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Question

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?

Solution

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

Time taken by the spaceship to reach the planet as measured by an observer on the spaceship is t0=9.2yt_0 = 9.2\:y

These two time intervals are related as

t=t01v2c2t = \dfrac{t_0}{\sqrt{1 - \dfrac{v^2}{c^2}}} (or) 12y=9.2y1v2c212\:y = \dfrac{9.2\:y}{\sqrt{1 - \dfrac{v^2}{c^2}}}

where vv is the speed of the spaceship.

This equation can be rewritten as

v2c2=19.22122\dfrac{v^2}{c^2} = 1 - \dfrac{9.2^2}{12^2}

v=19.22122c=0.642c=0.642×3×108=1.926×108m.s1v = \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×td = {v} \times {t}

1y=3.16×107s1\:y = 3.16 \times 10^7\:s

So, t=12×3.16×107=37.92×107st = 12 \times 3.16 \times 10^7 = 37.92 \times 10^7\:s

d=1.926×108×37.92×107=73×1015md = 1.926 \times 10^8 \times 37.92 \times 10^7 =73 \times 10^{15}\:m

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