Percent Length Contraction (check solution)

[Quelsita]

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$$\sqrt{1-(v/c)^2}$$

If you divide over the L to get:

L'/L=$$\sqrt{1-(v/c)^2}$$=% length contraction

Plug-n-chug from here to get:

L'/L=$$\sqrt{1-(680.58/c)^2}$$
=$$\sqrt{1-(5.15X10^-12)}$$
=$$\sqrt{1}$$
=1

Is this correct? It only contracts 1%?

 

[LowlyPion]

I'd recheck your calculation and be careful in taking your square root.

680/300,000 is what you were intending I trust?

 

[Quelsita]

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?

 

[LowlyPion]

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.

 

[davieddy]

I'd recheck your calculation and be careful in taking your square root.

680/300,000 is what you were intending I trust?

c = 300,000,000 m/s