Chapter 25, Problem 90QAP

To determine

(a)

What is the minimum speed the spaceship will need for the captain to be no more than 60 years old at arrival in December 2009 the discovery was announced of a planet that may have a large amount of water and hence would be a good candidate for possible life. The planet, GJ 1214b, orbits a small star that is 42 light years from Earth. In the future we might decide to send some astronauts to explore the planet. When they arrive there, we want them to be young enough to perform tests. Suppose that the captain is 25 years old at launch time?

Explanation of Solution

Calculation:

At rest, the half-life of a certain radioisotope is $\Delta t_{proper}=2.25\times {10}^{-6}s$. The half-life of the same radioisotope traveling in a high-speed spaceship as measured from Earth is $\Delta t=3.15\times {10}^{-6}s$. Since the radioisotope is at rest relative to the spaceship, an astronaut in the ship will measure the same lifetime as the radioisotope at rest on Earth. The speed of the spaceship can be calculated from the proper and observed lifetimes, $\Delta t=\frac{\Delta t_{proper}}{\sqrt{1-\frac{v^2}{c^2}}}$

A 25-year-old captain pilots a spaceship to a planet that is $\Delta x=42ly$ from Earth. The captain needs to be no more than 60 years old when she arrives at the planet, which means the proper time interval, as observed by the spaceship, is $\Delta t_{proper}=35ly$. The speed of the spaceship relative to Earth v is equal to $\Delta x$ divided by the time interval as observed by Earth $\Delta t$, where $\Delta t=\frac{\Delta t_{proper}}{\sqrt{1-\frac{v^2}{c^2}}}$ In order to make the math easier, we'll use the fact that 1 ly = (1 y)c. The captain sends a radio signal back to Earth as soon as it reaches the planet. The total time after the launch when the signal arrives is equal to the time it takes the ship to reach the planet plus the time it takes the signal to travel back to Earth. We can relate these times to the distance between Earth and the planet as well as the speeds of the spaceship and the speed of light.
$\begin{array}{l}\Delta t=\frac{\Delta t_{proper}}{\sqrt{1-\frac{v^2}{c^2}}}\\ v=\frac{\Delta x}{\Delta t}=\frac{\Delta x}{\Delta t_{proper}}\sqrt{1-\frac{v^2}{c^2}}\\ v^2={\left(\frac{\Delta x}{\Delta t_{proper}}\right)}^2-{\left(\frac{\Delta x}{\Delta t_{proper}}\right)}^2\frac{v^2}{c^2}\\ v=\frac{\Delta x}{\Delta t_{proper}\sqrt{1+{\left(\frac{\Delta x}{c\Delta t_{proper}}\right)}^2}}=\frac{c\Delta x}{\sqrt{{\left(c\Delta t_{proper}\right)}^2+{\left(\Delta x\right)}^2}}=\frac{c\times 42}{\sqrt{{\left(c\times 35\right)}^2+{\left(42\right)}^2}}=0.77c\end{array}$

Conclusion:

Minimum speed the spaceship will need for the captain to be no more than 60 years old at arrival in December 2009 = 0.77c

To determine

(b)

How many years after launch from Earth will it be when the signal arrives at Earth as soon as the spaceship arrives at the planet, the captain has orders to send a radio signal to Earth to notify Mission Control that the trip was successful.? You can ignore acceleration times and any motion of Earth and GJ 1214b.

Explanation of Solution

Calculation:

  $\begin{array}{l}\Delta t_{total}=\Delta t_{spaceship}+\Delta t_{signal}=\frac{\Delta x}{v}+\frac{\Delta x}{c}=\Delta x\left(\frac{1}{v}+\frac{1}{c}\right)=\Delta x\left(\frac{1}{0.77c}+\frac{1}{c}\right)=\frac{\Delta x}{c}\left(\frac{1}{0.77}+1\right)\\ \Delta t_{total}=\frac{42\times c}{c}\left(2.30\right)=97ly\end{array}$
Years after launch from Earth will it be when the signal arrives at Earth as soon as the spaceship arrives at the planet = 97 ly

Conclusion:

Years after launch from Earth will it be when the signal arrives at Earth as soon as the spaceship arrives at the planet = 97 ly