Final answer:
To find the probe's ejection speed from the spaceship traveling at 0.88c so that its speed relative to the planet is 0.95c, the relativistic velocity addition formula is employed to solve for the probe's velocity relative to the spaceship.
Step-by-step explanation:
The question asks for the calculation of the probe's speed as it leaves a spaceship traveling at 0.88c so that its relative speed to the planet is 0.95c. This problem is situated within the physics concept of relative velocity in special relativity. According to the relativistic velocity addition formula:
v = (u + v') / (1 + u*v'/c^2)
Where:
- v is the desired relative speed to the planet (0.95c),
- u is the spaceship's speed relative to the planet (0.88c), and
- v' is the probe's speed relative to the spaceship.
To find v', the speed the probe must leave the spaceship, we solve the equation for v':
v' = (v - u) / (1 - u*v/c^2)
Inserting the values gives us:
v' = (0.95c - 0.88c) / (1 - 0.88*0.95/c^2)
After simplifying, we calculate the probe's speed relative to the spaceship, which would then allow it to have a relative speed of 0.95c as seen from the planet.