A person is travelling on a spaceship that moves at a speed of 5c/13 . Determine the time interval calculated by this person between consecutive birthday celebrations of his friend on the earth.

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Correct answer: D. (13)/(12) year

The proper time interval between consecutive birthdays on Earth is

\Delta t_0 = 1 \text{ year}

.

The speed of the spaceship is

v = \dfrac{5c}{13}

.

The time interval measured by the person on the spaceship is given by the time dilation formula:

\Delta t = \dfrac{\Delta t_0}{\sqrt{1 - \dfrac{v^2}{c^2}}}

Substituting the given values:

\Delta t = \dfrac{1}{\sqrt{1 - \left(\dfrac{5}{13}\right)^2}}

\Delta t = \dfrac{1}{\sqrt{1 - \dfrac{25}{169}}}

\Delta t = \dfrac{1}{\sqrt{\dfrac{144}{169}}}

\Delta t = \dfrac{13}{12} \text{ year}

A person is travelling on a spaceship that moves at a speed of...

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