For this experiment we have four spacecrafts S1, S2, S3 & S4, each one of them has a watch capable of measuring time up to millisecond accuracy which displays time in 24hrs format. (Let’s say S1 has W1, S2 has W2, S3 has W3 & S4 has W4). S1 travels at 20% of Light Speed (C), S2 at 40%, S3 at 60% and S4 at 80%.
For this experiment they have to go to Planet A & return to Earth, but when they return they have to exchange their watches when they cross each other (i.e. When S4 returns from Planet A it will cross S3 at that time S4 will give its watch W4 to S3 & will take watch W3 from S3 & so on)
All of them Starts at :: hrs & when S1 returns to Earth, the time on Earth is :: (i.e. one full day.)
Question:
1) Find out distance between Earth & Planet A
2) Find out time on W1,W2,W3 & W4 when they returned to earth.
2
✅ Answers
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EDIT: Due to carelessly missing a number early on and copy-pasting mistakes, I retracted everything I had before. Sigh.
Assuming no acceleration and that they instantaneously “rebound” from Planet A once they reach it (no time for sight-seeing–what a waste) and swap watches instantly…
Total time for S1, one day (measured from Earth):
60²·24 s = 864 s
So the distance (D) is (as measured from the ‘unmoving’ Earth):
2·D = r·t
D = r·t/2 = (0.20c)·(864s)/2 = 8640 c·s
(Answer 1) 8640 c·s
— — — — — — — — — — — — — — — —
To just do W1 requires looking at just S1 and S4:
W1: S1→S4→Earth
– – – –
The time (from Earth) for S4 to get to the other side (and S1 to get 1/4 the way, etc):
t = d/r = ( 8640 c·s ) / ( 0.80 c ) = 108 s
As S4 reaches the planet, S1 is at 1/4 (0.20/0.80) the way there.
Solving for their intersection gives a time (from Earth) of:
D – 0.80c·t = .2/.8 D + 0.20c·t
t = ( 3D/4 )/c = 6480s
W4 undergoes 108s + 6480s of time (from Earth’s perspective) before being switched.
– – – –
Distance (from Earth) when S4 and S1 to meet:
d = r·t = ( 0.20c )·( 108s + 6480s ) = 3456 c·s
Remaining time (from Earth) for S4 to travel:
t = d/r = ( 3456 c·s )/( 0.80 c ) = 4320s
– – – –
Proper time felt by W1 before swapping:
τ = t/γ = ( 108s + 6480s )·√( 1 – 0.20² ) = 6912·√(6) s ≈ 16930.8731 s
Proper time felt by W1 in S4’s possession:
τ = t/γ = ( 4320s )·√( 1 – 0.80² ) = 2592s
– – – –
Time experienced by W1:
6912·√(6) s + 2592s ≈ 19522.8731 s
≈ 5 hours, 25 minutes, 22.873 seconds
(Answer 2A) W1 would read: 05:25:22.873
— — — — — — — — — — — — — — — —
The rest of the watch-times require more steps. An (A) indicates the ship “rebounds” from planet A with the watch.
W4: S4(A)→S3(A)→S2(A)→S1(A)→Earth
W3: S3→S4→S2→S3→S1→S2→Earth
W2: S2→S4→S1→S3→Earth
I have bumbled over the numbers already, so I probably shouldn’t try the rest.
1. D= v* t/2 = .2c * day/2 = .1 lightdays
2.
Find meetings
r = D (1- (vf-vs)/(vf+vs))
vf = vfast,
vs = vslow
S4-S3 = R43= D(1- .2/1.4)
S4-S2 = R42= D(1-.4/1.2)
S4 – S1 = R41= D (1-.6/1)
S3-S2 = R32 = D (1- .2/1)
S3-S1 = R31= D (1-.4/.8)
S2-S1 = R12 = D (1 – .2/.6)
Find Time dialations
T4 = â(1 – .8^2) = .6
T3 = â (1-.6^2) = .8
T2 = â (1- .4^2) = 0.9165151
T1 = â (1 – .2^2) = 0.9797959
Watch W1, Given to S4 at R41= D (1-.6/1)
TD1 = (1-T1) *R41/ V1 + (1-T4)*R41/V4 = .024days = 34min 37.127s
Watch 1 is returned to earth with Ship 4 at 06:.
It will read 05:25:22.873
W2 ==> given to S4 at R42 Given to S1 at R41, Given to S3 at R31, goes home
T = (1-T2) R42/V2 + (1-T4) (R42-R41)/V4 + (1-T1) (R31 – R41)/V1 + (1-T3)(R31)/V3
TD2= .0449 days = 1 hour, 4min, 41 sec
Watch 2 will return to earth with ship 3 at 08::
It will read 06:55:19
W3==> given to S4 @R43, Given to S2 @R42, Given to S3 at R32, Given to S1@ R31, Given to S2 @R21
T= (1-T3)R43/V4 + (1-T4) (R43-R42)/V4+ (1-T2)(R32-R42)/V2+ (1-T3)(R31- R32) + (1-T1)(R31-R21)/V1 + (1-T2)R21/V2
TD3= .06648days = 1 hour, 35 min, 44 sec
Watch 3 will return to earth at 12::
It will read 10:24:16
W4==> Gets to planet then given to S3 @R43, Rounds Planet given to S2 at R32, Rounds Planet given to S1 at R21, Rounds planet then home
T = (1-T4) (2D-R43)/V4 + (1-T3)(2D -R43 -R32)/V3 + (1-T2)(2D-R32-R21)/V2 + (1-T1)(2D – R21)
TD4 = .0932 days = 2 hours, 14 min, 10 secs
Watch 4 will return at 24::
It will read 21:45:50
*********************
gravitative is correct — I just did this to check his work — give him his well deserved ten points.