Question

A starship voyages to a distant planet $10ly$away. The explorers stay 1 year, return at the same speed, and arrive back on earth $26years$, as measured on earth after they left. Assume that the time needed to accelerate and decelerate is negligible.

a. What is the speed of the starship?

b. How much time has elapsed on the astronauts’ chronometers?

Short answer

a. the speed of the starship is $v=0.8C$.

b. $t=16years$time has elapsed on the astronauts’ chronometers.

Step-by-step solution

Part (a)Step 1: Given Information

We need to find the speed of the starship .

Part (a)step 2: Simplify

The total time taken by the spaceship in travel is$26-1=25years$.

The total distance covered by the spaceship is $2\times 10=20lightyears.$

So the velocity of the ship is $\frac{20ly}{25ly}=0.8C$

Note:$C=1\frac{ly}{ly}$

Part (b) step 2: Given Information 

We need to find how much time has elapsed on the astronauts’ chronometers .

Part (b) step:2 Simplify

For one year the spaceship was stationary (in earth's frame of reference ) the time will be the same for that one year.

During travel, the time in the spaceship's chronometer is :

$\delta t\text{'}=\delta t\left(\gamma \right)=25\times \sqrt{1-{\left(\frac{0.8C}{C}\right)}^2}=15y$

So the total time that will be shown in spaceships chronometers is $15+1=16years$.