BrettCollins

I am using: t'/t = √((1- v^2/c^2 )) Actually I did make a mistake when v = 3/4 (0.75) the ratio is √(7/16) but the two way to calculate time dilation of missile still come up with different values. ​to combine the time dilations you need to be multiplied, so the earth observer observing the astronauts clock that is synchronised with missile clock when the missile is moving at 0.75c is √(7/16) * √(7/16) = (7/16)

I am interested in Special relativity and have noticed something i do not understand to do with velocity addition.If a spacecraft takes of from earth and is travelling at 3/4 the speed of light and launches a missile at a speed of 3/4 the speed of light in the same direction as the spacecraft. I used the relativity velocity addition formula W = (u + v)/(1 + uv/c^2) and this gives the speed of the missile as 24/25c (or 0.96c) if measured by an observer on earth. Now if the missile has a clock the special relativity time dilation equation say the the clock on the missile will be running at 16/25 slower. the clock on the spacecraft is running 4/5 slower to the observer on earth. An astronaut on the spacecraft will see the clock on the missile running 4/5 slower than the clock on the spacecraft. If the astronaut set their clock to run at the same speed as the clock on the missile and the observer on earth would then see the clock on the spacecraft as running (4/5)*(4/5) = 16/25 slower. I do not understand why this is not the same as observing the clock on the missile directly?Similarly the astronaut and the observer on earth would view the mass of the missile is different?