Question

One cosmic-ray particle approaches Earth along Earth’s north-south axis with a speed of $\text{0.80c}$toward the geographic north pole, and another approaches with a speed of $0.60c$ toward the geographic south pole (Fig. 37- 34). What is the relative speed of approach of one particle with respect to the other?

Short answer

The speed of one particle relative to another is$0.95c$.

Step-by-step solution

Relativistic velocity addition:

Suppose an object is moving with the velocity ${\mathit{u}}^{\mathbf{\text{'}}}$ with respect to ${\mathit{S}}^{\mathbf{\text{'}}}$ frame, which is moving with the velocity $\mathit{v}$ with respect to frame $\mathit{S}$ then the velocity of the object $u$ with respect to $\mathit{S}$ frame is,

role="math" localid="1663136206277" $\mathit{u}\mathbf{=}\frac{\mathbf{u}\mathbf{\text{'}}\mathbf{+}\mathbf{v}}{\mathbf{1}\mathbf{+}\frac{\mathbf{u}\mathbf{\text{'}}\mathbf{v}}{{\mathbf{c}}^{\mathbf{2}}}}$

Determine the relative velocity:

Let’s consider the rest frame of $0.6c$particle moving towards the south-pole to be $S$frame. And the earth frame is called $S^{\text{'}}$frame.

In S frame, earth is moving towards it with velocity $v=0.6c$ and relative to this $S^{\text{'}}$ frame a particle is moving towards the North pole with speed $u\text{'}=0.8c$.

Then the velocity of this same particle relative to $S$frame is,

$$\begin{gathered} u=\frac{0.8c+0.6c}{1+\left(0.8\right)\left(0.6\right)} \\ =\frac{1.4c}{1.48} \\ =0.946c \end{gathered}$$

Hence, the speed of one particle relative to other is $0.95c$.