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Escape velocity from the Earth is \SI{40000}{\kilo\meter\per\hour}. What would be the percent decrease in length of a \SI{95.2}{m} long rocket traveling at that speed?
The percentual change in the length is: \begin{align} \frac{\Delta \ell}{\ell_0} &= \frac{\ell_0\ell}{\ell_0}\\ &= \frac{\ell_0\frac{\ell_0}{\gamma}}{\ell_0}\\ &= 1\frac{1}{\gamma}\\ &\approx \frac12\frac{v^2}{c^2}\\ &= \numprint{6.868e10} \end{align} The last equation is valid because $v\ll c$ (Taylor expansion).
00:09, 5. June 2017  lsg  Urs Zellweger (urs)  Current Version 
11:20, 4. June 2017  Initial Version.  Urs Zellweger (urs)  Compare with Current 