Question

An alien spaceship traveling at 0.600c toward the Earth launches a landing craft with an advance guard of purchasing agents and physics teachers. The lander travels in the same direction with a speed of 0.800c relative to the mother ship. As observed on the Earth, the spaceship is 0.200 ly from the Earth when the lander is launched.(a) What speed do the Earth observers measure for the approaching lander?(b) What is the distance to the Earth at the time of lander launch, as observed by the aliens?

Answer 1

Step-by-step explanation:

Let the velocity of lander with respect to earth be v .

In relativistic mechanism the expression for relative velocity is

v_r =
`(v -u)/(1-(uv)/(c^2))`

Given u = .6c , v_r = .8 c

Substituting the values

.8c =
`(v -0.6c)/(1-(.6c* v)/(c^2) )`

.8c-.48v = v - .6c

v = .946c

b )

Distance in terms of time = .2 ly

In relativistic mechanism , expression for relativistic time is given by the following relation

t =
`\frac{t_0}{\sqrt{1-(v^2)/(c^2) }}`

Substituting v = .946c

t₀ = .2

t =
`\frac{.2}{\sqrt{1-(0.946* .946c^2)/(c^2)}}`

.2 / √.1050

= .62 ly

distance to the Earth at the time of lander launch, as observed by the aliens will be .62 ly.