A small asteroid is 1 light year (in the earth’s frame) from earth and is travel
ID: 1884935 • Letter: A
Question
A small asteroid is 1 light year (in the earth’s frame) from earth and is traveling towards the earth at 0.85c. At this time, scientists launch a missile towards the rocket at a speed of 0.75c. The missile is equipped with a timer that should cause the missile to explode when it reaches the asteroid.
(a) Taking the direction towards the asteroid to be the x-axis, with the earth at x = 0, find the trajectory x(t) of the asteroid in the frame of the earth.
(b) What is the trajectory x(t) of the asteroid in the missile’s frame?
(c) What should the timer on the missile be set to for the earth to be saved? In other words,
at what value of t does the the asteroid meet the missile at x = 0?
(d) What is the velocity of the missile in the asteroid’s reference frame?
Please answer all parts, thank you so much :)
Explanation / Answer
given
do = 1 LY ( in earths frame)
speed of asteroid towards earth wrt earth = vo = 0.85c
speed of rocket topwards the asteroid = vr = 0.75c
a. in the frame of earth
trajectory of asteroid x(t) = do - vo*t = 1 - 0.85*t LY ( where x is in LY and t is in years)
b. in missile's frame
relatiove speed of asteroid = u' = (-vo - vr)/(1 + vovr/c^2)
u' = -(1.6c)/(1 + 0.75*0.85) = 0.6432 c ( -ve sign for approaching asteroid)
hence
x'(t) = do*sqrt(1 - u'^2/c^2) - u't = 0.7656982 - 0.6432t' ( x' in LY and t' in years)
c. in missile's frame
when x'(t) = 0
t' = 1.19045 years when the miossile meets asteroid and should explode
d. velocity of missile in asteroid's refernce frame is the same as velocity of asteroid in missiles reference frame
u' = -0.6432c