Copernicus used his model of the solar system to obtain the distances to the
inferior planets in units of the Earth's distance from the Sun (also known as
an astronomical unit or AU),
as shown in the lefthand panel of
Figure 30. By carefully measuring
the elongation
(angle SEP, i.e. the angular distance between
the Sun and the planet in the sky) on a series of nights around
greatest elongation,
it is possible to obtain a value for the greatest
elongation, SEP_{max}. At this time, the angle between
the line joining the planet to the Sun and the line joining the planet
to the Earth is 90°. Hence
SP / SE = sin (SEP_{max}),
where the value of angle SEP_{max} is known. The quantity
SP / SE is therefore the distance of the planet
from the Sun in units of the Earth's distance from the Sun, i.e.
in AU. There is an
example problem which shows how
Copernicus deduced the distances to Mercury and Venus in this way.
figure 30:
Measurement of the distance of an inferior planet (lefthand
panel) and a superior planet (righthand panel).
If the planet is a superior
one, the situation is slightly more
complicated. Kepler was the first to realise it could be solved
if the planet's synodic period, S, is known.
If the planet's sidereal
period is P days and the Earth's
sidereal period is E days, then the planet moves at a rate of
360°/P degrees per day in its orbit and the Earth moves
at a rate of 360°/E, as viewed from the Sun.
Let the planet be in opposition
at position 1, as shown in the
righthand panel of Figure 30. After
t days have elapsed, the planet will have moved through
an angle 360°t/P, as viewed from the Sun, to
position 2. In this time, the Earth will have moved through an angle
360°t/E from position 1 to position 2. The
angle between the planet and the Earth at position 2, as viewed from
the Sun,
,
is hence given by
= (360°t/E) 
(360°t/P) =
360°t(1/E  1/P).
Recalling that, for a superior planet,
1/S = 1/E  1/P
we obtain
= 360°t/S.
Since t and S are both known,
can be evaluated.
The elongation of the planet at position 1 is 180°, i.e. it is in
opposition. At position 2, the elongation of the planet is reduced to

this value can be measured in the sky. Hence the
angle
can be found from the relation
= 180° 

.
Using the sine formula of plane trigonometry, we have
sin
/ SP = sin / SE
or
SP / SE = sin
/ sin ,
giving the distance of the planet from Sun in units of the Earth's distance
from the Sun. Proceeding in this way, it is thus possible to construct an
accurate model of the solar system in terms of astronomical units without
knowing the scale absolutely, as shown in the
example problems.
Such a model was first obtained by Kepler.
It was a far more difficult problem, only satisfactorily solved with the
advent of radar this century, to measure the scale (i.e. the distances
of the planets in kilometres). The radar method of
distance determination involves timing the interval between
transmission of a radar pulse by a powerful radio telescope on Earth and
the reception of its echo from the planet. Thus if EP, c
and t are the Earthplanet distance, the velocity of radio waves and
the time interval, respectively,
EP = ½ct.
Various corrections have to be made to derive the true distance.
For example, the distance actually measured is the distance
from the radio telescope to the surface of the planet, and not the distance
between the Earth's centre and the planet's centre. The effect of the change
in the speed of the radar pulse when passing through the Earth's
atmosphere also has to be taken into account. With a knowledge of the
distance in kilometres from the Earth of just one planet in the solar system
(e.g. Venus), it is possible to determine the distance to all of the rest
(and the size of the astronomical unit) in kilometres.