Solution

The separation simultaneous in the mothership frame at launch is the length contraction of the Earth-frame simultaneous separation:

D^{'} = \frac{2 ly}{γ _{m}} = 1.60 ly, γ_{m} = \frac{1}{1 - 0. 6 ^{2}} = 1.25.

(1)

In that frame Earth approaches the launch position at

0.6 c

while the lander travels toward Earth at

0.8 c

. Their coordinate separation closes at

1.4 c

, so

t_{arrival}^{'} = \frac{1.60 ly}{1.4 c} = \frac{8}{7} yr = 1.14 yr .

(2)

A closing rate between two objects can exceed

c

; it is not either object's speed in the other's rest frame. Simultaneity matters: simply transforming the Earth event at Earth-frame launch time would not give the simultaneous mothership-frame separation.

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / Section B / 7 / b / Solution

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