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Relativistic bullet (Posted on 2003-11-28) Difficulty: 5 of 5
We all know about the ultimate speed limit... the speed of light.

If person A stands on Earth and shoots his pistol, he observes the bullet to fly directly away at 1000 mph. Person B is standing right next to him (not in front) and watches this event, and agrees that the bullet flies directly away at 1000 mph.

Let's change the situation and say that B is in a spaceship, and A is in a different (and very long) spaceship with lots of windows. B's ship is hovering in space (no thrusters/acceleration). A's ship is approaching from a distance and is going to pass B's ship (very close) but at incredible speed. Make careful note that A's ship is NOT thrusting or accelerating at all, it is "coasting". In fact, A's ship is moving, relative to B's ship at 10 mph less than the speed of light. WOW!

A stands in the middle of his ship and points his gun directly forward (in the direction of travel), and fires the same pistol at the exact moment that he is passing B.

The questions are: How fast does he observe the bullet leave the gun? How fast does B observe the bullet leave the gun?

How do your answers change (if at all) if A aims backwards when he fires?

See The Solution Submitted by SilverKnight    
Rating: 3.5000 (8 votes)

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Some Thoughts What a doppler effect! | Comment 2 of 20 |
(In reply to A's observations by rerun141)

Let us consider all this happening in a one dimensional x - axis number line sort of thing where B stands at the origin and A begins at say x = -1 light hours at t = 0 hours. One light-hour is about 600 million miles. At t =1 hour the very first wave of light will hit B's eyes, but by this time A will already be physically 10 miles away. So in the following 1/60 millionth of an hour B will witness A's entire 600 million mile arrival. Bullets aside, A will physically arrive at point x = 1 light-hour a hair after t = 2 hours. It wont be until t = 3 (plus another 1/60 millionth of an hour) for B to finally see A cover the same distance away from him/her.

Since all these times are from B's point of view we must now introduce the most significantly relevant element of the theory of relativity. When mass is accellerated close to the speed of light, that mass' energy becomes devoted to the travel itself. This occurs in such a way that for a passenger on a near-light speed craft, the trip seems to be near instantaneous.
Theoretically, if A were actually travelling the speed of light, his/her trip would be completely instantaneous giving him/her no time to even pull the trigger.
But A is travelling 10 mph less than light speed so let us say that the bullet leaves the gun precisely when A passes B (ignoring the acceleration of the bullet within the chamber) The bullet will then appear to be travelling (10mph*1000mph/600millionmph) = 1/60thousandths mph relative to A. The apparent speed of the bullet fired in the other direction would I believe be the same - although I'll return with some equations.
  Posted by Eric on 2003-11-28 19:06:26

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