Special Relativity: Time Dilation

AI Thread Summary
The discussion focuses on calculating the fractional change in frequency of a GPS satellite signal due to time dilation as described by special relativity. The satellite orbits with a period of 11 hours and 58 minutes, and its oscillator frequency is 1575.42 MHz in its reference frame. The attempted solution uses the formula for fractional change, resulting in a value of 8.36 x 10^-11. Participants are encouraged to verify the relationship between the fractional frequency change and time dilation. The conversation emphasizes the importance of understanding the underlying principles of special relativity in this context.
JennV
Messages
21
Reaction score
0

Homework Statement



An Earth satellite used in the Global Positioning System moves in a circular orbit with period 11
hours and 58 min. The satellitle contains an oscillator producing the principal nonmilitary GPS signal. Its frequency is 1575.42 MHz in the reference frame of the satellite. When it is received on the Earth's surface, what is the fractional change in this frequency due to time dilation, as described by special relativity?

Homework Equations



Fractional change = 1/2 (v/c)^2

The Attempt at a Solution



Answer: 1/2 ( 3880m/s / c )^2 = 8.36*10^-11


Not sure if I'm even on the right track at all... some guidance would be much appreciated.
 
Physics news on Phys.org
Well, you can check your logic: the fractional frequency change is \frac{\Delta\nu}{\nu} ... so do you know how that relates to the formula you used? To time dilation?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top