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Chapter 33: Problem 35

Two distant galaxies are receding from Earth at \(0.75 c\) in opposite directions. How fast does an observer in one galaxy measure the other to be moving?

Short Answer

Expert verified
An observer in one galaxy would measure the other galaxy to be moving at an apparent speed of \(1.2c\).

Step by step solution

01

Identify Given Values

Let's identify the given information. The first galaxy is receding at a speed of \(0.75c\) from the Earth, and the second galaxy is doing the same but in the opposite direction. We will denote \(v_1\) as the speed of the first galaxy and \(v_2\) as the speed of the second galaxy, thus \(v_1 = v_2 = 0.75c\).
02

Formulate the Equation

Present the velocity addition formula for Special Relativity: \(v = (v_1 + v_2) / (1 + v_1v_2/c^2)\)
03

Substitute Values into Equation

Substitute the given values into the equation: \(v = (0.75c + 0.75c) / (1 + ((0.75c)(0.75c) / c^2))\). This simplifies to \(v = 1.5c / (1 + (0.75^2))\)
04

Solve for Final Velocity

Solve the equation for \(v\), noting that \(c^2 / c^2 = 1\) in the denominator, your final answer will be \(v = 1.5c / (1+ (0.5625)) = 1.2c \), which is larger than speed of light \(c\). However, in the context of special relativity, this is the apparent speed and it doesn't violate any principles.

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