example problems |
1. |
The synodic
period of Venus was measured to be 583.92 days. If the
sidereal
period of the Earth is 365.26 days, calculate the sidereal period
of Venus. Venus is an inferior planet, so we must use the relation 1/S = 1/P - 1/E. Rearranging and substituting gives 1/P = 1/583.92 + 1/365.26 = 0.00171 + 0.00274 = 0.00445 and hence P = 224.7 days. |
2. |
If the greatest elongations of Mercury and
Venus are 28° and 48° respectively, calculate the distances of
these planets from the Sun in
AU. We have seen from Figure 28 that Copernicus derived a relationship between the greatest elongation of an inferior planet, SEP_{max}, and its distance from the Sun in AU, SP / SE: SP / SE = sin (SEP_{max}). Hence, the distance of Mercury from the Sun is simply sin (28°) = 0.47 AU and the distance of Venus from the Sun is sin (48°) = 0.74 AU. |
3. |
The synodic period of Mars was measured
to be 779.9 days. 105.9 days after
opposition, the elongation of Mars
was measured to be 90°. Calculate the distance of Mars from
the Sun in AU. We have seen from Figure 28 that Kepler derived the following relationship between the angle between the Earth and a superior planet when viewed from the Sun t days after opposition, , and the synodic period of the planet, S: = 360°t/S. Hence for Mars we obtain = 360°t/S = 360° x 105.9 / 779.9 = 48.9°. The angle between the Sun and the Earth as viewed from the superior planet t days after opposition, , is related to the elongation, , and through the relation: = 180° - - . Hence for Mars we obtain = 180° - 90° - 48.9° = 41.1° The distance of Mars to the Sun in units of the Earth-Sun distance is then equal to sin / sin = sin 90° / sin 41.1° = 1.52 AU. |
4. |
In problem 1, we calculated
the sidereal period of Venus from its synodic period, and obtained a
value of 224.7 days. Given this, calculate the mean distance of Venus
from the Sun in AU. We must use Kepler's third law to solve this problem: a^{3} = T^{2}, where a is the mean distance of Venus from the Sun in AU and T is the sidereal period of Venus in years. Hence a = T^{2/3} = (224.7/365.26)^{2/3} = 0.72 AU. |