# Spaceship Question

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On slide 20 of lecture L1402 could you explain each step towards getting the final solution of v=0.857c, I can't follow the calculations you made.

Thanks!!

The observer on the earth sees the space ship passing by with velocity v. The earth observer also measures a different length of the spaceship than its proper length which is $L=L_p/\gamma$.

Now let's find the velocity v. The length of the spaceship measured by earth observer is L. It takes $\Delta t=0.8\times 10^{-6}s$ to pass a point on the earth. Therefore, the speed is $v=L/\Delta t$. So we have
$$v=\frac {L}{\Delta t}=\frac {L_p/\gamma}{\Delta t}=\frac {400\sqrt{1-v^2/c^2}}{0.8\times 10^{-6}}$$

Now you can solve for v: first move the $\sqrt{1-v^2/c^2}$ term to the left of the equation, so we have $$\frac{v}{\sqrt{1-v^2/c^2}}=\frac{400}{0.8\times 10^{-6}}=5\times 10^8$$

Square both sides and then divide both sides with c2, so we have $$\frac {v^2/c^2}{1-v^2/c^2}=\frac {25\times 10^{16}}{9\times 10^{16}}=\frac {25}{9}$$

Take $v^2/c^2$ as one term, then you can solve it to have $v^2/c^2=\frac{25}{34}$.

Therefore, $v/c=\sqrt{25/34}=0.857$ and $v=0.857c$.

answered Dec 10, 2013 by (21,750 points)