A planet is approaching the Earth at a speed of 0.51C. From the opposite side a rocket approaching the Earth with a speed of V=0.51C. So planet and rocket are a distance d from Earth but on opposite sides of Earth. Both approaches the Earth at the speed of V=0.51C. All three will be in the same place after a time t = d/0.51C This "see" those who are on Earth. What "see" those who are on the planet? They see the Earth which is at distance d and approaching it with speed V = 0.51C. They see the rocket, the same direction as the Earth but at a distance 2d. All three will be in the same place after a time t. So the rocket had a speed of 2d / t = 1.02C.
Nope, they see dialation effects. They see eachother approach at approx. .81c This was similar to my question in the Special Relativity thread, except for some reason I kept getting different times for the objects closing on each other. In any case, the objects moving that fast will see length contraction along the route they take, and will see time moving at a different rate. Because of this they will see the other ship approaching not at twice the velocity they are at, but at .81c. Nothing will violate speed of light from any reference point.
Let's analyze what happens between the planet and Earth, neglecting the rocket. They have speed V = 0.51C between them, and the distance between them is d. Regardless of where the look, they meet after a time t = d/0.51C?
Well, not exactly. As I understand it, the observer from the earth sees that they meet at t = d/.51c. However for the rocket, the distance is not the same, even though the velocity of approach is. So from the rockets perspective they meet at t' = Gamma (t - V * d/c^2) where Gamma = 1/SQRT(1-v^2/c^2) Gamma can also be used to determine the distance he observes the earth (and the other rocket) as being at.
Im sorry, but did not get my little experiment. It about what "see" those on the Earth and what "see" those on the planet. Also velocity effects (compression) are on the rocket and not from the rocket or distance. If you want to be convincing then to come with the exact calculations show that V = 0.81C as you affirmed.
