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# A spaceship travels at a constant speed from earth to a planet orbiting another star. When the spacecraft arrives, 12 years have elapsed on earth, and 9.2 years have elapsed on board the ship. How far away (in meters ) is the planet, according to observers on earth?

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Time taken by the spaceship to reach the planet as measured by an observer on earth is $t = 12\:y$

Time taken by the spaceship to reach the planet as measured by an observer on the spaceship is $t_0 = 9.2\:y$

These two time intervals are related as

$t = \dfrac{t_0}{\sqrt{1 - \dfrac{v^2}{c^2}}}$ (or) $12\:y = \dfrac{9.2\:y}{\sqrt{1 - \dfrac{v^2}{c^2}}}$

where $v$ is the speed of the spaceship.

This equation can be rewritten as

$\dfrac{v^2}{c^2} = 1 - \dfrac{9.2^2}{12^2}$

$v = \sqrt{1 - \dfrac{9.2^2}{12^2}} c =0.642 \:c = 0.642 \times 3 \times 10^8 = 1.926 \times 10^8\:m.s^{-1}$

Now the distance between the earth and the planet as measured by an observer in earth is given by

$d = {v} \times {t}$

$1\:y = 3.16 \times 10^7\:s$

So, $t = 12 \times 3.16 \times 10^7 = 37.92 \times 10^7\:s$

$d = 1.926 \times 10^8 \times 37.92 \times 10^7 =73 \times 10^{15}\:m$

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