Chapter 11 Gravity

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224 Chapter 11 Using Kepler’s third law, relate the period of the Sun T to its mean distance R0 from the center of the galaxy: Substitute numerical values and evaluate MG/Ms: M M G s galaxy T 4π M 4π R 2 2 3 2 = GM G 3 G R0 ⇒ M S 0 = 2 GM ST 15 2⎛ 4 9. 461× 10 m ⎞ 4π ⎜ ⎜3.00× 10 c ⋅ y× c y ⎟ ⎝ ⋅ = ⎠ 2 ⎛ −11 N ⋅ m ⎞ 30 ⎛ 6 ⎜ ⎜6. 6742× 10 ( 1. 99× 10 kg) ⎜250× 10 y× 3. 156× 10 2 kg ⎟ ⎝ ⎠ ⎝ 11 ≈ 1. 1× 10 ⇒ M ≈ 11 1. 1× 10 M Kepler’s Laws 25 •• [SSM] One of the so-called ″Kirkwood gaps″ in the asteroid belt occurs at an orbital radius at which the period of the orbit is half that of Jupiter’s. The reason there is a gap for orbits of this radius is because of the periodic pulling (by Jupiter) that an asteroid experiences at the same place in its orbit every other orbit around the sun. Repeated tugs from Jupiter of this kind would eventually change the orbit of such an asteroid. Therefore, all asteroids that would otherwise have orbited at this radius have presumably been cleared away from the area due to this resonance phenomenon. How far from the Sun is this particular 2:1 resonance ″Kirkwood″ gap? Picture the Problem The period of an orbit is related to its semi-major axis (for circular orbits this distance is the orbital radius). Because we know the orbital periods of Jupiter and a hypothetical asteroid in the Kirkwood gap, we can use Kepler’s third law to set up a proportion relating the orbital periods and average distances of Jupiter and the asteroid from the Sun from which we can obtain an expression for the orbital radius of an asteroid in the Kirkwood gap. Use Kepler’s third law to relate Jupiter’s orbital period to its mean distance from the Sun: T s 2 Jupiter 2 4π = GM Sun r 3 3 Jupiter 7 s y ⎞ ⎟ ⎠ 2
Use Kepler’s third law to relate the orbital period of an asteroid in the Kirkwood gap to its mean distance from the Sun: Dividing the second of these equations by the first and simplifying yields: Solving for rKirkwood yields: Because the period of the orbit of an asteroid in the Kirkwood gap is half that of Jupiter’s: T 2 Kirkwood T 2 Kirkwood 2 TJupiter 2 4π = GM Sun 2 4π GM = 4π GM Sun 2 r Sun 3 Kirkwood r 3 Kirkwood r 3 Jupiter Gravity r = 2 3 Kirkwood Kirkwood Jupiter ⎟ Jupiter ⎟ ⎛ ⎞ ⎜ T = r ⎜ T r Kirkwood = = ⎝ 10 ( 77. 8× 10 m) 4. 90× 10 11 ⎠ ⎛ 1 2 T ⎜ ⎝ T m 3 Kirkwood 3 rJupiter Jupiter Jupiter ⎞ ⎟ ⎠ 11 1AU = 4. 90× 10 m× 1.50× 10 = 3. 27 AU Remarks: There are also significant Kirkwood gaps at 3:1, 5:2, and 7:3 and resonances at 2.5 AU, 2.82 AU, and 2.95 AU. 29 •• Kepler determined distances in the Solar System from his data. For example, he found the relative distance from the Sun to Venus (as compared to the distance from the Sun to Earth) as follows. Because Venus’s orbit is closer to the Sun than is Earth’s orbit, Venus is a morning or evening star—its position in the sky is never very far from the Sun (Figure 11-24). If we suppose the orbit of Venus is a perfect circle, then consider the relative orientation of Venus, Earth, and the Sun at maximum extension, that is when Venus is farthest from the Sun in the sky. (a) Under this condition, show that angle b in Figure 11-24 is 90º. (b) If the maximum elongation angle a between Venus and the Sun is 47º, what is the distance between Venus and the Sun in AU? (c) Use this result to estimate the length of a Venusian ″year.″ Picture the Problem We can use a property of lines tangent to a circle and radii drawn to the point of contact to show that b = 90°. Once we’ve established that b is a right angle we can use the definition of the sine function to relate the distance from the Sun to Venus to the distance from the Sun to Earth. r 2 3 11 m 225
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Chapter 11 Gravity

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