A rocket fires a projectile at a speed of 0.95c (relative to the rocket) while traveling past the Earth. An earthbound scientist measures the projectile's speed to be 0.90c (relative to earth). What was the rocket's speed (relative to earth) as a fraction of c?
To determine the rocket's speed relative to Earth, we can use Einstein's theory of special relativity. According to this theory, the addition of velocities in different reference frames can't be done simply by adding them directly. Instead, we need to use the relativistic velocity addition formula:
V' = (V + u) / (1 + (V * u / c^2))
Where:
V' = velocity of the projectile relative to Earth
V = velocity of the projectile relative to the rocket (0.95c)
u = velocity of the rocket relative to Earth (unknown)
c = speed of light in a vacuum (299,792,458 m/s)
Substituting the given values into the formula, we have:
0.90c = (0.95c + u) / (1 + (0.95c * u / c^2))
Now, let's solve for u, the velocity of the rocket relative to Earth.
To simplify the equation, divide both sides by c:
0.90 = (0.95 + u/c) / (1 + (0.95u / c))
Rearrange the equation:
0.90 * (1 + 0.95u / c) = 0.95 + u/c
Multiply through by (1 + 0.95u / c):
0.90 + 0.855u / c = 0.95 + u / c
Rearrange again:
0.855u / c - u / c = 0.95 - 0.90
Combine the terms on both sides:
(0.855 - 1) u / c = 0.05
Simplify:
-0.145u / c = 0.05
Multiply through by -c / 0.145:
u = (0.05 * -c / 0.145)
Now, let's calculate the value of the rocket's speed relative to Earth.
Substituting the given values for c and computing the equation:
u = (0.05 * - 299,792,458 m/s) / 0.145
u ≈ -0.344c
Therefore, the rocket's speed relative to Earth is approximately -0.344 times the speed of light (c).