12 
SURVEY OF THE PROBLEM 
for this kind of observation; but in 1920 Michelson’s interferometer 
method of observation used in connection with the 100-inch reflector at 
Mount Wilson achieved its first striking success by measurements of the 
separation and position angle of Capella. These observations, begun by 
Anderson and continued by Merrill, have yielded a good visual orbit. It 
should be understood that the spectral lines of both components can be 
observed in only a small proportion of the spectroscopic binaries; and it 
is extremely rare for visual and spectroscopic orbits to be determined for 
the same star. Thus our knowledge of the system of Capella is unusually 
complete. 
The line-of-sight velocity of the brighter component varies between 
+ 4 and + 56 km. per sec., a range of 52; the fainter varies between + 63 
and — 3 km. per sec., a range of 66. Neglecting eccentricity the mean velocity 
+ 30 (necessarily the same for both components) is the motion of recession 
of the centre of mass of the system from the sun. The masses must be in 
the inverse ratio of the respective velocity ranges so that the brighter 
component has 66/52 or 1-26 times the mass of the fainter. 
The eccentricity is found to be very small (-0086), so that we may treat 
the orbits as circular and the orbital speeds as constant. If the line of 
sight were in the plane of the orbit the orbital speed would be equal to 
the half-range of velocity, 26 and 33 km. per sec. respectively; to allow 
for projection these values must be multiplied by cosec i, where i is the 
inclination of the orbit plane to the plane of the sky. The circumference 
of the orbit is at once found by multiplying the orbital speed by the period. 
In this way we find that if a x and a 2 are the radii (or semiaxes) of the two 
orbits 
a x sin i = 36,800,000 km., a 2 sin i = 46,400,000 km. 
We now turn to the visual observations. These treated in the usual 
way determine the inclination i = 41°-1. Hence 
a x = 56,000,000 km. 
a 2 = 70,600,000 km. 
a x + a 2 = 126,600,000 km. = 0-847 astronomical units. 
By Kepler’s third law the mass of the system is 
M x + M 2 = (a x + a 2 ) 3 /P 2 , 
the unit of mass being the sun’s mass, a x + a 2 being in astronomical units 
(i.e. in terms of the earth’s distance from the sun), and the period P in 
years. We thus have 
M\ + M 2 = (-847) 3 /(-285) 2 = 7-5, 
and dividing this between the two components in the ratio 1-26 : 1 already 
found 
M x = 4-18, M 2 = 3-32.