Question

The starship Enterprise leaves a distant earth colony, which is nearly at rest with respect to the earth. The ship travels at nearly constant velocity toward earth. When the Enterprise reaches earth 1.03 y has elapsed, as measured on the ship. Clocks in the earth's reference frame show that the voyage lasted 3.26 y.

How far from earth is the colony, as measured in the reference frame of the earth?


(b) How far from earth is the colony, as measured in the reference frame of the ship?

Answer 1

t = 3.26 years, and

t′ = 1.03 years

The time dilation factor depends on the speed:
γ = 1/√(1−(v/c)²)
This is also the ratio of the two time intervals (t/t′ or 3.26/1.03):
t/t′ = 1/√(1−(v/c)²)
Solve for v:
v = c√(1−(t′/t)²)

(v/c)² = 1- (1.03/3.26)² =1-0.099824=0.900175

v/c= 0.948

v= 0.948 c
The distance in the earth frame is therefore:
d = vt = ct√(1−(t′/t)²)

= 0.948 * 3.26

=3.0927 Light years
And the distance in the ship frame is therefore:
d′ = vt′ = ct′√(1−(t′/t)²)

= 0.948 * 1.03

=0.97644 Light years