Solve Time Dilation: Get 24hr Solution in 105hrs

AI Thread Summary
The problem involves a space probe traveling at 0.9c with a bomb set to explode after 24 hours. The initial calculation yields 55 hours for Earth observers to see the explosion, but the textbook states 105 hours. The discrepancy arises from not accounting for the time it takes for light to travel back to Earth after the explosion. When factoring in the light travel time of approximately 49.5 hours, the total time observed on Earth becomes about 104.5 hours, aligning closely with the textbook answer. This highlights the importance of considering both time dilation and the speed of light in such calculations.
EngageEngage
Messages
203
Reaction score
0
[SOLVED] Time Dialation Help

Homework Statement


A bomb is placed on a space probe just before it's launched. The timer is set to trigger the bomb after exactly 24hrs. The probe travels away from Earth on a straight line at v=.9c. How long after launch will the observers on the Earth see the flash of light from the exploding bomb?


Homework Equations



\Delta\tau=\Delta t \sqrt{1-\frac{v^{2}}{c^{2}}}

The Attempt at a Solution


\Delta\tau = 24hrs.

\Delta t = \frac{24hrs}{\sqrt{1-.9^{2}}} = 55hrs
But, the textbook gives a time of 105hrs. Can anyone please tell me why what I did is wrong? This seems like a straightforward problem and I have no clue where I mesed up. Any help is greatly appreciated.
 
Physics news on Phys.org
You're half way,
I think you need to add time for the light from the explosion to get back to Earth - which is when people on Earth see the explosion
Calc how far away it is.
Then calc the time for light to travel that distance.
 
Then,
\Delta x = .9c(55hrs)
t = \frac{.9c(55hrs)}{c} = 49.5 hrs
55+49.5=104.5

Thank you for the help, I completely blanked that part!
 
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