Question

# A spaceship moving $0.80 \: \mathrm{c}$ direction away from Earth fires a missile that the spaceship measures to be moving at $0.80 \: \mathrm{c}$ perpendicular to the ship's direction of travel. Find the velocity components and speed of the missile as measured by Earth.

Solution

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K system is system attached to Earth and K' is attached to spaceship. Let the $v=0.8 \: c$ be relative speed beetween the system and let's say that it is in x direction. In K' only speed that is measured is in y direction, $u_{\text{y}}'=0.8 \: c$. Then velocity components measured from system K (from Earth) are:

\begin{align*} & \text{x direction:} \\ u_{\text{x}} &= \frac{u_{\text{x}}' + v }{ 1 + \frac{v}{c^2}u_{\text{x}}'} \\ u_{\text{x}} &= \frac{0 + 0.8 \: c}{1 + 0} \\ u_{\text{x}} &= \boxed{0.8 \: c}, \\ & \text{y direction:} \\ u_{\text{y}} &= \frac{u_{\text{y}}'}{\gamma \left( 1+ \frac{v}{c^2} u_{\text{x}}' \right)} \\ u_{\text{y}} &= \frac{0.8 \: c}{ \frac{1}{\sqrt{1 - (0.8)^2}} \left( 1 + 0 \right)} \\ u_{\text{y}} &= \boxed{0.48 \: c}, \\ & \text{z direction:} \\ u_{\text{z}} &= \frac{u_{\text{z}}'}{\gamma \left( 1+ \frac{v}{c^2} u_{\text{x}}' \right)} \\ u_{\text{z}} &= \frac{0 }{ \frac{1}{\sqrt{1 - (0.8)^2}} \left( 1 + 0 \right)} \\ u_{\text{z}} &= \boxed{0}, \\ \end{align*}

and speed of projectile is:

\begin{align*} u &= \sqrt{u_{\text{x}}^2 + u_{\text{y}}^2} \\ u &= \sqrt{(0.8 \: c)^2 + (0.48 \:c)^2} \\ u &= \boxed{0.93 \:c \: < \: c}. \end{align*}

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