A B1-B bomber is a 44.5 m long airplane capable of flying at supersonic speeds.

Question

A B1-B bomber is a 44.5 m long airplane capable of flying at supersonic speeds. It takes off from an airbase to carry out a mission. It flies 8405 km at an average speed of 419.5 m/s to reach its destination. It drops its payload - medication and equipment - and then turns around. It returns to the base on the same path at the same constant speed.
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a), What is the time difference between two identical atomic clocks, one on the airplane and one at rest at the airbase after the airplane lands?
b).By how much does the length of the airplane change due to relativistic effects?
c).By how much does the length of the airplane change due to thermal contraction or expansion when it takes off or lands? Assume the airplane is built from aluminum. The coefficient of linear expansion for aluminum is 2.39×10-5 1/C°. The temperature at the airbase is 37.1 °C and -48.6 °C at a height of 12.8 km where the airplane flies.

Explanation / Answer

a) First, we need to know the time interval that it was flying for (this is the proper time, because it is measured with respect to the plane). It flew 8405km two ways at 419.5 m/s, so

tp = (2 * 8405*10^3 m)/(419.5 m/s) = 40,072 s

So, the amount of time that the clock ticks off on the ground is

t = tp = (1/(1 - (419.5 m/s / 3*10^8 m/s)^2) ) * (40,072 s) = 40,072.000000039 s

So, the difference between the two clocks is 0.000000039 s

b) While in flight, someone outside of the airplane at rest to the Earth would measure the length of the airplane to be

L = Lp/ = (1 - (419.5 m/s / 3*10^8 m/s)^2 ) * 44.5m = 44.5 (the online calculator, wolframalpha.com, isn't powerful enough to calculate the 1/ factor to anything more accurate than 1, sorry).

c) The change in the length due to thermal expansion is

L = L(0)T = (2.39×10^-5 1/C)*(44.5m)*(-48.6 C - 37.1 C) = - 0.091 m

So, it shrinks by 0.091m.

Something to note, 419.5m/s is so ridiculously small for any relativistic effect to be apparent, so don't be suprised by how small the numbers are. Don't forget that = 1/(1 - ^2), where = v/c = 1.4*10^-6, so ^2 is going to be of the order 10^-12, and 1 - 10^-12 is very, very small.

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