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That atomic clock on board of a deep space probe launched decades ago, if turns around back and safely lands on earth, will show the same or different time than that other atomic clock that never traveled ?

After decades, an atomic clock that has always been at the equator will show equal or different time than one that has always been at 10 metres from the north pole ?

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That atomic clock on board of a deep space probe launched decades ago, if turns around back and safely lands on earth, will show the same or different time than that other atomic clock that never traveled ?

 

After decades, an atomic clock that has always been at the equator will show equal or different time than one that has always been at 10 metres from the north pole ?

The first will show a different time.

 

The second I'm not sure about. The north pole has a higher gravitational field than the equator, but the radial speed at the equator is higher than at the pole. I'm guessing that they would cancel each other out, and the clocks would show the same time, but only calculation could confirm that.

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The first will show a different time.

 

The second I'm not sure about. The north pole has a higher gravitational field than the equator, but the radial speed at the equator is higher than at the pole. I'm guessing that they would cancel each other out, and the clocks would show the same time, but only calculation could confirm that.

I don't think there is any particular reason why they should exactly cancel out, but which would have the greater effect in that situation, I'm unsure.
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That atomic clock on board of a deep space probe launched decades ago, if turns around back and safely lands on earth, will show the same or different time than that other atomic clock that never traveled ?

I guess you're talking about the Voyager satellites. They didn't have atomic clocks. As far as I know, the only satellites outfitted with atomic clocks are the GPS satellites.

 

Suppose some spacecraft outfitted with atomic clocks goes to the far reaches of the solar system and then returned. On return, the spacecraft clock would probably show a different time than Earth clocks. The spacecraft goes far from any gravitational sources, so general relativistic time dilation would make their clocks tick faster. However, it also went away from the Earth at significant speed (at least initially), so special relativistic effects would have made their clocks tick slower. Which is the dominant effect depends on how fast the spacecraft moves relative to the Earth and how much time it spends far away from the Sun.

 

After decades, an atomic clock that has always been at the equator will show equal or different time than one that has always been at 10 metres from the north pole ?

Atomic clocks at mean sea level all tick at the same rate. The geoid, which is what mean sea level is, is an equipotential surface of the gravitation force and the centrifugal force due to Earth rotation.

Edited by D H
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I guess you're talking about the Voyager satellites. They didn't have atomic clocks. As far as I know, the only satellites outfitted with atomic clocks are the GPS satellites.

Or other navigation system satellites (Glonass, Galileo, and the Japanese quasi-zenith system). It's possible communications satellites have them as well.

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The second I'm not sure about. The north pole has a higher gravitational field than the equator, but the radial speed at the equator is higher than at the pole. I'm guessing that they would cancel each other out, and the clocks would show the same time, but only calculation could confirm that.

Your intuition is correct. The effects cancel each other out completely because the geoid surface is equipotential. So, a clock at sea level at the equator and one at the pole are subjected to the same exact gravitational potential. Hence, same ticking rate. See here, for example.

After decades, an atomic clock that has always been at the equator will show equal or different time than one that has always been at 10 metres from the north pole ?

Vessot ran such an experiment . See here

Edited by xyzt
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Your intuition is correct. The effects cancel each other out completely because the geoid surface is equipotential. So, a clock at sea level at the equator and one at the pole are subjected to the same exact gravitational potential. Hence, same ticking rate. See here, for example.

Ah, I suppose that would make sense.
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Thanks, gentlemen.

From responses, see emphasis in gravity effects on the clocks, more than the speed they move along decades, which was the relationship I though would have more effect.

 

Yes, deep space probes as Voyager have no atomic clocks on board; but I wrongly used that as an 'if' example because of the long distance it as traveled at high speed.

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Your intuition is correct. The effects cancel each other out completely because the geoid surface is equipotential. So, a clock at sea level at the equator and one at the pole are subjected to the same exact gravitational potential. Hence, same ticking rate. See here, for example.

 

Vessot ran such an experiment . See here

 

Sorry for the Out Of Subject but the description of the vessot experiment in your link makes me raise some question:

 

The effect of gravity on clocks was tested to great precision by Vessot etal (1980, PRL, 45 2081) who launched a hydrogen maser straight up at 8.5 km/sec, and watched its frequency change as it coasted up to 10,000 km altitude and then fell back to Earth. The frequency shift due to gravity was 400 parts per trillion at 10,000 km altitude, and the experimental result agreed with the expected shift to within 28 parts per quadrillion. The space-time diagram on the left above shows the rocket going up and then falling back down. The black dots on its worldline are clock ticks, and a flash of light (red) is emitted at each tick. The observed frequency changes dramatically during the flight due to the Doppler shift as well as the gravitational redshift. The red curve on the plot at right shows the observed frequency of flashes from the rocket clock arriving at the the ground observer. In this exaggerated diagram, with a rocket velocity of 0.8c, the frequency changes from 1/3 to 3 times the actual clock frequency. To separate the Doppler shift from the gravitational effect, a transponder was used: the ground station transmitted pulses (blue) at a fixed rate, which were then echoed by the rocket (green) and detected on the ground. The rate of transponded pulses changed from 1/9 to 9 times the transmitted rate due to the double Doppler shift on the round trip. By dividing the rate of pulses from the rocket clock (red curve) by the square root of the rate of transponded pulses (green curve), one gets the blue curve which just shows the gravitational effect on clock rates. Note that shaking during launch may change the frequency of the rocket clock, which will move the blue curve up and down, but will not change its shape. The clock was in free fall during the entire time that data was taken.

