Special Relativity and Airplane Problem

[harrietstowe]

Homework Statement

An airplane whose rest length is 46m is moving at uniform velocity with respect to Earth, at a speed of 575 m/s
(a) By what fraction of its rest length is it shortened to an observer on Earth?
(b) How long would it take, according to Earth clocks, for the airplane's clock to fall behind by 1.00 µs?

Homework Equations

time dilation: delta t=gamma * delta t_o
length contraction: L=L_o/gamma

The Attempt at a Solution

For part (a) I used the length contraction equation. gamma= 1/(1-(575/3e8)^2)=
1.0000000000018
I then divided 45 by this to get 44.999999999919m.
The questions asks though for what fraction is it shortened and so I divided 44.999999999919m by 45m to get .9999999999982 but this is not the right answer.

For part (b) I used the time dilation equation and multiplied gamma by 1.00 µs and I got 1.0000000000018 µs but this was not right. The weird thing for this part is they want the answer in days which would seem to be a very small answer

 

[JesseC]

EDIT: whoops, did reply but then i realized i'd cocked up! Shows me not to speak to soon... haha

You seem to have the right idea for the first part, just make sure you're using the right length of the plane. ( You said 46m somewhere and 45m somewhere else)

For the second part, you can't just use time dilation blindly! What you get when you use time dilation, is the time an observer on the ground thinks has gone by, when a clock on the plane has ticked 1µs. So for every 1µs tick on the plane, an observer on the ground thinks 1.0000000000018 µs has passed. Thus, every 1µs, the two clocks become out of sync by 0.0000000000018 µs. How long would it this tiny error every µs to add up to a total error of 1µs?

By my possibly erroneous calculations, I get about 6.3 days...

 

[vela]

You can say the contracted length of the plane will be equal to L'=L₀-δL. In part (a), the problem is asking you to find δL/L₀.

 

[harrietstowe]

JesseC you are right I did it your way and I got 6.4 days which was correct. Funny how you said you cocked up. Thats some funny **** right there. Vela, that is what I did before and I got .9999999999982 which it wouldn't take. As a side question when dealing with gamma our physics professor says he gets the value for it by using taylor expansion. I have just been plugging into my calculator. In addition to finding out what I am doing wrong in part a I was also wondering if someone could give me a quick overview of how to do that taylor expansion.

 

[vela]

Vela, that is what I did before and I got .9999999999982 which it wouldn't take.

No, you didn't. You calculated L'/L, not δL/L.

 

[harrietstowe]

vela, that just gives you δ which is 1.0000000000018 which unfortunately isn't the right answer.

 

[vela]

Perhaps you're mistaking δ for γ. δL ("delta L") is one variable, the change in length of the plane due to length contraction.

 

[harrietstowe]

I see. Would it kill you to use Δ like a normal functioning human being? jk Anways I did that and ΔL was 8.3e-11. I divided this by 46m to get 1.8e-12 which was incorrect. I also tried -1.8e-12 which was also incorrect.

 

[vela]

Well, that's the right answer. If you're entering this into a computer, perhaps you need more sig figs?

As far as the Taylor approximation goes, your professor is using the expansion

(1-x)^{-1/2} = 1+\frac{1}{2} x + \frac{3}{8} x^2 + \cdots

When x is small, like in this problem, 1+x/2 gives a very good approximation for γ.

 

[harrietstowe]

Ok I added another sig fig and it took the answer. Thanks for the help