91影视

Consider a distance Planet $\text{Y}$ to Earth is${喂}_0=20\text{鈥刲测}$.

Consider the time taken is $t=20\text{鈥墆}$.

Write the formula of speed of spaceship.

$谓=\frac{喂}{t}$ 鈥︹�� (1)

Here, $喂$ is distance travelled by the spaceship and $\mathbf{t}$ is time taken.

The connection between distance and time can be used to characterise velocity:

$谓=\frac{d}{t}$ 鈥︹�� (2)

The following equation may be used to show how a relativistic effect causes an object's length to contract:

$喂={喂}_0\sqrt{1鈭�\frac{{谓}^2}{c^2}}$ 鈥︹�� (3)

A lightyear may be stated as $cy$, which is the speed of light multiplied by the years, as a lightyear is equal to the speed of light in a year. Using Eq. (2), we can describe how far the spaceship has travelled as a result of relativistic effects:

$\begin{array}{c}喂={喂}_0\sqrt{1鈭�\frac{{谓}^2}{c^2}}\\ =\left(20\text{ly}\right)\sqrt{1鈭�\frac{{谓}^2}{c^2}}\\ =\left(20\text{cy}\right)\sqrt{1鈭�\frac{{谓}^2}{c^2}}\end{array}$

It is possible to link the distance and speed using Eqs. (2) and (3). We choose and establish the following phrase to represent the spaceship's speed:

Determine the speed of spaceship.

Substitute $\left(20\text{cy}\right)\sqrt{1鈭�\frac{{谓}^2}{c^2}}$ for $喂$ and $20\text{鈥�}y$ for $t$ into equation (1).

$\begin{array}{c}谓=\frac{\left(20\text{cy}\right)\sqrt{1鈭�\frac{{谓}^2}{c^2}}}{20\text{鈥�}y}\\ =c\sqrt{1鈭�\frac{{谓}^2}{c^2}}\end{array}$

We further simplify the derived expression to obtain $谓$ as follows:

$\begin{array}{l}谓=c\sqrt{1鈭�\frac{{谓}^2}{c^2}}\\ \frac{谓}{c}=\sqrt{1鈭�\frac{{谓}^2}{c^2}}\\ \frac{{谓}^2}{c^2}=1鈭�\frac{{谓}^2}{c^2}\\ \frac{{谓}^2}{c^2}+\frac{{谓}^2}{c^2}=1\end{array}$

Solve further as:

$\begin{array}{l}\frac{2{谓}^2}{c^2}=1\\ \frac{{谓}^2}{c^2}=\frac{1}{2}\\ {谓}^2=\frac{c^2}{2}\\ 谓=\frac{c}{\sqrt{2}}\end{array}$

Therefore, the value of speed of spaceship is$谓=\frac{c}{\sqrt{2}}$.