Show the relative difference in clock time between plane/ground is

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
akennedy
Messages
12
Reaction score
0

Homework Statement


http://puu.sh/7vC46.png
http://puu.sh/7vBG5.png
Note: In this image the ohm symbol represents the angular frequency of the Earth's rotation

Homework Equations


Gamma = 1/SQRT(1-v^2/c^2)
V (relative to centre of earth) = angular frequency(R+h) +/- v
v= plane velocity dependent on direction
R= radius of earth
h= height plane is flying


The Attempt at a Solution


The first part about showing it isn't dependent was easy and I just did some basic algebra. However, the second part is problematic.I can't seem to get to the equation it's asking for using the 2 tricks suggested. I think possibly because I'm using the wrong velocities when putting in Gamma(earth) and Gamma(plane)

The velocities I used were
V(plane) = Angular frequency*(R+h)+/-v
V(Earth) = Angular frequency*(R+h)
I put those into the gamma formula and went from there. I think I might be using the wrong velocities and have mucked up the inertial frames or something.
 
Last edited by a moderator:
Physics news on Phys.org
For what it's worth, I was able to derive the equation using
V(Earth) = Angular frequency*R

I figured since that clock is on the Earth, there's no reason to include "h" for that particular clock (The "h" is still present in the planes' velocities though).

But, and this is a big "but," I also had to assume that [itex]\frac{\Omega^2 h^2}{c^2}[/itex] and [itex]\frac{\Omega h v}{c^2}[/itex] were negligible compared to other, similar terms and can be neglected (contrasted with terms like [itex]\frac{R \Omega^2 h}{c^2}[/itex] and [itex]\frac{R \Omega v}{c^2}[/itex] which are obviously much larger). That particular approximation wasn't spelled out in the problem statement, so I'm not sure if my method is "correct" either. :redface:
 
  • Like
Likes   Reactions: 1 person
Omg... Wow thank you so much. I can't believe I overlooked that haha :P