Question

An alien spaceship traveling at 0.610c toward the Earth launches a landing craft. The landing craft travels in the same direction with a speed of 0.710c relative to the mother ship (as measured by aliens on the mother ship). As measured on the Earth, the spaceship is 0.280 ly from the Earth when the landing craft is launched.

(a) What speed do the Earth-based observers measure for the approaching landing craft?
c

(b) What is the distance to the Earth at the moment of the landing craft's launch as measured by the aliens?
ly

(c) What travel time is required for the landing craft to reach the Earth as measured by the aliens on the mother ship?
yr

(d*) What travel time is required for the landing craft to reach the Earth as measured by Earth-based observers?
yr

(e*) What travel time is required for the landing craft to reach the Earth as measured by those on the landing craft?
yr

Answer 1

a) Take the spaceship as the primed frame, moving towards the right at v = +0.610c.

Then u'x = +0.710c and

ux = u'x + v / (1+(u'x*v)/c^2)
= 0.710c + 0.610c / 1+(0.710*0.610)
= 0.921c

b) L = Lp/y
L = (0.280 ly) *sqrt(1-(0.710)^2)
L = 0.1972 ly

c)The aliens observe the 0.1972-ly distance closing because the probe nibbles into it from one end at 0.610c and the earth reduces it at the other end at 0.710c

Thus, time = 0.1972ly / 0.710c + 0.610c
= 0.149 yr

d) From the earth based observer, the landing craft is traveling at a velocity u found in part (a). The distance between the landing craft and the earth is given as 0.28 ly when the landing craft is released. Therefore, the time takes the landing craft to meet the earth calculated in earth frame is

t = 0.28/0.921
= 0.304 yr

e) For those on the landing craft, the earth's velocity is v=−u. Here u is the velocity found in part b. Now we need to find the distance between the earth and the landing craft as observed by the landing craft.

We can use this velocity to calculate γ for the observer on the landing craft to observe earth events. Then we can calculate the distance with Δx′′=Δx/γ. Then the time observed on the landing craft is

Δt′′ = Δx′′/v

= Δx/γ*v
= 0.280ly / (0.710*0.610)