 #1
John SpaceY
 49
 4
 TL;DR Summary

Calculations on the time inside the spacecraft turning at 330 km from our Earth a a speed of 7700 m/s : effects of the speed on the time and effect of the gravity on the time.
Analysis if the speed goes towards the speed of light.
I have some questions linked to the following experimentation:
In 1985, the space shuttle Challenger took on board a cesium atomic clock. The duration of the mission, around our Earth, was seven days and the parameters of the trajectory were as follows: speed equal to 7700 m/s and altitude h equal to 330 km in practically circular orbit.
The onboard atomic clock lagged its counterpart on Earth by about
295 ps/s (1 ps = 1 pico seconde = 10^12s).
330 ps/s is attributable to the special relativity (effect of the speed on the time inside the moving spacecraft ).
The effect attributable to general relativity is +35 ps/s, it is linked, among other things, to the fact that, the gravitational field being weaker there, the clock sees its rate increase.
I have checked these values with the SR and GR Theories and I found the following values :
With the SR for 1 second seen on the Earth, Challenger see
(1 – 330.10^12) and so there is a difference of  330 pico seconds per second (ps/s)
With the GR I have found an increase of time of + 32 ps/s inside Challenger: there is an increase of time because the gravity at 330 km is lower (8,87 m/s^2) than on the surface of our Earth (9,81 m/s^2).
And so  330 + 32 =  298 ps/s
And this is very near about the 295 ps/s measured in this experimentation.
The error I have made for the effect of gravity on the time inside the moving spacecraft is 9,375% in order to go from 32 ps/s to 35 ps/s.
Now, I will do the same experimentation on the Moon, and I have found the following values:
With the SR for 1 second seen on the Moon, the spacecraft will see
(1 – 330.10^12) and so there is a difference of  330 pico seconds per second (ps/s)
With the GR I have found an increase of time of + 96,26 ps/s inside the spacecraft : there is an increase of time because the gravity at 330 km is lower than on the surface of our Moon.
The gravity on the surface of the Moon is 1,62 m/s^2
The gravity at 330 km from the surface of the Moon is 1,148 m/s^2
I will add 9,375% of error due to my calculation to 96,26 ps/s (I suppose the same error than the error I have made for the experimentation done on our Earth: see above) and so I will have 105,28 ps/s
And so  330 + 105,28 =  224,72 ps/s
If the time spent on the Moon will be 1 s, the time seen inside the spacecraft will be: 1s – 224,72 ps (which is > 0)
This experimentation, by turning around the Moon, even if it is theoretically possible with our current technology, will not add more information than in 1985, with the space shuttle Challenger turning around our Earth
Now, I will do another experimentation on the Moon:
I have calculated the distance under the surface of the Moon where the gravity will be the same than the gravity at 330 km from the surface of our Earth, which is 8,87 m/s^2
The distance is 993,49 km under the surface of the Moon.
It is not possible to turn at 7700 m/s around the Moon at 993,49 km under the surface, but it is possible to turn around our Earth at 330 km from the surface of our Earth and the time inside the spacecraft , seen from the surface of the Moon, will be the same: indeed, the speed of the spacecraft will be the same (same effect of the speed on the time inside the spacecraft (SR Theory) and same effect of the gravity on the time inside the spacecraft (GR Theory)
I have found the following values:
With the SR for 1 second seen on the Moon, the spacecraft , turning around our Earth at 330 km at a speed of 7700 m/s, will see
(1 – 330.10^12) and so there is a difference of  330 pico seconds per second (ps/s)
With the GR I have found a decrease of time of  1476 ps/s inside the spacecraft : there is a decrease of time because the gravity at 330 km from the surface of our Earth is higher than on the surface of our Moon.
The gravity on the surface of the Moon is 1,62 m/s^2
The gravity at 330 km from the surface of the Earth is 8,87 m/s^2
I will add 9,375% of error due to my calculation to  1476 ps/s (I suppose the same error than the error I have made for the experimentation done on our Earth: see above) and so I will have 1614,375 ps/s
And so  330 – 1614,375 =  1944,375 ps/s
If the time spent on the Moon will be 1 s, the time seen inside the spacecraft will be: 1s – 1944,375 ps (which is > 0)
This experimentation is interesting to be done and should be easy to do: just put a reference clock on the surface of the Moon and turn again (like in 1985) around our Earth at 330 km.
For the 2 experimentations above, the time inside the spacecraft will be reduced, but less than 1 second, for 1 second spent on the Moon. And so, the time inside the spacecraft , for the 2 experimentations will be positive. It is because the speed of the spacecraft is low compared to the speed of light.
