Question

. A rocketship of length 100 m when at rest, traveling at v/c 0.6, carries a radio receiver at its nose. A radio pulse is emitted from a stationary space station just as the tail of the rocket passes the station (a) From the point of view of the rocket reference frame, how far from the tail does the radio pulse move before being detected by the receiver in the nose? What is the velocity of the radio pulse in the reference frame of the rocket? (b) From the point of view of the rocket reference frame, what is the time interval between the the emission of the radio signal at the rocket's tail to it's detection at the rocket's nose? -(c) From the point of view of the space station reference frame, how far from the space station is the nose of the rocket at the instant of arrival of the radio signal at the nose? Hint: Lorentz transformations, using previous results, can be very helpful. (d) By space-station time, what is the time interval between the arrival of this sig nal and its emission from the station?

Answer 1

(a) The velocity of the radio pulse with respect to the space ship is given by,

$\nu = \frac{v-u}{1-\frac{vu}{c^2}}$

Substituting values we get,

$\Rightarrow \nu = \frac{0.6c-c}{1-\frac{(c)(0.6c)}{c^2}} = \frac{-0.4c}{1-0.6} = -c$

Therefore with respect to the space ship, the velocity of the radio pulse will be c itself. (-ve implies that it is approaching towards it).

Now with respect to the spaceship, the radio pulse will move 100m, before being detected at the receiver at the nose of the spaceship. (There will be no length contraction).

(b) With respect to the rocket frame the time taken for the signal to travel from tail to the nose is,

$t_r = \frac{100}{c} = 3.33 \times 10^{-7} \,s$

(d) Now, with respect to the space station, there will an effect of length contraction on the spaceship (relativistic effect). The length perceived by the space station will be,

$L = L_0 \sqrt{1-\frac{v^2}{c^2}}$

Substituting values we get,

$\Rightarrow L = 100 \sqrt{1-\frac{(0.6c)^2}{c^2}} = 80 \, m$

Distance traveled by rocket nose in time t (t=0 is time when the radio signal is emitted), is given by,

$s_r = 80 + (0.6c)t$

Distance traveled by the radio signal in time t is,

$s_s = ct$

Now, the signal is detected by space ship when both these distances are equal. This implies,

$\Rightarrow s_r = s_s \Rightarrow 80 + (0.6c)t = ct$

$\Rightarrow 80 = 0.4ct$

$\Rightarrow t = \frac{80}{0.4 \times 3 \times 10^8} = 6.67 \times 10^{-7} \, s$

(c) Now the distance traveled by the nose before the signal gets detected is,

$\Rightarrow s_r = 80 + (0.6c)(6.67 \times 10^{-7}) = 200.06 \, m$