What is the speed of gases from a star that is receding from Earth at 0.94c, moving back towards Earth at 0.67c, relative to Earth?

To find the speed of the gases from a star that is moving away from Earth and then back towards it, we need to apply the principles of relativistic velocity addition. In this scenario, the star is receding from Earth at a speed of 0.94c and then moving back towards Earth at a speed of 0.67c.

When considering velocities in the context of special relativity, the formula to use for the addition of velocities ( u ) and ( v ) (where ( u ) is the speed of one object as seen from another and ( v ) is the speed of that object as seen from a third reference point) is as follows:

[

w = \frac{u + v}{1 + \frac{uv}{c^2}}

]

In this case, we need to recognize that the star is moving away first, and then we look at the gases moving back toward Earth. Therefore, we set ( u = -0.94c ) (negative because it's moving away) and ( v = 0.67c ).

Applying the formula:

1. Calculate ( uv ):

[

uv = (-0.94c)(0.67c