The spaceship Enterprise 1 is moving directly away from earth at a velocity that an earth-based observer measures to be 0.66c. A sister ship, Enterprise 2, is ahead of Enterprise 1 and is also moving directly away from earth along the same line. The velocity of Enterprise 2 relative to Enterprise 1 is 0.34c. What is the velocity of Enterprise 2, as measured by the earth-based observer

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Answer:

The answer is "0.82 c".

Explanation:

Given:

Spacecraft speed 1 is [tex]u = + 0.66 \ c[/tex]

Space velocity 2 relative to spacecraft 1 is [tex]v = + 0.34\ c[/tex]

The spacecraft velocity 2 measured by the Earth observation

   [tex]\to u' = \frac{u +v}{1 + ( \frac{uv}{c^2})}[/tex]

            [tex]= \frac{0.66 \ c +0.34\ c}{ 1+ (\frac{0.66\ c \times 0.33\ c }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (\frac{0.2178\ c^2 }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (0.2178 )}\\\\ = \frac{1 \ c }{ 1.2178 }\\\\=0.82\ c[/tex]

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