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A spaceship is travelling with a velocity of 0.7c away from a space station. The spaceship ejects a probe with a velocity 0.59c opposite to its own velocity. A person in the space station would see the probe moving at a speed Xc, where the value of X is ________ (up to three decimal places).
Correct answer is '0.185 to 0.189'. Can you explain this answer?
Most Upvoted Answer
A spaceship is travelling with a velocity of 0.7c away from a space st...
Explanation:

  • Velocity of spaceship (v) = 0.7c

  • Velocity of probe ejected from spaceship (u) = -0.59c (opposite direction to spaceship's velocity)

  • We need to find the velocity of the probe as observed by a person in the space station (let's call it w)

  • We can use the relativistic velocity addition formula to find w:



Relativistic velocity addition formula:
w = (u + v)/(1 + uv/c^2)
where c = speed of light

Substituting values:
w = (-0.59c + 0.7c)/(1 - 0.59c*0.7c/c^2)
w = 0.11c/0.5923
w = 0.185c

Therefore, the velocity of the probe as observed by a person in the space station is 0.185c.

Note:
The answer given in the question is a range (0.185 to 0.189c), which is due to the fact that the velocity addition formula involves square roots and rounding off errors may occur.
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A spaceship is travelling with a velocity of 0.7c away from a space station. The spaceship ejects a probe with a velocity 0.59c opposite to its own velocity. A person in the space station would see the probe moving at a speed Xc, where the value of X is ________ (up to three decimal places).Correct answer is '0.185 to 0.189'. Can you explain this answer?
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