, like a person.

So like I said, where is the explanation?  All you have given me is an observation and an equation that has to be adjusted so that it will work with that observation.  You still have yet to explain the 'magic' part.

I've a better idea: Can you give an explanation for the Cavendish experiment.

Can you duplicate that "experiment"?

Can you explain it?
I haven't done this experiment but some of my friends did this experiment.
Where I studied physics we must do a lot experiments this was one of the experiments we could do but I've done different experiments.

Oh so when you studied physics... in high school? In university?

Sounds a bit too complex for just an "experiment."

Please go into detail though.

I can but you wouldn't understand it.
Maybe instead of avoiding the question you can answer it.

Quote from: Raist on January 21, 2009, 08:53:50 AM

Quote from: GeertD on January 21, 2009, 08:52:14 AM

Quote from: Raist on January 21, 2009, 08:38:35 AM

Quote from: GeertD on January 21, 2009, 05:29:29 AM

Quote from: TheEngineer on January 21, 2009, 04:43:44 AM

Quote from: GeertD on January 21, 2009, 12:59:19 AM

It's equivalent with the cavendish experiment. There's a force between tow objects, this can be the earth and a person or 2 masses like in the cavendish experiment. How can you explain it?
Gravitation is the force between the earth and an object on the earth, like a person.

So like I said, where is the explanation?  All you have given me is an observation and an equation that has to be adjusted so that it will work with that observation.  You still have yet to explain the 'magic' part.

I've a better idea: Can you give an explanation for the Cavendish experiment.

Can you duplicate that "experiment"?

Can you explain it?
I haven't done this experiment but some of my friends did this experiment.
Where I studied physics we must do a lot experiments this was one of the experiments we could do but I've done different experiments.

Oh so when you studied physics... in high school? In university?

Sounds a bit too complex for just an "experiment."

Please go into detail though.

I can but you wouldn't understand it.
Maybe instead of avoiding the question you can answer it.

I doubt it.

I will help him out a bit
http://www.leydenscience.org/physics/gravitation/cavend.htm

I will help him out a bit
http://www.leydenscience.org/physics/gravitation/cavend.htm

indeed that's it. Not very complex or is it?
Actually this is not a difficult experiment (some of us did this in their first year in university) you just need some accurate equipment.