If a spaceship approaches a rapidly spinning planet, would the planet's inhabitants , the inhabitants of the planet where the spaceship came from , and the spaceship's occupants observe time dilation between all each other due to their relative motion?
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3 Answers
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Modern atomic clocks keep track of time so accurately that velocities well within reach of our modern technogies are sufficient for detection of relativistic effects.
The following is sourced from a discussion that is available on the website mathpages.com, titled: Proper time for intersecting orbits
Two factors combine to give rise to an overall difference in amount of proper time that elapses for an object in motion around a gravitating body.
- How deep down in the gravitational well the object is
- the velocity of the object relative to the gravitating body
For a spacecraft in low Earth orbit a smaller amount of proper time elapses than for an object located on the Earth's surface, co-rotating with the Earth.
How that difference comes about:
The spacecraft in low Earth orbit is higher up in the gravitational well, but only several hundreds of kilometers. Compared to the radius of the Earth that is only a very small height difference.
The spacecraft in low Earth orbit has a very high velocity relative to the Earth. (Period of low Earth orbit: about 90 minutes.)
Overall: for the spacecraft in low Earth orbit a smaller amount of proper time elapses than for an object on the surface of the Earth.
For an orbit at a very high altitude the circumnavigating velocity is much slower than for low Earth orbit. (It's not just that the angular velocity is slower; the velocity itself is much slower than in low Earth orbit.) Example: geostationary orbit. For satellites in geostationary orbit the difference in gravitational potential is very large. Overall the amount of proper time that elapses is larger than for an object located on the Earth's surface.
Another example: the GPS satellites orbit at about 26.000 kilometers away from the Earth's center, such that the period of the orbit is 12 hours. For GPS satellites too the amount of proper time that elapses is larger than for objects located on the Earth's surface.
There is a crossover altitude, and Kevin Brown offers a calculation of that altitude. He arrives a a value of about 3000 kilometers above the Earth's surface. For a satellite orbiting at that altitude the same amount of proper time elapses as for an object located on the surface of the Earth.
Finally, the surface of the Earth itself. An object located on the Equator is moving at a velocity (relative to the Earth's center) of about 1600 km/h. Does it follow that for a object located on the Equator a smaller amount of proper time will elapse than for an object located at polar latitude?
Interestingly: no.
Due to its rotation the Earth has an equatorial bulge. That means that an object located at polar latitude is closer to the Earth's center of gravity than an object located on the Equator is, hence deeper in the gravitational well.
The surface of the rotating Earth's is a surface of equal geopotential height. The usual expression is: the Earth's surface is an equipotential surface. In terms of GR: gravitational potential and the amount of proper time that elapses are correlated. At the relatively slow rotation rate of the Earth: to a very good approximation the correspondence is 1-to-1.
Therefore: on any latitude: for objects positioned at the same geopotential height the same amount of proper time elapses.
(The reference of geopotential height is sea level. The land mass of the continents is considerably above sea level of course, but it is always possible to work out the height above sea level, so that it can be accounted for.)
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$\begingroup$ You are talking about a completely different case of orbiting satellites and different heights , geopotential , etc . My question was simple as just a approaching spaceship not a satellite , and I talked about what the following reference frames . The planet, the planet where the spaceship came from, and the spaceship itself would see each other or observe. $\endgroup$ Commented Sep 17, 2023 at 17:37
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$\begingroup$ @ACuriousMind The thing is: if your question was just about approaching then it's not clear why you added the detail of a spinning planet. The approaching case is about linear velocity. Resources about that case are abundant on the internet. So I opted to concentrate on the detail for which there are far less resources. Anyway, you have just confirmed that the rotation of the planet is not relevant for your question. $\endgroup$– CleonisCommented Sep 17, 2023 at 18:10
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Yes, but you don't even need the planet to be spinning. If you have two observers with a relative velocity between them, they both see the other with the normal time dilation described by special relativity
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Let's say the ship approaches the north pole of the planet along the spin axis of the planet. The speeds of clocks on the planet are constant and easy to calculate according to the ship. So time dilations of clocks are easy to calculate and constant by using the basic time dilation formula. If linear speed of a clock increases, time dilation of the clock increases. If spinning speed of a clock increases, time dilation of the clock increases.
Let's say ship approaches the equator of the planet transversely to the spin axis of the planet. The speed of a clock on the ship is not constant according to a guy standing on the equator. But he can use the average speed to calculate the average time dilation.
But there is an effect that causes the current time to change at the ship according to the guy. This thing: https://en.wikibooks.org/wiki/Special_Relativity/Simultaneity,_time_dilation_and_length_contraction#The_Andromeda_Paradox
The time changes forwards when the guy accelerates towards the ship and backwards when the guy accelerates away from the ship. The forward change is larger than the backwards change.