 

(emphasis mine)

Where was the clock?

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A clock launched straight up at 8.5km/s was atop a multistage rocket. There is little choice for that: a hydrogen gun might achive the speed under the best conditions, but the clock wouldn't survive, and that speed isn't sustainable through the atmosphere.

 

8.5km/s is nearly Earth's liberation speed (11km/s). Zero gravity would result from the extinction of the rocket and the lack of atmosphere above a few 10km.

 

Hydrogen masers can be rugged, more easily than atomic clocks, and serve in (some, not all) GPS satellites for instance.

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on the rocket

But the rocket was launched. That is not "free fall".

And 8.5km/s is not "free fall"

On the other hand If the engines stopped at 10 km as Enthalpy suggests, then the first part of the diagram must be erased and only the return phase could be considered in free fall. I suppose.

And on the third hand, there must have been another clock at rest in order to get the measurement. I suppose- again.

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But the rocket was launched. That is not "free fall".

Once the engines shut down, it's free fall. The only acceleration is that of gravity.

 

And 8.5km/s is not "free fall"

Why?

 

On the other hand If the engines stopped at 10 km as Enthalpy suggests, then the first part of the diagram must be erased and only the return phase could be considered in free fall. I suppose.

Why? A rocket going up after the engines shut off is still in free fall.

 

And on the third hand, there must have been another clock at rest in order to get the measurement. I suppose- again.

Yes. All clock measurements are comparisons between clocks. A clock that runs fast/slow does so when compared to some other clock.

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But the rocket was launched. That is not "free fall".

Once the engines shut down, it's free fall. The only acceleration is that of gravity.

 

I am O.K. with that.

 

 

 

 

 

 

And 8.5km/s is not "free fall"

Why?

 

because free fall is accelerated motion.

 

 

 

 

 

On the other hand If the engines stopped at 10 km as Enthalpy suggests, then the first part of the diagram must be erased and only the return phase could be considered in free fall. I suppose.

Why? A rocket going up after the engines shut off is still in free fall.

 

Yes but then the engines were stopped before reaching the 10 km, let's say at 5 km altidude, and then the resulting diagram would be asymmetric since the part from launching till 5 km would not have been in free fall.

 

 

 

 

 

And on the third hand, there must have been another clock at rest in order to get the measurement. I suppose- again.

Yes. All clock measurements are comparisons between clocks. A clock that runs fast/slow does so when compared to some other clock.

 

So there must have been another clock at rest upon the surface of the Earth, which was not in free fall condition.

 

 

 

 

 

 

 

 

 

Edited by michel123456
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because free fall is accelerated motion.

 

So why can't a rocket get to 8.5 km/sec and then go into free fall?

 

 

 

Yes but then the engines were stopped before reaching the 10 km, let's say at 5 km altidude, and then the resulting diagram would be asymmetric since the part from launching till 5 km would not have been in free fall.

That's 10,000 km, and the powered part of the launch didn't need to go very high, since it reached an acceleration of 18g. Launch probably went up a few hundred km. On return, if it's traveling ~7 km/s at the point where the rockets cut off, it means the return is less than a minute longer than the upward free fall part of the flight, with each half taking almost an hour. Not a big asymmetry.

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Oh yes, I missed the 10,000. My bad.

 


So why can't a rocket get to 8.5 km/sec and then go into free fall?

 

 

It can, it was unclear to me that 8,5 km/s was initial velocity.

 

At return , at the point where the rocket cut off, the rocket in free fall should travel at the same velocity 8 km/s. If the diagram shows the situation from cut off and after, it must be perfectly symmetric.


If i understand correctly, the spacetime diagram on the link simply does not show the departure of the rocket- added by me on the left down part of the diagram. See below

post-19758-0-70165900-1373907982_thumb.jpg

Edited by michel123456
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At return , at the point where the rocket cut off, the rocket in free fall should travel at the same velocity 8 km/s. If the diagram shows the situation from cut off and after, it must be perfectly symmetric.

 

If i understand correctly, the spacetime diagram on the link simply does not show the departure of the rocket- added by me on the left down part of the diagram. See below

The diagram is not experimental data.

"In this exaggerated diagram, with a rocket velocity of 0.8c, the frequency changes from 1/3 to 3 times the actual clock frequency."

(emphasis added)

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