# Special Relativity - Finding Velocity

1. Dec 27, 2013

### Insolite

A spaceship is measured to be 50m long in it's own rest frame takes $1.50 \times 10^{-6} s$ to pass overhead, as measured by an observer on earth. What is its speed relative to earth?

My attempt at the solution involves the use of the equation for length contraction, $l = l_{0}\sqrt{1 - \frac{u^{2}}{c^{2}}}$. I have calculated an answer that seems reasonable, but I don't know if the approach that I've taken is correct.

Solution:

$l = l_{0}\sqrt{1 - \frac{u^{2}}{c^{2}}}$, where $l_{0} = 50m$, $u = \frac{l}{t}$ with respect to earth.
$l = l_{0}\sqrt{1 - \frac{u^{2}}{c^{2}}}$

$l^{2} = l_{0}^{2}(1 - \frac{u^{2}}{c^{2}})$

$\frac{1}{l_{0}^{2}}l^{2} + \frac{u^{2}}{c^{2}} = 1$

$\frac{l^{2}}{l_{0}^{2}} + \frac{l^{2}}{(tc)^{2}} = 1$

$\frac{l^{2}c^{2}t^{2} + l_{0}^{2}l^{2}}{l_{0}^{2}t^{2}c^{2}} = 1$

$l^{2}t^{2}c^{2} + l^{2}l_{0}^{2} = l_{0}^{2}t^{2}c^{2}$

Substituting in the known values of $l_{0} = 50m$ and $t = 1.50 \times 10^{-6}$:

$l = 49.694m$

Therefore, the spaceship travels this length (which is its length with respect to the earth) in $1.5 \times 10^{-6} s$ with respect to the earth:

Speed = $\frac{49.694m}{1.5 \times 10^{-6} s}$

$= 3.31 \times 10^{7}ms^{-1}$

If this is in fact correct, are there any other methods that I may have used to arrive at the same answer? (Those using very basic special relativity concepts, including time dilation and Lorentz transformations).

Thanks for any help and advise.

2. Dec 27, 2013

### Staff: Mentor

Looks right but complicated.
As seen from earth, the spaceship has a speed of $\beta c$ and a length contraction factor $\gamma$, so we get $t = \frac{l_0}{\beta \gamma}$ or
$$t \beta c = l_0 \sqrt{1-\beta^2}$$
$$\frac{\beta}{\sqrt{1-\beta^2}} = \frac{l_0}{tc}$$
The right side is a known value (0.111) and you can directly solve the equation for β.

The result is the same.