91Ó°ÊÓ

The velocity of the Bob’s spaceship, v=0.8 c

The velocity of carl’s spacecraft, u=0.9 c

The expression for the Lorentz velocity transformation is given as follows.

$\mathbf{u}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{u}\mathbf{-}\mathbf{v}}{\mathbf{1}\mathbf{-}{\displaystyle \frac{\mathbf{uv}}{{\mathbf{c}}^{\mathbf{2}}}}}$ …… (i)

Here, u' is the velocity of the object in frame s' , u is the velocity of object in frame s , v is the velocity of frame s' relative to s and c is the speed of the light.

Calculate the speed of Carl relative to Bob’s spaceship.

Substitute 0.8c for V and for 0.9c for u into equation (i).

$$\begin{gathered} u\text{'}=\frac{0.9c-0.8c}{1-{\displaystyle \frac{\left(0.9c\right)\left(0.8c\right)}{c^2}}} \\ u\text{'}=\frac{0.1c}{1-0.72} \\ u\text{'}=\frac{0.1c}{0.28} \\ u\text{'}=0.357c \end{gathered}$$

Hence the velocity of the speed of Carl relative to Bob’s spaceship is 0.357c .

The velocity Carl with respect to Earth is given by,

u'=u-v …… (ii)

Here, u is the apparent velocity of Bob to Carl and v is the apparent velocity of the Earth with respect to Carl.

Substitute 0.357c for u and -0.8c for V into equation (ii).

u'=0.357c-(-0.8c)

u'=0.357c+0.8c

u'=1.157c

Hence according, the speed of the Carl relative to Earth is 1.157c.