Time Dilation & 2 Atomic Clocks (one in orbit)

[Physics345]

Homework Statement

Two atomic clocks are synchronized. One is placed on a satellite, which orbits around the Earth at a high speed for a whole year. The other is placed in a lab and remains at rest, with respect to the earth. You may assume that both clocks can measure time accurately to many significant digits.

a) Will the two clocks still be synchronized after one year? Explain your reasoning.

b) Imagine that the speed of light was much slower than its actual value. How would the results of this experiment change if the speed of light was only twice the average speed of the satellite? Explain your reasoning, using a calculation.

Homework Equations

The Attempt at a Solution

a) The clock on Earth is on Earth's frame of reference, while the satellite is in the satellites reference frame, which is slightly different from the Earth's therefore no the clocks will not be synchronized after one year, by a very small margin, which can only be seen by a clock that is capable of measuring such a small amount of time dilation that is in the satellite.
b) The speed of the satellite will be half the speed of light which is twice as fast as the speed of the satellite. Therefore, much more time is dilated considering the satellite is approaching the speed of light.
Calculations done for one year:
v=7777.8 m⁄s (speed of the satelite)
c=15555.6 m⁄s (speed of light)
∆t_s=3.154×10^7 s
∆t_m=(∆t_s)/√(1-v^2/c^2 )
∆t_m=(3.154×10^7 s)/√(1-(7777.8 m⁄s)^2/(15555.6 m⁄s)^2 )
∆t_m=36418786.76 seconds=10116 hours=421.5 days
Therefore from Earth's reference frame 421.5 days will have past and in the satellites frame of reference only 365 days will have past.

 

[Physics345]

What do you guys think? Opinions are very much appreciated.

 

[razidan]

What do you guys think? Opinions are very much appreciated.

looks good (this of course doesn't take into account GR).
I would recommend using $\beta = \frac{v}{c}$ to ease the calculations. you then don't really need to manage all those pesky numbers.
here's what i mean. in you example, the speed is half the speed of light, so $\beta=\frac{v}{c}=\frac{0.5c}{c}=0.5$, which makes $\gamma=\frac{1}{\sqrt{1-\beta^2}}=\frac{1}{\sqrt{1-0.5^2}}=1.15$.
seems more tidy.

 

[Physics345]

looks good (this of course doesn't take into account GR).
I would recommend using $\beta = \frac{v}{c}$ to ease the calculations. you then don't really need to manage all those pesky numbers.
here's what i mean. in you example, the speed is half the speed of light, so $\beta=\frac{v}{c}=\frac{0.5c}{c}=0.5$, which makes $\gamma=\frac{1}{\sqrt{1-\beta^2}}=\frac{1}{\sqrt{1-0.5^2}}=1.15$.
seems more tidy.

I'll be sure to do that, thank you very much, for your help :).