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A satellite orbiting the earth experiences time faster than a person on earth as its in a weaker gravitational field (which has a greater effect than the slowing down of time caused by its motion) does this mean that if someone was on the satellite they would disagree with someone on earth about the satellite's velocity around the earth saying it was slower because it would seem to move less distance per second?

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  • $\begingroup$ If you read about this experiment hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html which was performed using a flight around the world in 1971 and atomic clocks, it's pretty much the same idea as you ask. But don't mix up gravitional time effects (GR) with velocity induced effects (SR). If you read about GPS satellites you will see how they need to compensate for both effects. $\endgroup$
    – user108787
    Commented Sep 11, 2016 at 10:43

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Consider a satellite in a geostationary orbit which means that it's relative velocity with the earth is equal to 0, there is an astronaut Nick on satellite and Bob on earth observing Nick. Bob will observe that time for Nick is going faster due to gravitational time dilation so they will disagree about time but they will not disagree about speed of the satellite relative to earth. This is due to a phenomenon of length contraction, Bob observes that time goes faster for Nick but also he observes that Nicks length is "contracted" so that velocity of Nick remains constant in both frames of reference.

enter image description here

A clock which is on top of a mountain(farther from center of the earth) will run faster relative to clock which is closer to the earths center, in the same way two clocks moving at relative velocity v from each other will also disagree about time. the difference between time is represented by following equation:

$$\Delta t'= \frac{\Delta t}{\sqrt {1-\frac{v^2}{c^2}}}$$

And gravitational time dilation is represented by following equation:

$$\Delta t'= \frac{\Delta t}{\sqrt {1-\frac{2GM}{rc^2}}}$$

enter image description here

(Source: http://www.leapsecond.com/great2016a/)

To explain why is this the case consider a light clock consisting of two mirrors A and B at a distance of L from each other and a particle of light bounces from A mirror to B mirror continuously. Now let's consider what time it takes for a light particle to complete one pulse, this is represented by following formula:

(1) $$\Delta t= \frac{2L}{c}$$

enter image description here

(Source: Wikipedia)

Now let's look at this system from different frame of reference which has velocity v relative to our light clock. Consider the following - Bob standing on earth and Nick flying in a rocket with a light clock with a velocity v relative to Bob, Nick will observe the light clock same way it is illustrated above but Bob will observe light particle covering longer distance - D than Nick - L.

enter image description here enter image description here

the first picture is light clock as observed from Nick's (who's in a rocket) frame of reference and second picture is light clock observed from Bob's frame of reference. Now let's consider what time it takes for a light particle to complete one pulse but now from Bob's(on earth) frame of reference, this is represented by the following formula:

(2) $$\Delta t'= \frac{2D}{c}\Rightarrow D = \frac{\Delta t' c}{2}$$

Now we can apply Pythagorean theorem to solve for path covered by particle of light - D (from Bob's frame of reference), one of the Cathetus is equal to L and the other one we can derive that it is equal to:

$$\frac{\Delta t'}{2}v$$ (Half the time because light has covered by our definition half the pulse)

Now we can solve for the Hypotenuse - D using Pythagorean theorem and we get:

$$D = \sqrt{(\frac{1}{2}\Delta t'v)^2+L^2}$$

Now we can substitute D using (2) equation

$$\frac{\Delta t' c}{2} = \sqrt{(\frac{1}{2}\Delta t'v)^2+L^2}\Rightarrow \Delta t'=\frac{\sqrt{(\Delta t'v)^2+(2L)^2}}{c}$$

simplify:

$$(\Delta t')^2 = \frac{v^2}{c^2}(\Delta t')^2+\left (\frac{2L}{c} \right )^2\Rightarrow(1-\frac{v^2}{c^2})(\Delta t')^2 = \left (\frac{2L}{c} \right )^2$$

More simplification:

$$\Delta t'^2 = \frac{\left (\frac{2L}{c} \right )^2 }{(1-\frac{v^2}{c^2})} \Rightarrow \Delta t' = \frac{\left (\frac{2L}{c} \right ) }{\sqrt{1-\frac{v^2}{c^2}}}$$

Now we can substitute (1) equation and we have derived equation of time dilation:

$$\Delta t' = \frac{\Delta t }{\sqrt{1-\frac{v^2}{c^2}}}$$

from this observation we can conclude that In order for speed of light to remain constant Bob must observe the frame of reference of Nick slower so that:

$$\frac{2L}{\Delta t}=\frac{2D}{\Delta t'}$$

L and D are constants therefore times must be different so that this equality holds.

Now to calculate gravitational time dilation we have to substitute escape velocity equation, escape velocity is defined as the speed an object must reach to escape the gravitational pull of another object. to have an intuitive understanding of why we have to substitute escape velocity equation, Think of it as a speed the earth pulls you to its center and it is different at different distances from earth and because of this we can observe that time for a satellite goes faster relative to a person standing on earth. escape velocity is represented as following formula:

$$v_{escape}=\sqrt{\frac{2GM}{r}}$$

Where r is distance from an object(earth) and M is the mass of an object

When we substitute this escape velocity into time dilation equation we calculated before we get this formula:

$$\Delta t'= \frac{\Delta t}{\sqrt {1-\frac{2GM}{rc^2}}}$$

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  • $\begingroup$ Does this mean that after the clock is slowed down on the satellite so that the GPS works correctly someone on the satellite would now disagree with someone on earth about the satellites velocity? $\endgroup$ Commented Sep 22, 2016 at 3:15
  • $\begingroup$ When you say that nicks length is contracted do you mean that nick and the satellite are shortened or that nicks journey around the earth is shortened by space being shortened as well? It seems to me that if nick is shortened it would have no effect on his velocity and that if space is shortened we should now disagree on nicks distance from earth. $\endgroup$ Commented Sep 22, 2016 at 3:37
  • $\begingroup$ It seems strange to me that a satellite can have a relative velocity with some point on earth that is equal to 0 and yet we know its real velocity is faster than the point on earth as it has to travel further to maintain its relative velocity position. Could this mean that true velocity is never known and that velocity is a dodgy component when its used in a maths equation? $\endgroup$ Commented Sep 22, 2016 at 21:22
  • $\begingroup$ @StewartCheckley There is a specific orbit at which satellites have velocity equal to zero relative to earth, This orbit is called Geostationary orbit, which is at a specific distance away from earth (different for other planets) and at that specific orbit angular velocity of a satellite is equal to the angular velocity of Earth's rotation, As a result a satellite at this specific orbit has velocity equal to 0 relative to earth. $\endgroup$ Commented Sep 28, 2016 at 15:05

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