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 1.10 y has elapsed, as measured on the ship. Clocks in the earth's reference frame show that the voyage lasted 3.29 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

(a) Relationship between the time dilation factor and the speed is expressed as -  

γ = 1/√(1−(v/c)²)

where, time dilation γ = t' / t (t' = 3.29 y and t = 1.10 y)

So -

t/t′ = 1/√(1−(v/c)²)

Solve for v:

v = c√(1−(t/t')²)

Therefore, the distance of the colony from the earth reference frame -  

d = vt = ct√(1−(t/t')²)

Put the values -

d = (3.0 x 10^8 x 1.10 x 365 x 24 x 3600)*√(1−(1.10/3.29)²)

= (1.04 x 10^16) x 0.94 = 9.78 x 10^15 m

(b) And the distance of the colony in the ship frame is therefore -  

d′ = vt′ = ct′√(1−(t′/t)²)

= (3.0 x 10^8 x 3.29 x 365 x 24 x 3600)*√(1−(1.10/3.29)²)

= (3.11 x 10^16) x 0.94 = 2.92 x 10^16 m (Answer)