Now, I would like to go to the limits and these experimentations are not possible today but theoretically can be calculated.
Indeed, I would like to turn around our Earth at a speed v which will tend towards the speed of light, c, and this is not possible today (we have not the technology for such a spacecraft !).
If v tends towards c, If I do again the experimentations above, I will have the following values:
First experimentation:
The spacecraft will turn around the Moon at a speed very near to c, at 330 km from the surface of the Moon:
With the SR for 1 second seen on the Moon, the spacecraft will see
a time going towards 0, because the speed is very near of the speed of light c. The time inside the spacecraft will tend towards 0 in these conditions.
With the GR I have found an increase of time of 105,28 ps/s
inside the spacecraft (see before): there is an increase of time because the gravity at 330 km is lower than on the surface of our Moon.
The gravity on the surface of the Moon is 1,62 m/s^2
The gravity at 330 km from the surface of the Moon is 1,148 m/s^2
And so 1  1 + 105,28 = + 105,28 ps/s
If the time spent on the surface of the Moon will be 1 s, the time seen inside the spacecraft , turning around our Earth, will be: + 105,28 ps
The time inside the spacecraft will be reduced, but less than 1 second.
Second experimentation:
The spacecraft will turn around our Earth at a speed very near to c, at 330 km from the surface of the Earth:
With the SR for 1 second seen on the Moon, the spacecraft will see
a time going towards 0, because the speed is very near of the speed of light c. The time inside the spacecraft will tend towards 0 in these conditions.
With the GR I have found a decrease of time of  1614,375 ps/s inside the spacecraft : there is a decrease of time because the gravity at 330 km from the surface of our Earth is higher than on the surface of our Moon.
The gravity on the surface of the Moon is 1,62 m/s^2
The gravity at 330 km from the surface of the Earth is 8,87 m/s^2
And so 1  1  1614,375 =  1614,375 ps/s
If the time spent on the surface of the Moon will be 1 s, the time seen inside the spacecraft , turning around our Earth, will be:
 1614,375 ps
The time inside the spacecraft will be reduced, and more than 1 second.
And now come my questions:
What do you think about these calculations ?
If it is not correct, where are the mistakes ?
If it is correct, what could be the interpretations of these results ?
Maybe the cumulative effects of speed and gravity on the time inside a moving object could create the conditions where the time could become negative and this negative time could be used as a parameter in some calculations ?
In 1985, the space shuttle Challenger took on board a cesium atomic clock. The duration of the mission, around our Earth, was seven days and the parameters of the trajectory were as follows: speed equal to 7700 m/s and altitude h equal to 330 km in practically circular orbit.
The onboard atomic clock lagged its counterpart on Earth by about
295 ps/s (1 ps = 1 pico seconde = 10^12s).
330 ps/s is attributable to the special relativity (effect of the speed on the time inside the moving spacecraft ).
The effect attributable to general relativity is +35 ps/s, it is linked, among other things, to the fact that, the gravitational field being weaker there, the clock sees its rate increase.
I have checked these values with the SR and GR Theories and I found the following values :
With the SR for 1 second seen on the Earth, Challenger see
(1 – 330.10^12) and so there is a difference of  330 pico seconds per second (ps/s)
With the GR I have found an increase of time of + 32 ps/s inside Challenger: there is an increase of time because the gravity at 330 km is lower (8,87 m/s^2) than on the surface of our Earth (9,81 m/s^2).
And so  330 + 32 =  298 ps/s
And this is very near about the 295 ps/s measured in this experimentation.
The error I have made for the effect of gravity on the time inside the moving spacecraft is 9,375% in order to go from 32 ps/s to 35 ps/s.
Now, I will do the same experimentation on the Moon, and I have found the following values:
With the SR for 1 second seen on the Moon, the spacecraft will see
(1 – 330.10^12) and so there is a difference of  330 pico seconds per second (ps/s)
With the GR I have found an increase of time of + 96,26 ps/s inside the spacecraft : there is an increase of time because the gravity at 330 km is lower than on the surface of our Moon.
The gravity on the surface of the Moon is 1,62 m/s^2
The gravity at 330 km from the surface of the Moon is 1,148 m/s^2
I will add 9,375% of error due to my calculation to 96,26 ps/s (I suppose the same error than the error I have made for the experimentation done on our Earth: see above) and so I will have 105,28 ps/s
And so  330 + 105,28 =  224,72 ps/s
If the time spent on the Moon will be 1 s, the time seen inside the spacecraft will be: 1s – 224,72 ps (which is > 0)
This experimentation, by turning around the Moon, even if it is theoretically possible with our current technology, will not add more information than in 1985, with the space shuttle Challenger turning around our Earth
Now, I will do another experimentation on the Moon:
I have calculated the distance under the surface of the Moon where the gravity will be the same than the gravity at 330 km from the surface of our Earth, which is 8,87 m/s^2
The distance is 993,49 km under the surface of the Moon.
