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1 August, 12:04

A rocket moves forward with velocity v = 3c/4, with respect to the earth. Inside this rocket, a second, smaller rocket is launched. In the inertial frame of reference where the larger rocket is at rest, the smaller rocket has velocity v = 3c/4. What is the velocity of the smaller rocket with respect to the earth? You may express the answer as a fraction times c, if you wish (but simplify as much as possible).

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  1. 1 August, 14:57
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    We shall apply the formula for relative velocity in relativistic mechanics

    Vr = v₁ - v₂ / (1 - v₁v₂/c²), v₁ is velocity of smaller rocket with respect to earth, v₂ is velocity of large rocket with respect to earth.

    Given Vr = 3c/4, v₁?, v₂ = 3c/4

    3c/4 = (v₁ - 3c/4) / (1 - v₁ 3c/4c²)

    3c/4 - (9v₁ / 16) = v₁ - 3c/4

    v₁ + 9v₁/16 = 3c/2

    = 25v₁ / 16 = 3c / 2

    v₁ = 16 / 25 x 3c / 2

    = 24/25 x c
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