Question

As you pilot your space utility vehicle at a constant speed toward the moon, a race pilot flies past you in her space racer at a constant speed of 0.850 c, relative to you. At the instant the space racer passes you, both of you start timers at zero.
a) At the instant when you measure that the space racer has traveled 1.22 x 108 m past you, what does the race pilot read on her timer?
b) When the race pilot reads the value calculated in the previous part on her timer, what does she measure to be your distance from her?
c) At the instant when the race pilot reads the value calculated in part a) on her timer, what do you read on yours?

Answer 1

Answer:

a) At the instant when you measure that the space racer has traveled 1.22 x 108 m past you the time read by the race pilot is 0.478 s

b) When the race pilot reads the value of her time calculated,  in part on her timer(0.478 s), she measure the distance of the space utility vehicle from her to be 6.43 × 10⁷ m

c) The instant when the race pilot reads the 0.478 s value calculated in part a) on her timer, the pilot of the space utility vehicle reads 0.478 s

Explanation:

To solve the question, we take account of the theory of relativistic motion as follows

Let Δt be the time duration observed by the race pilot and

Let Δt₀ is the time interval measured by the pilot in the space utility vehicle.

With u, being the velocity of the space racer and c, the speed of light.

Therefore by Lorentz factorization, we have

$\Delta t = \frac{\Delta t_0}{\sqrt{1-\frac{u^2}{c^2} } }$

(a) When the space racer has traveled a distance of 1.22 x 10⁸ m past then we have

$\Delta t = \frac{d}{0.850\cdot c}= \frac{1.22\times 10^8 \hspace{0.09cm} m}{0.85 \times 3 \times 10^8 \hspace {m/s} }$ = 0.478 s

Since $\Delta t = \frac{\Delta t_0}{\sqrt{1-\frac{u^2}{c^2} } }$ , we have

$\Delta t_0 = {\Delta t} \times {\sqrt{1-\frac{u^2}{c^2} } }= 0.478\times\sqrt{1-\frac{(0.8\cdot c)^2}{c^2} }$ = 0.252 s

(b) The distance measured by the racer is given as

d = Δt₀ × 0.850 × c =  0.252 × 0.850 × 3.0 x 10⁸ m/s = 6.43 × 10⁷ m

(c) The time red from within the utility vehicle is given as;

$\frac{1.22 \times 10^8 m}{0.85 \times 3\times 10^8 m/s} = 0.478 s$.