True/False: Theory of Relativity Questions

AI Thread Summary
The discussion explores the implications of the Theory of Relativity during a flight from Amsterdam to Detroit at 900 km/hr. It concludes that the walk to the restroom does not become shorter, and a clock will not remain synchronized with Amsterdam time after the flight. Observers will perceive another 747 as shorter due to relativistic effects, but time zones do not affect the actual time difference upon landing. The importance of calculating Frequent Flyer miles from a ground reference frame rather than the airplane's frame is emphasized, as it yields a more advantageous distance measurement. Overall, the effects of relativity, while subtle at such speeds, can still be empirically demonstrated through experiments with atomic clocks.
dietcokemaste
Messages
2
Reaction score
0
You fly in a 747 across the Atlantic from Amsterdam to Detroit at about 900 km/hr. Effects of the Theory of Relativity are of course unnoticeable at such speeds, but just as a thought experiment—which of the following are true?

1. The walk from your seat to the restroom actually becomes shorter.
2. If you don’t adjust your clock at all and fly back a few days later, your clock is again going to be exactly synched with the Amsterdam time.
3. From your window you see another 747 traveling in the opposite direction. The other747 is shorter than yours.
4. You are crossing six time zones, but your clock is actually going to be off by less than six hours.
5. You really want Frequent Flier miles calculated from a reference frame on the ground, not from one attached to the airplane.
 
Physics news on Phys.org
Gotta start at the beginning:
dietcokemaste said:
Effects of the Theory of Relativity are of course unnoticeable at such speeds
False! If you carry an atomic clock in your carryon, you will notice the difference.
1. The walk from your seat to the restroom actually becomes shorter.
False (though vague on who is doing the measuring).
2. If you don’t adjust your clock at all and fly back a few days later, your clock is again going to be exactly synched with the Amsterdam time.
False.
3. From your window you see another 747 traveling in the opposite direction. The other747 is shorter than yours.
As viewed by you, true.
4. You are crossing six time zones, but your clock is actually going to be off by less than six hours.
Time zones are irrelevant. When you land, your clock will be ahead of a ground-based clock synchronized to the international standard.
5. You really want Frequent Flier miles calculated from a reference frame on the ground, not from one attached to the airplane.
False.

Here's the actual experiment: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html
"During October, 1971, four cesium atomic beam clocks were flown on regularly scheduled commercial jet flights around the world twice, once eastward and once westward, to test Einstein's theory of relativity with macroscopic clocks. From the actual flight paths of each trip, the theory predicted that the flying clocks, compared with reference clocks at the U.S. Naval Observatory, should have lost 40+/-23 nanoseconds during the eastward trip and should have gained 275+/-21 nanoseconds during the westward trip ... Relative to the atomic time scale of the U.S. Naval Observatory, the flying clocks lost 59+/-10 nanoseconds during the eastward trip and gained 273+/-7 nanosecond during the westward trip, where the errors are the corresponding standard deviations. These results provide an unambiguous empirical resolution of the famous clock "paradox" with macroscopic clocks."
 
dietcokemaste said:
5. You really want Frequent Flier miles calculated from a reference frame on the ground, not from one attached to the airplane.

It’s desirable to get more number of FF miles. Measuring it wrt the plane frame will make the distance shorter than as measured wrt the ground frame. Measuring it wrt the ground frame will thus be advantageous, and I should want it. (Not considering the starting and stopping of the plane here.)
 
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