Question

The radius of our galaxy is about $3\times {{10}^{20}}m$. With what speed should a person travel so that he can reach from the center of the galaxy to its edge in 20 years of his lifetime?

Answer 1

Hint: This problem can be solved by Einstein’s theory of special relativity. Since the person will be moving at a very high speed, the length of the radius of the galaxy as observed by him will be shorter than what would be if he was at rest. This is because when moving near the speed of light, time will pass slowly for him and it will appear to him that he has covered the distance faster and hence it will seem that the distance that had to be covered became smaller.

Formula used:
For a body moving at speed $v$, comparable to the speed of light and observing a length, the length apparent $\left( l \right)$ to the body will be
$l={{l}_{0}}\left( \sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}} \right)$
where ${{l}_{0}}$ is the length when the body is at rest and $c=3\times {{10}^{8}}m/s$ is the speed of light.

Complete step-by-step answer:
Since, the person travels such a large distance in a relatively short amount of time, it can be said that he is moving close to the speed of light. Hence, here Einstein’s theory of special relativity will come in. Going by this theory, the length of the radius of the galaxy as observed by the person will be shorter than the actual length if he were at rest. This is because when moving close to the speed of light, time passes slower and hence he will feel that he has reached his destination quicker. This can also be considered as the time passed at the same rate but the length of the radius became shorter by the same factor.
For a body moving at speed $v$, comparable to the speed of light and observing a length, the length apparent $\left( l \right)$ to the body will be
$l={{l}_{0}}\left( \sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}} \right)$ --(1)
where ${{l}_{0}}$ is the length when the body is at rest and $c=3\times {{10}^{8}}m/s$ is the speed of light.
Hence, now let us analyze the question.
Let the rest length of the radius of the galaxy be ${{l}_{0}}=3\times {{10}^{20}}m$.
The apparent length of the radius of the galaxy be $l$
The speed of the person is $v$.
Now, as observed by the person, the edge of the galaxy comes towards him at a speed $v$.
The time taken by him to reach the edge of the galaxy is $t=20yr$.
$\therefore l=vt=v\times 20yr$ $\left( \because \text{Length=Speed}\times \text{time} \right)$ --(2)
Putting (2) in (1), we get,
$20\times 3.154\times {{10}^{7}}\times v=3\times {{10}^{20}}\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}$ $\left( \because 1yr=3.154\times {{10}^{20}}s \right)$
$\therefore \sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}=\dfrac{20\times 3.154\times {{10}^{7}}\times v}{3\times {{10}^{20}}}=2.103\times {{10}^{-12}}v$
Squaring both sides, we get,
${{\left( \sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}} \right)}^{2}}={{\left( 2.103\times {{10}^{-12}}v \right)}^{2}}$
$\therefore 1-\dfrac{{{v}^{2}}}{{{c}^{2}}}=4.42\times {{10}^{-24}}{{v}^{2}}$
$\therefore {{c}^{2}}-{{v}^{2}}=4.42\times {{10}^{-24}}{{v}^{2}}{{c}^{2}}$
$\therefore {{c}^{2}}={{v}^{2}}\left( 1+4.42\times {{10}^{-24}}{{c}^{2}} \right)$
${{c}^{2}}={{v}^{2}}\left( 1+4.42\times {{10}^{-24}}\times 9\times {{10}^{16}} \right)={{v}^{2}}\left( 1+3.978\times {{10}^{-7}} \right)=\left( 1.0000004 \right){{v}^{2}}$
Square rooting both sides, we get,
$\sqrt{{{c}^{2}}}=\sqrt{1.0000004{{v}^{2}}}$
$\therefore c=1.0000002v$
$\therefore v=\dfrac{c}{1.0000002}=0.99999998c$
$\therefore v=0.99999998\times 3\times {{10}^{8}}=2.99999994\times {{10}^{8}}m/s$
Hence, the required speed of the person is $2.99999994\times {{10}^{8}}m/s$.

Note: It might be clear to students at first glance that this is a problem involving the theory of special relativity and they may proceed simply by finding out the speed by calculating the ratio of the distance travelled to the time taken. However, they must realize that where speeds in the region of the speed of light are concerned, then the effect of time dilation enters and then concepts like rest length and apparent length come in. This is especially a tricky question, the likes of which are often set to check whether the students have the presence of mind to factor in these concepts or not.