Question

Is anyone familiar with Special relativistic time dilation?

Answer 1

A li'll bit

Answer 2

We have many physics teachers here on OS, so ask away..

Answer 3

what is your question ?

Answer 4

Thankyou! I'm pretty new to the course so I'm still trying to get my head around it, and keep trying to follow examples then getting myself confused!

"A GPS satellite typically orbits at an altitude of 26000 km and orbits the Earth in about 12
hours. The radius of the Earth is about 6400 km. The strength of the Earth’s gravitational
field at the surface is about 10 ms^(−2).

Calculate the Special Relativistic time dilation for the satellite relative to the Earth’s
surface?

Answer 5

Try to relate special relativity with doppler's effect (for waves).
For example, consider this question from the point of view of waves.

Answer 6

I'll be honest, I'm not familiar with the Doppler Effect. I'm a maths major, and I've just started a General Relativity and Spacetime module, so sorry if I take a while to get my head around it. ( But I appreciate the help!)

Answer 7

Ok, so I'll try to use as elaborate as possible

Answer 8

*be

Answer 9

Let T be the time taken by the satellite to compute one revolution.
The clock on the satellite will slow down as observed from the earth.
If the time elapsed on the satellite's clock is 't' as the satellite completes one revolution (this is proper time and T is improper time)
\[t=\sqrt{1-(\frac{ v }{ c })^2}*T\]

Finally, find t - T

Answer 10

^^ This will give you the time dilation in one revolution

Answer 11

Oops sorry, we also need to find the velocity..
We can assume that non-relativistic analysis can be made to compute the speed of the satellite.
so \[v = \sqrt{\frac{ GM }{ R }}\]

Answer 12

so is the formula for velocity as stated above, for the velocity of the satellite or the Earth? Is the mass of the earth a constant? (that seems like a stupid question!)

Answer 13

I'm so sorry, I'm completely lost with this!

Answer 14

It is the velocity of the satellite with respect to the earth (can be found by simple force balance).

As I did all the calculation from the frame of reference of earth, its mass can be considered a constant.

Answer 15

Its not your fault, I did unnecessary calculations. Can you describe time dilation please? (I am no good with remembering definitions :D )

Answer 16

As per my understanding, time dilation is the difference in time of two events occuring from different frames of reference?

Answer 17

Can you take it as though I have no prior knowledge of physics (which is kind of true bar the essentials!) - it's a long time since I did physics!!

Answer 18

If time dilation is the difference of time - you can simply find t - T
where T = 12 hours (given) and we have a formula for t

If time dilation is γ (factor by which time slows down) - then use equation
\[γ=\frac{ 1 }{ \sqrt{1-(\frac{ v }{ c })^2} }\]

Answer 19

So gamma = t-T ?

And how do I calculate 'v' for the satellite? If I am not given a mass or radius for it?

Answer 20

Actually \[γ=T/t\]

Answer 21

Consider the notation -
a = altitude (given)
R = radius (given)
x = a + R

Calculating velocity -
\[v=\sqrt{\frac{ GM }{ x }}\]

Calculate GM using the information given - "... The strength of the Earth’s gravitational field at the surface is about 10 ms^(−2)..."

Answer 22

So, try and tell me the value of GM

Answer 23

So if x=a+R = 26000+6400 = 32400

Then if: \[v=\sqrt{\frac{ GM }{ 32400 }}\] then (32400v)^2= GM ?

Answer 24

No, GM should be calculated from the equation -
\[\frac{ GM }{ R^2 }=10\]

Btw, there is also an easier method to calculate velocity. Consider -
\[T=\frac{ 2\pi x }{ v }\]
As T and x are given, we can find v

Answer 25

Okay so using that I get
GM=4.096x10^8 ?

Answer 26

and I get velocity = 2*pi*32400 (/12) = 5400 pi

Answer 27

Nearly correct. Time was 12 hour, so you should convert first it into seconds.

Answer 28

*first convert

Answer 29

@sarahusher I hope you got your answer.

Answer 30

right,
so \[v=\frac{ 2 \pi * 32400 }{ 12*60*60 }= \frac{ 3 \pi }{ 2 }m/s\]

Is that correct?

Answer 31

Yes

Answer 32

I don't know how but the velocity calculated by the two methods is not equal. I think 2 hours implies time t (the question is not clear with whose frame of reference time has been calculate)

Answer 33

(Thank God!) lol Right okay, then:
\[\gamma = \frac{ T }{ t }= \frac { 1 }{ \sqrt{1-(\frac{v}{c})^2} }\]
and I just substitute everyhting in?

Answer 34

To find γ, you have to use \[v=\frac{ 320\sqrt{10} }{ 9 }\] which is derived from the earlier formula

Answer 35

So, is everything clear know ??

Answer 36

I think so!! Thank you very much for your help, I really appreciate it!