A spaceship is traveling at two-thirds of the speed of light directly toward a stationary asteroid. If the spaceship turns on it headlights, what will be the speed of the light traveling from the spaceship to the asteroid as observed by a) someone on the spaceship? b) someone on the asteroid?

Since the speed of light is a fundamental constant and is always \(3.0 \times 10^8\,m/s\) in any inertial frame, the observer on the spaceship will see the light beam traveling at the speed of light (\(c\)). Speed of light for spaceship observer = \(c = 3.0 \times 10^8\,m/s\).

To find the speed of light observed by someone on the asteroid, we'll first find the velocity of the light according to the spaceship (i.e., the source frame of reference). Velocity of the spaceship, \(v = \frac{2}{3}c = \frac{2}{3}(3.0 \times 10^8\,m/s) = 2.0 \times 10^8\,m/s\). Next, we use the relativistic velocity addition formula to determine the velocity of light observed by the person on the asteroid, which is: \(u' = \frac{u+v}{1+\frac{uv}{c^2}}\), where \(u' = \) velocity of light observed by asteroid observer, \(u = c\) (speed of light), and \(v = 2.0 \times 10^8\,m/s\) (velocity of the spaceship). Now we plug in the values: \(u' = \frac{3.0 \times 10^8 + 2.0 \times 10^8}{1+\frac{(3.0 \times 10^8)(2.0 \times 10^8)}{(3.0 \times 10^8)^2}}\) Simplifying the equation: \(u' = \frac{5.0 \times 10^8}{1+\frac{2}{3}} = \frac{5.0 \times 10^8}{\frac{5}{3}}\) Thus, the speed of light observed by someone on the asteroid is also equal to the speed of light (\(c\)): \(u' = 3.0 \times 10^8\,m/s\).