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Chapter 33: Problem 33

Twins \(A\) and \(B\) live on Earth. On their 20 th birthday, twin \(B\) climbs into a spaceship and makes a round-trip journey at \(0.95 c\) to a star 30 light years distant, as measured in the Earth-star reference frame. What are their ages when twin B returns to Earth?

Short Answer

Expert verified
When twin B returns to Earth, twin A will be 60 years old because in his frame, the spaceship took 60 years for a round-trip to the star. But due to the effects of time dilation, twin B is approximately 18.7 years older. So, their ages when twin B returns to Earth are, respectively, 60 and 18.7 years older than their age when twin B left Earth.

Step by step solution

01

Identifying the Given Information

It's given in the problem that twin B travels to a star 30 light years away at a speed of \(0.95c\), where \(c\) is the speed of light. Because one light-year is the distance that light travels in one year, the time taken for the journey in the frame of reference of twin A on Earth, who is stationary relative to the star, is \(2*30 = 60\) years as light speed is constant.
02

Understanding Time Dilation

According to the principle of time dilation in special relativity, time slows down or dilates when in motion relative to an observer. This can be calculated using the formula \(\Delta t = \Delta t_0 / \sqrt{1 - (v^2 /c^2)}\) where: \(\Delta t\) is the dilated time, \(\Delta t_0\) is the time in the stationary frame, \(v\) is the relative velocity of the frame, \(c\) is the speed of light.
03

Calculating Time Experienced by Twin B

In this case we are looking for \(\Delta t_0\), the time experienced by twin B during the journey in his own frame. Using the time dilation formula and rearranging we find \(\Delta t_0 = \Delta t * \sqrt{1 - (v^2 / c^2)}\) . Substituting the given values: \(\Delta t_0 = 60 years * \sqrt{1 - (0.95)^2)} = 60 years * \sqrt{1 - 0.9025} = 60 years * \sqrt{0.0975} \)\(\Delta t_0 = 60 years * 0.312\) approximately. So, \(\Delta t_0 = 18.7\) years approximately.

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