An alien spaceship traveling at 0.590c toward the Earth launches a landing craft

Question

An alien spaceship traveling at 0.590c toward the Earth launches a landing craft. The landing craft travels in the same direction with a speed of 0.740c relative to the mother ship (as measured by aliens on the mother ship). As measured on the Earth, the spaceship is 0.270 ly from the Earth when the landing craft is launched. (a) What speed do the Earth-based observers measure for the approaching landing craft? 938 (b) What is the distance to the Earth at the moment of the landing craft's launch as measured by the aliens (mother ship)? 217998876 ly (c) What travel time is required for the landing craft to reach the Earth as measured by the aliens on the mother ship? 1639 (d*) What travel time is required for the landing craft to reach the Earth as measured by Earth-based observers? 2878 (e*) What travel time is required for the landing craft to reach the Earth as measured by those on the landing craft? Your response differs from the correct answer by more than 100%, yr yr

Explanation / Answer

To find the landing craft's velocity relative to the earth, using the Lorentz transformation on velocity:

V = (v + u)/(1 + vu/c^2)

V = (0.590 + 0.740)/(1 + 0.590*0.740)

V = 0.9257c = 0.9257 LY/Y

time measured from earth = 0.270/0.9257 = 0.292 year

Now

time dilation = y = 1/sqrt (1 - v^2/c^2)

y = 1/sqrt (1 - 0.9257^2) = 2.644

So required travel time

t'' = t/y = 0.292/2.644 = 0.1104 yr

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