Special Relativity - Proper Length

AI Thread Summary
The discussion revolves around calculating time intervals and distances related to a meteoroid approaching Earth at 0.800c. Participants clarify the concept of proper length, noting that it is the length measured when at rest relative to the object in question. The initial 20 light-years distance measured by the Earth-bound astronomer is debated as potentially being the proper length, given that the meteoroid is in motion. It is concluded that the tourist on the meteoroid would perceive the distance to Earth as contracted due to relativistic effects, estimating it to be about 60% of the original distance. The conversation emphasizes understanding how relative motion affects measurements in special relativity.
eprparadox
Messages
133
Reaction score
2

Homework Statement


An astronomer on Earth observes a meteoroid in the
southern sky approaching the Earth at a speed of 0.800c.
At the time of its discovery the meteoroid is 20.0ly from
the Earth. Calculate (a) the time interval required for the
meteoroid to reach the Earth as measured by the Earth-
bound astronomer, (b) this time interval as measured by a
tourist on the meteoroid, and (c) the distance to the Earth
as measured by the tourist.



Homework Equations





The Attempt at a Solution



I just need help with how to figure out what is the proper length. I don't know if the 20 ly measured from astronomer on Earth is the proper length because it seems I could be on the meteoroid and calculate the same 20 ly and then that could be the proper length. But it seems that someone has to see some length contraction.

I know proper length is the length of something measured when at rest relative to that something. And here, the length we're measuring is the distance to the meteoroid, but that is changing constantly if the meteor is moving towards the Earth (or from the meteors perspective, if the Earth is moving towards the meteor). So how do we know what the proper length is?

Thanks ahead of time for the help.
 
Physics news on Phys.org
my guess would be that you're thinking a little bit too much. just find the percentage of length that something moving at .8c is, and then multiply that percentage by 20 light years.
 
and who is the tourist?
 
sorry about that. the tourist would be someone on the meteoroid.
 
so someone on the meteoroid sees Earth coming to them at .8c and we on Earth see the meteoroid coming to us at .8c. But for the tourist the distance to Earth I believe is 60% of what it would be if he was not moving at all.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top