Calculating Time Dilation: Apollo's Record-Breaking Speed and Earth's Clocks

[Quelsita]

Question:

The record for the fastest speed at which anyone has ever traveled, relative to the Earth, is held by the Apollo X modue at 24,791 mi/h on their return trip from the moon.
At this speed, what is the percent difference between the clocks on the Apollo and the clocks on Earth?

OK, I think I understand how to do the problem, but I'm not getting the answer that our text gives.

I set the moving IRF as the Apollo and the stationary frame as the Earth.
I then found the time elapsed in the frame of the spaceship by using d=vt and the given velocity and the distance from the Earth to the moon.
So:

V_Apollo=V=24,791mi/h=11082.3m/s
d_{(Earth to Moon)}=3.84x10^8 m

With no time dilation for the ship, the time elapsed then is:
from d=vt: t_Apollo=t'=3.84x10^8 m/11082.3m/s = 34649.85 s

I then found the time dilation for the time elapsed in Earth's reference frame using Lorentz tranformation:
t=t^'/$$\sqrt{1-(V/c)^2}$$

gives an elapsed time in Earth' IRF of 346449.85s

and 346449.85s/34649.85s = 9.985%

The correct answer is 6.82x10^-8 s

Could someone help me figure out where I went wrong? I went over the math a few times, so it must be my logic...

Thanks!

 

[Delphi51]

I don't have a calculator handy, but something went wrong in that square root calc. Did you note that the spacecraft speed is in mph, so in the same units c = 186000*3600 ?

Doubtless that square root is well within 1% of 1.

 

[Quelsita]

Yeah, I actually converted the velocity to m/s I just skipped the conversion work.