Question

Part A

Astronaut Spud Nick is space-travelling from planet X to planet Y at a speed of 0.84c. When he is precisely halfway between the planets, a distance of 1 light-hour from each in the planets' frame, nuclear devices are detonated. The explosions are simultaneous in the frame of the planets. What is the difference in time of arrival of the flashes from the explosions as observed by Spud?

Astronaut Spud Nick is space-travelling from planet X to planet Y at a speed of . When he is precisely halfway between the planets, a distance of 1 light-hour from each in the planets' frame, nuclear devices are detonated. The explosions are simultaneous in the frame of the planets. What is the difference in time of arrival of the flashes from the explosions as observed by Spud?

190 min

93 min

0 min

46 min

Answer 1

difference in time of arrival of the flashes from the explosion will be:

dt = time taken by flash of planet X to arrive - time taken by flash of planet Y to arrive

Since Spud is going towards planet Y, So speed of flash will be = c + u

And Spud is going away from planet X, So speed of flash will be = c - u

Now Since spud is at the halfway between planets, So suppose total distance between X and Y = 2L, then flash has to travel L distance on both side, So

dt = L/(c - u) - L/(c + u)

dt = (L*(c + u) - L(c - u))/(c^2 - u^2)

dt = [Lc + Lu - Lc + Lu]/(c^2 - u^2)

dt = 2Lu/(c^2 - u^2)

Now since distance is measured w.r.t planet's frame, So according to length contraction, actual distance will be:

L = L0/ = L0*sqrt (1 - u^2/c^2)

= Lorentz factor

So,

dt = 2*L0*sqrt (1 - c^2/u^2)*u/(c^2 - u^2)

Now given that

L0 = 1 light-hr = c*1 hr = c*60 min

u = speed of spud = 0.84*c

So,

dt = 2*(1 li-hr)*sqrt (1 - (0.84c)^2/c^2)*0.84c/(c^2 - (0.84c)^2)

dt = 2*(c*60 min)*sqrt (1 - 0.84^2)*0.84*c/(1 - 0.84^2)*c^2

dt = 2*60*sqrt (1 - 0.84^2)*0.84/(1 - 0.84^2)

dt = 185.77 min = 190 min

Correct option is A.

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