Question

A 481-m long spaceship passes by an observer at the speed of 2.70×10^8 m/s. What length does the observer measure for the spaceship?

Answer 1

Answer:

209.66 m

Explanation:

Given:

Original length of the spaceship, L = 481 m

Speed of the spaceship, v = 2.70 × 10⁸ m/s

Now,

using the concept of length contraction, we have

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

where,

L' is the observed length

c is the speed of the light

Thus,

on substituting the respective values, we get

$481=\frac{L'}{\sqrt{1-\frac{(2.70\times10^8)^2}{(3\times10^8)^2}}}$

or

L'  = 209.66 m