Question 39.6: Relative Velocity of Two Spacecraft Two spacecraft A and B a......

Conceptualize There are two observers, one (O) on the Earth and one \left(O^{\prime}\right) on spacecraft \mathrm{A}. The event is the motion of spacecraft \mathrm{B}.

Categorize Because the problem asks to find an observed velocity, we categorize this example as one requiring the Lorentz velocity transformation.

Analyze The Earth-based observer at rest in the S frame makes two measurements, one of each spacecraft. We want to find the velocity of spacecraft \mathrm{B} as measured by the crew on spacecraft A. Therefore, u_{x}=-0.850 c. The velocity of spacecraft A is also the velocity of the observer at rest in spacecraft A (the S^{\prime} frame) relative to the observer at rest on the Earth. Therefore, v=0.750 c.

Obtain the velocity u_{x}^{\prime} of spacecraft \mathrm{B} relative to spacecraft A using Equation 39.16:

u_{x}^{\prime}=\frac{u_{x}-v}{1-\frac{u_{x} v}{c^{2}}}=\frac{-0.850 c-0.750 c}{1-\frac{(-0.850 c)(0.750 c)}{c^{2}}}=-0.977 c

Finalize The negative sign indicates that spacecraft B is moving in the negative x direction as observed by the crew on spacecraft A. Is that consistent with your expectation from Figure 39.14? Notice that the speed is less than c. That is, an object whose speed is less than c in one frame of reference must have a speed less than c in any other frame. (Had you used the Galilean velocity transformation equation in this example, you would have found that u_{x}^{\prime}=u_{x}-v=-0.850 c- 0.750 c=-1.60 c, which is impossible. The Galilean transformation equation does not work in relativistic situations.)