Solutions To Special Relativity Problems (3)

Solutions to the problems from the Oct. 17 instalment:

1) Let t _A be the time on the astronauts' clock and  t _E be the time recorded on an Earth-based clock.

Then, we have t _E = 4.4 yrs.

And:

t _A  = t _E  [1 - v²/c²]^½

t _A = (4.4 yrs.) [1 - (0.95c)²/c²]^½ = 1.37 yrs.


(b)  Since we know: t _A  = 1.37 yrs.

then the distance D _A = (0.95c) (1.37 yrs) = 1.31 Ly


2) In this case, v = c = 3 x 10⁸ m/s

d = v/H = (3 x 10⁸ m/s)/ (2.26 x 10^-18 s^-1)

d = 1.32 x 10²⁶ m

Converting to light years:

d = (1.32 x 10²⁶ m)/ (9.5 x 10¹⁵ m /Ly) = 1.4 x 10¹⁰ Ly


b) Would it be observable from Earth?

Given that modern telescopes can penetrate to about 1.8 x 10¹⁰ Ly, the galaxy should easily be observable to the Hubble but might be more problematical for land-based scopes.

3) We know the recessional velocity v = 6 x 10⁴ km/s

By Hubble's law: v = Hd so the distance d = v/H

Then, attending to the proper units for v, H:

d = (6 x 10⁷ m/s)/(2.26 x 10^-18 s^-1)= 2.6 x 10²⁵ m

and d = (2.6 x 10²⁵ m)/(9.5 x 10¹⁵ m /Ly) = 2.8 x 10⁹ Ly

(b) z = v/c = (6 x 10⁷ m/s)/(3 x 10⁸ m/s) = 0.2

(c) T = d/v = (2.6 x 10²⁵ m)/(6 x 10⁷ m/s)

= 4.3 x 10¹⁷ s

But 1 yr. = 3.15 x 10⁷ s

so T = (4.3 x 10¹⁷ s)/(3.15 x 10⁷ s/ yr)

T = 1.36 x 10¹⁰ years, or 13.6 billion years


4) Let  l_o  be the normal wavelength = 1200 Å  and l be the red-shifted value.

We know v = 0.8c so we must use the modified Doppler version, viz.

l/l_o = (1 - v/c) ^½ /(1 + v/c) ^½

l/l_o = (1 + 0.8) ^½/ (1 - 0.8) ^½ = (1.8/0.2) ^½

l/l_o = Ö(9) = 3

then:

l = 3 l_o = 3 (1200 Å) = 3600 Å


(b) The red shift of the quasar is found from:

1 + z = (1 + v/c) ^½ /(1 - v/c) ^½

1 + z = (1.8/0.2) ^½ = Ö(9)  = 3

Then: z = 3 - 1 = 2

c)  The corrected velocity, v =  c [(z ² + 2z) / (z ² + 2z + 2)]

Then:  v =  c[4 + 4]/ [4 + 4  + 2]  =   8c/10  = 4c/5 = 0.8c