1)
Let
t = 106 yrs be the quasar's lifetime in its own rest frame. Then the
total Earth-based time that its radiation will be received is:
t' = t [1 - v2/c2] ½
where v = 0.8c
so:
t' = (106 yrs.) [1 - (0.8c)2/c2]½
t' = (106 yrs) [1 - 0.64]½ = 106 yrs (0.36) ½
t' = 0.6 (106 yrs) = 6 x 105 yrs.
t' = t [1 - v2/c2] ½
where v = 0.8c
so:
t' = (106 yrs.) [1 - (0.8c)2/c2]½
t' = (106 yrs) [1 - 0.64]½ = 106 yrs (0.36) ½
t' = 0.6 (106 yrs) = 6 x 105 yrs.
(2) Let D t' = (t2' - t1') = 9 s
t2' = (t2 - 3c/4)(1 - 9/16))-½
t1' = (t1 - 3c/4)(1 -
9/16)-½ t2' = (t2 - 3c/4)(1 - 9/16))-½
or:
t2' = (t2 - 3c/4) (7/16)-½
t1' = (t1 - 3c/4) (7/16)-½
whence:
(t2' - t1') = [(t2 - 3c/4) - (t1 - 3c/4)]/0.661
or: 9 sec = (t2 - t1)/0.661 and
D t = 5.95 sec
(3) The speed of the astronaut is given by:
v = (2 gr)½
and r = 7 x 106 m, g = 10 ms-2
v = 11, 832 m/s = 3.94 x 10-5 (c)
In the astronaut's frame, 1 week - 86,400 s x (7) = 604, 800 s
For the twin on Earth:
t = t'/ [1 - t2/c2] ½
t = 604, 800.0002 s
and so the time difference = 0.0002s, or the twin in orbit will be (t' - t) or 0.0002 s younger on his return.
(4) The speed of A observed in B = 0.7c, exactly equal to the speed of B observed in A, by principle of relative velocities.
To find the speed of C observed in B, we use relativistic addition of velocities or:
u = (u' + v)/ 1 + u'v/c2
u = (0.7c + 0.7c)/ [1 + (0.7c)(0.7c)/c2]
u = (1.4c)/ 1 + 0.49 = 1.4c/1.49 = 0.94c
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