Question: What is the percent length contraction of an aircraft traveling at Mach 2? So, we know that Mach 2= 680.58 m/s and that L'=L[tex]\sqrt{1-(v/c)^2}[/tex] If you divide over the L to get: L'/L=[tex]\sqrt{1-(v/c)^2}[/tex]=% length contraction Plug-n-chug from here to get: L'/L=[tex]\sqrt{1-(680.58/c)^2}[/tex] =[tex]\sqrt{1-(5.15X10^-12)}[/tex] =[tex]\sqrt{1}[/tex] =1 Is this correct? It only contracts 1%?
I'd recheck your calculation and be careful in taking your square root. 680/300,000 is what you were intending I trust?
Yep. That's why I thought it was off because it's such a small number that you'll have 1 under the radical... Or is the percent contraction supposed to be very small since the aircraft, compared to the speed of light, is going extremely slow?
Yes it is a small number. And when you take the square root it gets closer to 1. Use a calculator, and don't approximate or round until you have an expression for the percentage.