Question

A spacecraft with a proper length of 350 m passes by an observer on the Earth. According to this observer, it takes 0.780 µs for the spacecraft to pass a fixed point. Determine the speed of the spacecraft as measured by the Earth-based observer.

Please show how you arrive at your answer!

Answer 1

length,l=350 m

time=0.780 µs

Wew know that the apparent length = speed x time = (0.78*10^-6)s

We can assume that the speed is close to the speed of light,

So we can make an approximation as,

The apparent length = 3*10⁸ x 0.78*10^-6 = 234 m

Y = 1/sqrt(1–V²/c²) = 350/234 = 1.49

(1–V²/c²) = 1/1.49² = 0.45

V²/c² = 0.55

V = 0.741c

We can make another approximation as

apparent length = 0.741 x 3*10⁸ x 0.78*10^-6 = 173 m

Y = 1/sqrt(1–V²/c²) = 350/173 = 2.01

(1–V²/c²) = 1/2.01² = 0.247

V²/c²=1-0.247

V = 0.867c

We can make another approximation as

apparent length = 0.867 x 3*10⁸ x 0.78*10^-6 = 202 m

Y = 1/sqrt(1–V²/c²) = 350/202 = 1.73

(1–V²/c²) = 1/1.73² = 0.333

V = 0.816c

Next approximation as

apparent length = 0.816 x 3*10⁸ x 0.78*10^-6 = 191 m

Y = 1/sqrt(1–V²/c²) = 350/191 = 1.83

(1–V²/c²) = 1/1.83² = 0.297

V = 0.837c

The next approximation

apparent length = 0.837 x 3*10⁸ x 0.78*10^-6 = 196 m

Y = 1/sqrt(1–V²/c²) = 350/196 = 1.78

(1–V²/c²) = 1/1.78² = 0.313

V = 0.828c

So we can find that the values are starting to converge

Next approximation is

apparent length = 0.828 x 3*10⁸ x 0.78*10^-6 = 194 m

So I like to take 202 m as the correct answer.