Question

Two spaceships approach the earth from opposite directions. According to an observer on the earth, ship a is moving at a speed of 0.753c and ship b at a speed of 0.851c. What is the velocity of ship a as observed from ship b

Answer 1

As per the question two spaceships A and B are approaching towards  earth from opposite directions.

The velocity of ship A with respect to the observer on earth frame is 0.753c

where c is the velocity of light i.e 3×$10^{8} m/s$

Hence $v_{AE} =0.753$c

Again the velocity of spaceship B with respect to the observer on earth frame is 0.851c i.e $v_{BE} =0.851$c

Now we have to calculate the velocity of space ship A with respect to the space ship B.

We know $v_{AB} =v_{AE} +v_{EB}$

                                   =$v_{AE} -v_{BE}$ [$v_{XY} = -v_{YX}$]

                                    =0.753c-0.851c

                                    = -0.098c [ans]

                       

                                                       

Answer 2

Answer:

$v_{AB}=0.977c$    

Explanation:

It is given that,

Speed of ship A, $v_A=0.753c$

Speed of ship B, $v_B=0.851c$

Let $v_{AB}$ is  the velocity of ship A as observed from ship B. It can be calculated using the formula as :

$v_{AB}=\dfrac{v_A+v_B}{1+\dfrac{v_Av_B}{c^2}}$

Substituting all the values in above formula as :

$v_{AB}=\dfrac{0.753c+0.851c}{1+\dfrac{0.753c\times 0.851c}{c^2}}$

$v_{AB}=0.977c$

So, the velocity of ship A as observed from ship B is 0.977c. Hence, this is the required solution.