Question 39.3: A Voyage to Sirius An astronaut takes a trip to Sirius, whic......
A Voyage to Sirius
An astronaut takes a trip to Sirius, which is located a distance of 8 light-years from the Earth. The astronaut measures the time of the one-way journey to be 6 years. If the spaceship moves at a constant speed of 0.8 c, how can the 8 -ly distance be reconciled with the 6 -year trip time measured by the astronaut?
Conceptualize An observer on the Earth measures light to require 8 years to travel between Sirius and the Earth. The astronaut measures a time interval for his travel of only 6 years. Is the astronaut traveling faster than light?
Categorize Because the astronaut is measuring a length of space between the Earth and Sirius that is in motion with respect to her, we categorize this example as a length contraction problem. We also model the astronaut as a particle moving with constant velocity.
Analyze The distance of 8 ly represents the proper length from the Earth to Sirius measured by an observer on the Earth seeing both objects nearly at rest.
Calculate the contracted length measured by the astronaut using Equation 39.9:
L=\frac{L_p}{\gamma}=L_p \sqrt{1-\frac{v^{2}}{c^{2}}} (39.9)
L=\frac{8 \text { ly }}{\gamma}=(8 \text { ly }) \sqrt{1-\frac{v^{2}}{c^{2}}}=(8 \text { ly }) \sqrt{1-\frac{(0.8 c)^{2}}{c^{2}}}=5 \text { ly }
Use the particle under constant velocity model to find the travel time measured on the astronaut’s clock:
\Delta t=\frac{L}{v}=\frac{5 \text { ly }}{0.8 c}=\frac{5 \text { ly }}{0.8(1 \mathrm{ly} / \mathrm{yr})}=6 \mathrm{yr}
Finalize Notice that we have used the value for the speed of light as c=1 \mathrm{ly} / \mathrm{yr}. The trip takes a time interval shorter than 8 years for the astronaut because, to her, the distance between the Earth and Sirius is measured to be shorter.