Astronomy 10 Homework SolutionsChapter 2

Review Questions:

1. In Copernicus’ heliocentric model of the solar system, each planet orbited aboutthe Sun. Planets closer to the Sun had greater orbital speeds than those fartherfrom the Sun. Thus, the Earth has a greater orbital speed than Mars. Mars appearsto move backward—in retrograde motion—in the sky as the Earth catches andpasses it.

2. For a planet to be at opposition, the planet, the Earth, and the Sun must all be ina straight line with the Earth between the planet and the Sun. For this to happen,the orbit of the planet must be farther from the Sun than the orbit of the Earth.Thus, the planets that cannot be at opposition are those whose orbits are closer tothe Sun than the Earth: Mercury and Venus. For a planet to be at inferior conjunc-tion, the planet, the Earth, and the Sun must all be in a straight line with theplanet between the Earth and the Sun. For this to happen, the orbit of the planetmust be closer to the Sun than the orbit of the Earth. Thus, the planets that cannotbe at inferior conjunction are those with orbits farther from the Sun than the orbitof the Earth: Mars, Jupiter, Saturn, Uranus, Neptune, and Pluto.

3. The sidereal period of a planet is the time for the planet to complete one orbitabout the Sun relative to the fixed stars. The synodic period is the time for theplanet to return to a given orientation relative to the Sun and the Earth. For in-stance, it is the time for a planet to go from one opposition to the next.

4. Kepler’s three laws are Kepler’s three laws of planetary motion. The laws describedifferent characteristics or aspects of the orbits of the planets about the Sun. Hisfirst law, which states, in part, that the planetary orbits are ellipses, is importantbecause it does away with the ancient and incorrect idea that the orbits of the plan-ets are circles. His second law accounts for the varying speed of a planet in its orbitby noting that a planet’s speed is greater when it is closer to the Sun and smallerwhen it is farther from the Sun. His third law, which gives the relation between aplanet’s period and its average distance from the Sun, is important because, asmodified by Newton, allows us to calculate the masses of distant bodies. All threelaws are important because they aided Newton in the discovery of the law of gravityand allowed him to test this law and his laws of motion.

5. Galileo’s observations of the Sun and Moon showed them to be imperfect, whichmeant that the ancients were wrong about the perfection of the heavens. Theseobservations did not directly support the heliocentric model of the solar system, butdid indicate that it was possible for the ancients to be wrong. However, his observa-tions of Venus and Jupiter did provide direct proof of the validity of the heliocentricsolar system. He saw Venus go through a complete cycle of phases. For Venus toshow a gibbous phase, it must be able to get farther from the Earth than the Sun,which is possible heliocentric model but not for the geocentric model. His discoveryof the Galilean moons of Jupiter, four bodies that definitely did not orbit the Earth,showed that not all objects in the universe orbited the Earth. In addition, one of the

arguments against the Copernican model was that the moon would be torn awayfrom the Earth as the Earth orbited the Sun. Jupiter’s moons followed it around inits orbit.

6. The mass of a body is characteristic of the body; it is a measure of the inertia of abody, how hard it is to change the motion of the body. The mass of a body does notchange with location; the mass of a body is the same no matter where it might be.The weight of a body is the force of gravity on a body. If gravity is strong at somepoint, the weight of the body will be large; where gravity is weak, the weight of abody will be small. Thus, weight depends on the location of the body.

Problems:

1. Here we use Kepler’s third law as he originally stated it: P = a , where the period2 3

P is measured in years and the semimajor axis a is measured in AU. We are giventhe period and are asked to find the semimajor axis:

The orbits of comets are extremely eccentric, which means that their distance ofclosest approach to the Sun—perihelion—is much smaller than their greatest dis-tance from the Sun—aphelion. Thus, the maximum distance of this comet from theSun is approximately twice the semimajor axis of its orbit: maximum distance fromSun = 200 AU.

2. Here we are given both the perihelion and aphelion of the spacecraft. The sum ofthese two distances is the major axis of the elliptical orbit of the spacecraft (seefigure below), half of which is the semimajor axis of the ellipse. Thus,

The orbital period is now given by Kepler’s third law (as stated by Kepler):

3. To solve this problem, we must use Kepler’s third law a modified by Newton:

Here, a is the semimajor axis of the elliptical orbit of a body measured in astronomi-cal units, M is the mass of the body orbited about in solar mass units (assumingthat it is much larger than the mass of the body in orbit), and P is the period of theplanet measured in years. We are given period of Mars’ moon Deimos in days andits average distance from Mars in kilometers (which is the same as its semimajoraxis). We must covert these values to AU and years:

Substituting into the expression for the mass derived from Kepler’s third law, wefind

Multiplying by the mass of the Sun 1.99 × 10 kg, we get the mass of Mars in kilo-30

grams:

This value compares well to the mass of Mars given in the text, 6.42 × 10 kg.23