A spaceship moving away from the earth with velocity 0.6c fires a rocket in the direction of travel with a speed of 0.7c relative to the spaceship. What will be the velocity of the rocket, as observed from the earth? where c is the velocity of light.

Concept:

(Relativistic) Addition of velocities or velocity addition theorem:

The Lorentz transformation equations enable us to transform velocity from one frame of reference to another, in relative motion with respect to it.

Let S and S’ be the two inertial frames in relative motion so that S; moves with uniform velocity v to the right, along the X-axis, relative to S.

Let u and u’ be the velocities of a particle measured in the inertial frames S and S’ respectively.

\(u' = \frac{{u - v}}{{1 - \frac{{uv}}{{{c^2}}}}}\)

So, if a particle has a velocity u’ in a frame of reference S’ which is moving with velocity v relative to another frame of reference S, the velocity of the particle in the frame S will be:

\(u = \frac{{u' + v}}{{1 + \frac{{u'v}}{{{c^2}}}}}\)

Calculation:

u' = velocity of the rocket relative to space ship

v = velocity of the spaceship relative to the earth

u' = 0.7 c, v = 0.6 c

\(u = \frac{{u' + v}}{{1 + \frac{{u'v}}{{{c^2}}}}} = \frac{{\left( {0.7 + 0.6} \right)c}}{{1 + 0.7 \times 0.6}}\)

\(u = 0.92\;c\)