## Thursday, November 1, 2018

### Solutions To Special Relativity Problems (4)

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.

(2) Let D t' = (t2' - t1')  = 9  s

t2' = (t2 - 3c/4)(1 - 9/16))

t1' = (t1 - 3c/4)(1 - 9/16)

or:

t2' = (t2 - 3c/4) (7/16)

t1' = (t1 - 3c/4) (7/16)

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