Astronomy Principles and Practice Fourth Edition.pdf

164 Geocentric planetary phenomena
Expanding the right-hand side and rearranging, we obtain
tan φ =
a sin θ
b − a cos θ . (12.19)
Equation (12.19) gives φ when a, b and θ are known. The angle θ is readily calculated from the relation
θ = 360 × t S
(12.20)
where S is the planet’s synodic period and t is the time that has elapsed since inferior conjunction
(inferior planet) or opposition (superior planet).
Having found φ, equation (12.18) will then give the phase.
12.11 Improvement of accuracy
The expressions derived in previous sections in this chapter were obtained under the assumption that
the planets move about the Sun in circular, coplanar orbits. In fact, the actual planetary orbits are
ellipses of low eccentricity in planes inclined a few degrees to each other. The derivation of more
accurate expressions taking the eccentricities and inclinations of the orbits into account is beyond the
scope of this text and, in any case, gives values that are only a few per cent different from those given
by the simple expressions obtained here.
Problems—Chapter 12
Note: Take the length of the sidereal year to be 365·25 days. Assume all orbits are circular and coplanar, unless
otherwise stated.
1. A planet’s elongation is measured as 125 ◦ . Is it an inferior or superior planet
2. The sidereal period of Mercury is 88 days. What is its synodic period
3. What is the maximum possible elongation of Venus, given that its distance from the Sun is 0·723 AU
4. The synodic period of Jupiter is 398·9 days. What is its sidereal period
5. The heliocentric distance of Venus is 0·723 AU. On a certain day, the planet’s phase was
12 1 ; calculate its
elongation.
6. Calculate the ratio of the Earth’s orbital speed to that of the planet Neptune, given that the distance of
Neptune from the Sun is 30·06 AU.
7. The synodic period of Mars is 779·9 days and its heliocentric distance is 1·524 AU. Find its phase 85 days
after opposition.
8. An asteroid is orbiting the Sun in a circular orbit of radius 4 AU. Calculate the ratio of its angular diameters
at opposition and quadrature.
9. The planet Mars reaches a stationary point 36·5 days after opposition, its elongation being measured to be
136 ◦ 12 ′ . Given that the planet’s orbital period is 687·0 days, calculate the distance of Mars from the Earth
in astronomical units at the stationary point, also the planet’s phase.
10. During a synodic period of Venus, the elongation η at time t 1 is the same as the elongation at time t 2 (t 2 > t 1 ),
the planet being on the same side of the Sun at both times. If the phase at t 2 is three times the phase at t 1 ,
find
(i) the value of η,
(ii) the interval (t 2 − t 1 ) in days,
given that the heliocentric distance of Venus is 0·723 AU and its synodic period 583·9days.
11. Calculate the length of time during which Jupiter has retrograde motion in each synodic period, given that
Jupiter’s heliocentric distance is 5·2 AU and its sidereal period is 11·86 years.
12. At a point on the Moon’s equator, two astronauts notice that their Apollo spacecraft, in a circular equatorial
orbit, transits directly overhead with an observed sidereal angular velocity ω 1 , and sets a few minutes later