Question

A person travels on a spaceship moving at a speed of $\dfrac{{5c}}{{13}}$.
(a) Find the time interval calculated by him between the consecutive birthday celebrations of his friend on the earth.
(b) Find the time interval calculated by the friend on the earth between the consecutive birthday celebrations of the traveler.

Answer 1

Hint:We can find the time interval between the two events using the concept time dilation which states that it is the difference in the elapsed time as measured by two clocks placed in two different reference frames, one frame is moving with certain velocity with respect to the other.

Formula used:
The dilated time in terms of the proper time is given by
\[t' = \dfrac{t}{{\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} }}\]
Where, $t$ - proper time of object moving with velocity $v$ and $c$ - speed of light.

Complete step by step answer:
We are given that, the speed of the spaceship is $v = \dfrac{{5c}}{{13}}$.
(a) We have to calculate time interval calculated by him between the consecutive birthday celebrations of his friend on the earth i.e. $t = 1$ year. So, Using the formula, we get
\[t' = \dfrac{1}{{\sqrt {1 - \dfrac{{{5^2}{c^2}}}{{{{13}^2}{c^2}}}} }} \\
\Rightarrow t'= \dfrac{1}{{\sqrt {1 - \dfrac{{25}}{{169}}} }}\]
\[\Rightarrow t' = \dfrac{{13}}{{\sqrt {169 - 25} }} \\
\therefore t'= \dfrac{{13}}{{12}}\] year

(b) The friend on the earth also considers the same speed with which the person is moving in spaceship, so the time interval for him will also be same i.e.
\[\therefore t' = \dfrac{{13}}{{12}}\] year

Note: In time dilation, time observed by the person in the rest frame is less than the time observed by the person in the moving frame. So, while making calculations, keep it in mind. It is either due to a relative velocity between two bodies or to a difference in gravitational potential between locations of these bodies. Also, the time difference between the consecutive birthdays is 1 year.