A rocket speeding through space at 0.3c launches a small satellite. If the satellite is launched with its own velocity in the direction of the rocket’s motion at 0.4c, what will be the satellite’s observed velocity from the perspective of someone watching it from Earth?
To calculate the observed velocity of the satellite from the perspective of someone watching it from Earth, we need to apply the relativistic velocity addition formula.
The relativistic velocity addition formula is given by:
v' = (v1 + v2) / (1 + (v1 * v2) / c^2)
In this formula, v' represents the observed velocity, v1 represents the velocity of the rocket, v2 represents the velocity of the satellite, and c represents the speed of light (which is approximately 3 x 10^8 meters per second).
In this scenario, the velocity of the rocket (v1) is given as 0.3c. The velocity of the satellite (v2) is given as 0.4c.
Let's substitute these values into the formula:
v' = (0.3c + 0.4c) / (1 + (0.3c * 0.4c) / c^2)
Simplifying this equation, we have:
v' = (0.7c) / (1 + (0.12) / 1)
This further simplifies to:
v' = (0.7c) / (1 + 0.12)
Calculating the denominator:
v' = (0.7c) / (1.12)
Finally, simplifying the equation:
v' ≈ 0.625c
Therefore, from the perspective of someone watching the satellite from Earth, the observed velocity of the satellite will be approximately 0.625 times the speed of light (c).