Exercise 5 (Stars and the universe II) Suggested answers
1. (a) d = 1/p = 1/0.545 = 1.83 pc
   0 v
(b) By


0
0
c

v
 106.8  103
  
 656.28 = - 0.234 nm
0 c
3  108
Negative sign means that the observed wavelength is smaller than the wavelength observed if
the object is at rest. This corresponds to blue shift. Thus the star must be approaching.

(c) Speed of the star = 902  106.82  140 km s-1
(d) To find the change in angular position, transverse direction should be considered.
Observed change in position s = vt = 90 km s-1 x 365 x 24 x 3600 = 2.838 x 1012 m
Required change in angular position  = s/r
= 2.838 x 1012 / (1.83 x 3.09 x 1016) = 5.02 x 10-5 rad
2.
(a) (i) Doppler effect
(ii) The universe is expanding.
(b)  = 650  590 = 60 nm
By
λ v r

,
λ
c
 λ   60 
8
c = 
  3  10 = 3.05  107 m s1
 λ   590 
velocity of the galaxy relative to the Earth= 
(c) For diffraction grating,
d sin  = n
sin  =
nλ
= 2  (590  109)  (4.5  105)
d
 = 32.1
3. (a) Applying



vr
  
, we have v r  
c .
c
  
star A: v r   420.8  410.2   3  10 8   7.752  10 6  7.75  10 6 m s 1
410.2


Star A is receding from the Earth at a radial speed of 7.75 × 106 m s−1.
star B: v r   395.4  410.2   3  10 8   1.082  10 7  1.08  10 7 m s 1

410.2

Star B is approaching the Earth at a radial speed of 1.08 × 107 m s−1.
star C: v r   410.2  410.2   3  10 8   0

410.2

Star C has zero radial velocity and thus is neither approaching nor receding from the Earth.
(b) No. We need both the transverse and radial velocities of a star in order to calculate its
velocity, but by analyzing its spectrum, we can only obtain the latter.
4. (a) S rises and sets everyday. When it is below horizon, it cannot be observed.
(Or in day time, too much stray light from other stars would seriously affect the result).
(b)
There is relative motion between the source (i.e. the star) and the observer on
the Earth. This gives rise to the Doppler effect.
(c)
(i) From the plot, the peak values of the graph correspond to the speeds of S
when it is moving directly towards or away from the Earth.
Orbital speed of S
(ii) Orbital radius of S
(d)
2
By Kepler’s third law of planetary motion, T 2  4 a 3
GM
Orbital radius of P
MC
1-5 A D B C B
6-7 B C
Explanations to selected mc
1. Statement 2, the rotational velocities in the outer part of the galaxy are greater than expected.
Refer to the rotation curve.
5. If the star is approaching the Earth, blue shift occurs, i.e. the wavelength shifts to shorter
wavelengths. Since
v λ

, the longer the wavelength, the larger the shift in wavelength since v/c is
c λ0
a constant.
6.
GMm mv 2
GM

v
2
r
r
r
where M is the mass of the central massive part