It is not possible to turn at 7700 m/s around the Moon at 993,49 km under the surface, but it is possible to turn around our Earth at 330 km from the surface of our Earth and the time inside the spacecraft , seen from the surface of the Moon, will be the same: indeed, the speed of the spacecraft will be the same (same effect of the speed on the time inside the spacecraft (SR Theory) and same effect of the gravity on the time inside the spacecraft (GR Theory)
I have found the following values:
With the SR for 1 second seen on the Moon, the spacecraft , turning around our Earth at 330 km at a speed of 7700 m/s, will see
(1 – 330.10^12) and so there is a difference of  330 pico seconds per second (ps/s)
With the GR I have found a decrease of time of  1476 ps/s inside the spacecraft : there is a decrease of time because the gravity at 330 km from the surface of our Earth is higher than on the surface of our Moon.
The gravity on the surface of the Moon is 1,62 m/s^2
The gravity at 330 km from the surface of the Earth is 8,87 m/s^2
I will add 9,375% of error due to my calculation to  1476 ps/s (I suppose the same error than the error I have made for the experimentation done on our Earth: see above) and so I will have 1614,375 ps/s
And so  330 – 1614,375 =  1944,375 ps/s
If the time spent on the Moon will be 1 s, the time seen inside the spacecraft will be: 1s – 1944,375 ps (which is > 0)
This experimentation is interesting to be done and should be easy to do: just put a reference clock on the surface of the Moon and turn again (like in 1985) around our Earth at 330 km.
For the 2 experimentations above, the time inside the spacecraft will be reduced, but less than 1 second, for 1 second spent on the Moon. And so, the time inside the spacecraft , for the 2 experimentations will be positive. It is because the speed of the spacecraft is low compared to the speed of light.
Now, I would like to go to the limits and these experimentations are not possible today but theoretically can be calculated.
Indeed, I would like to turn around our Earth at a speed v which will tend towards the speed of light, c, and this is not possible today (we have not the technology for such a spacecraft !).
If v tends towards c, If I do again the experimentations above, I will have the following values:
First experimentation:
The spacecraft will turn around the Moon at a speed very near to c, at 330 km from the surface of the Moon:
With the SR for 1 second seen on the Moon, the spacecraft will see
a time going towards 0, because the speed is very near of the speed of light c. The time inside the spacecraft will tend towards 0 in these conditions.
With the GR I have found an increase of time of 105,28 ps/s
inside the spacecraft (see before): there is an increase of time because the gravity at 330 km is lower than on the surface of our Moon.
The gravity on the surface of the Moon is 1,62 m/s^2
The gravity at 330 km from the surface of the Moon is 1,148 m/s^2
And so 1  1 + 105,28 = + 105,28 ps/s
If the time spent on the surface of the Moon will be 1 s, the time seen inside the spacecraft , turning around our Earth, will be: + 105,28 ps
The time inside the spacecraft will be reduced, but less than 1 second.
Second experimentation:
The spacecraft will turn around our Earth at a speed very near to c, at 330 km from the surface of the Earth:
With the SR for 1 second seen on the Moon, the spacecraft will see
a time going towards 0, because the speed is very near of the speed of light c. The time inside the spacecraft will tend towards 0 in these conditions.
With the GR I have found a decrease of time of  1614,375 ps/s inside the spacecraft : there is a decrease of time because the gravity at 330 km from the surface of our Earth is higher than on the surface of our Moon.
The gravity on the surface of the Moon is 1,62 m/s^2
The gravity at 330 km from the surface of the Earth is 8,87 m/s^2
And so 1  1  1614,375 =  1614,375 ps/s
If the time spent on the surface of the Moon will be 1 s, the time seen inside the spacecraft , turning around our Earth, will be:
 1614,375 ps
The time inside the spacecraft will be reduced, and more than 1 second.
And now come my questions:
What do you think about these calculations ?
If it is not correct, where are the mistakes ?
If it is correct, what could be the interpretations of these results ?
Maybe the cumulative effects of speed and gravity on the time inside a moving object could create the conditions where the time could become negative and this negative time could be used as a parameter in some calculations ?