Question

In the twin paradox example, Alice boards a spaceship that flies to a space station 3.25 light-years away and then returns with a speed of \(0.650 c\). View the trip in terms of Alice's reference frame. a) Show that Alice must travel with a speed of \(0.914 c\) to establish a relative speed of \(0.650 c\) with respect to Earth when she is returning to Earth. b) Calculate the time duration for Alice's return flight to Earth at the speed of \(0.914 c\)

Short answer

Answer: Alice must travel with a speed of approximately 0.914 c on her return trip to Earth. Her return flight will take approximately 2.47 years.

Step-by-step solution

(a) Finding Alice's return speed with respect to Earth

To find Alice's speed on her return trip, we need to use the formula for relative speed in special relativity: \(u = \frac{v+w}{1+\frac{vw}{c^2}}\) Where u is the relative velocity, v is the velocity of Alice's spaceship, and w is the velocity of Earth. In our case, the relative velocity u is given as 0.650 c, and the velocity of Earth w is -0.650 c (negative since the spaceship is returning to Earth). We need to find the velocity of the spaceship, v. Plugging in the given values, we get: \(0.650 c = \frac{v-0.650 c}{1+\frac{v(-0.650 c)}{c^2}}\) Now let's solve for the velocity v: \(0.650 c (1+\frac{v(-0.650 c)}{c^2})=v-0.650 c\) \(0.650 c +0.650cv = v-0.650cv +0.650cv^2\) \(v = 0.650c + 0.650cv -0.650cv +0.650cv^2\) \(v(1 - 0.650c + 0.650c^2)= 0.650c\) \(v = \frac{0.650c}{1 - 0.650c + 0.650c^2}\) By calculating the value of v, we get: \(v \approx 0.914c\) So, Alice must travel with a speed of 0.914 c to establish a relative speed of 0.650 c with respect to Earth when she is returning to Earth.

(b) Calculating the time duration for Alice's return flight

Now, we need to compute the time duration for Alice's return flight at the speed 0.914 c. We can use the time dilation formula: \(\Delta t' = \Delta t \sqrt{1-\frac{v^2}{c^2}}\) We are given that the space station is 3.25 light-years away. So, the return distance is the same, 3.25 light-years. First, let's find the time in Earth's frame, \(\Delta t\), which can be obtained from the distance and velocity: \(\Delta t = \frac{distance}{speed} = \frac{3.25 \;lightyears}{0.914c}\) We know that light-years can be converted to years as follows: 1 light-year = \(c \times 1 \;year\). So, we get: \(\Delta t = \frac{3.25 \;years}{0.914} \approx 3.56\, years\) Now, we can use the time dilation formula to find the time duration in Alice's frame, \(\Delta t'\) \(\Delta t' = \Delta t \sqrt{1-\frac{v^2}{c^2}}\) \(\Delta t' = 3.56 \times \sqrt{1-\frac{(0.914c)^2}{c^2}}\) By calculating the value of \(\Delta t'\), we get: \(\Delta t' \approx 2.47 \;years\) So, the time duration for Alice's return flight to Earth at a speed of 0.914 c is approximately 2.47 years.