4. The length of a spaceship is measured to be exactly half its actual length.
Calculate i) the speed of the spaceship. ii) time dilation corresponding one second

To tackle this problem, we need to delve into concepts of special relativity, particularly how motion affects the measurement of length and time. When a spaceship moves at a significant fraction of the speed of light, its length appears shortened to an observer at rest due to an effect known as length contraction. Let's break down the two parts of your question step by step.

Understanding Length Contraction

According to Einstein's theory of special relativity, the length $L$ of an object moving at a speed $v$ is given by the formula:

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

Here, $L_0$ is the proper length (the length of the spaceship at rest), $L$ is the contracted length observed, $v$ is the speed of the spaceship, and $c$ is the speed of light (approximately $3\times {10}^8$ m/s).

Calculating Speed

You mentioned that the length of the spaceship is measured to be exactly half its actual (proper) length. This means:

$L=\frac{L_0}{2}$

Substituting this into the length contraction formula gives us:

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

We can simplify this equation by dividing both sides by $L_0$ (assuming $L_0\ne 0$):

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

Next, we square both sides:

$\frac{1}{4}=1-\frac{v^2}{c^2}$

Rearranging this gives:

$\frac{v^2}{c^2}=1-\frac{1}{4}=\frac{3}{4}$

Now, multiplying both sides by $c^2$ yields:

$v^2=\frac{3}{4}{c}^2$

Taking the square root provides us with the speed:

$v=c\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}c\approx 0.866c$

Examining Time Dilation

Time dilation is another fascinating consequence of special relativity. The time $t$ measured on the spaceship (proper time) and the time $t^{\prime }$ measured by an observer at rest are related by the formula:

$t^{\prime }=\frac{t}{\sqrt{1-\frac{v^2}{c^2}}}$

In this case, we want to find the time dilation corresponding to one second on the spaceship. If we assume $t=1$ second, we can substitute our previously calculated speed:

$t^{\prime }=\frac{1}{\sqrt{1-\frac{3}{4}}}=\frac{1}{\sqrt{\frac{1}{4}}}=2$ seconds

This implies that for every second that passes on the spaceship, two seconds elapse for an observer at rest. Hence, the time dilation effect is significant when the spaceship moves at such high speeds.

Summary of Results

• The speed of the spaceship is approximately $0.866c$.
• The time dilation corresponding to one second on the spaceship is two seconds for an observer at rest.

These results illustrate the profound implications of Einstein's theories on our understanding of space and time. The faster an object moves, the more pronounced these effects become, changing our perception of reality itself.