Angle changes in Relativity (Shooting the Deathstar on a flyby)

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SUMMARY

The discussion focuses on calculating the angle required to shoot a laser gun from the Millennium Falcon at the Death Star while traveling at 0.8c, where the Death Star is positioned at a 90° angle from the direction of motion. Using the equation cos(θ) = β, the calculated angle θ is 36.87°. This solution aligns with the principles of special relativity, where the angle decreases as the velocity approaches the speed of light, confirming that the approach is correct and not overly simplistic.

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Homework Statement


You are flying the Millenium Falcon and want to shoot the Death Star at the closeest range possible. You fly by the Death star at speed .8c. Just at the time of your closest approach, the death star is 90° from your direction of motion, you shoot the laser gun at it. What is the angle you must set between your direction of motion and the laser gun?


Homework Equations



cos(θ)=β

The Attempt at a Solution



All I did was use the above equation and got that θ=36.87°, which would make sense because the smaller β is, the closer to 90 the angle would become. But this just seems way too easy compared to our other relativity problems, am I doing something wrong?

Thanks in advance!
 
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