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Ch 13: Newton's Theory of Gravity
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 13, Problem 61

Comets move around the sun in very elliptical orbits. At its closet approach, in 1986, Comet Halley was 8.79 x 107 km from the sun and moving with a speed of 54.6 km/s. What was the comet’s speed when it crossed Neptune’s orbit in 2006?

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Step 1: Recognize that this problem involves the conservation of energy. The total mechanical energy of the comet (kinetic energy + gravitational potential energy) remains constant throughout its orbit, assuming no external forces act on it.
Step 2: Write the expression for the total mechanical energy at the closest approach (perihelion): \( E = \frac{1}{2} m v_1^2 - \frac{G M m}{r_1} \), where \( m \) is the mass of the comet, \( v_1 \) is the speed at perihelion (54.6 km/s), \( r_1 \) is the distance at perihelion (8.79 × 10⁷ km), \( G \) is the gravitational constant, and \( M \) is the mass of the Sun.
Step 3: Write the expression for the total mechanical energy at Neptune’s orbit (aphelion): \( E = \frac{1}{2} m v_2^2 - \frac{G M m}{r_2} \), where \( v_2 \) is the speed at Neptune’s orbit (what we are solving for) and \( r_2 \) is the distance of Neptune’s orbit from the Sun (approximately 4.50 × 10⁹ km).
Step 4: Set the total mechanical energy at perihelion equal to the total mechanical energy at aphelion, since energy is conserved: \( \frac{1}{2} m v_1^2 - \frac{G M m}{r_1} = \frac{1}{2} m v_2^2 - \frac{G M m}{r_2} \). Simplify by canceling \( m \) from all terms.
Step 5: Rearrange the equation to solve for \( v_2 \): \( v_2 = \sqrt{v_1^2 + 2 G M \left( \frac{1}{r_1} - \frac{1}{r_2} \right)} \). Substitute the known values for \( v_1 \), \( r_1 \), \( r_2 \), \( G \) (6.674 × 10⁻¹¹ N·m²/kg²), and \( M \) (1.989 × 10³⁰ kg) to calculate \( v_2 \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Elliptical Orbits

Elliptical orbits describe the path of celestial bodies, such as comets, around a focal point, typically a star like the sun. According to Kepler's laws of planetary motion, these orbits are elongated circles, with varying speeds depending on the distance from the sun. As a comet approaches the sun, it accelerates due to gravitational attraction, and as it moves away, it decelerates.
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Conservation of Energy

The principle of conservation of energy states that the total energy in a closed system remains constant. For a comet, this means that its kinetic energy (related to its speed) and potential energy (related to its distance from the sun) will interchange as it moves along its orbit. This principle allows us to calculate the comet's speed at different points in its orbit by equating the energy at those points.
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Gravitational Influence

Gravitational influence refers to the effect that a massive body, like the sun, has on other objects in its vicinity. The strength of this gravitational pull decreases with distance, affecting the speed and trajectory of orbiting bodies. In the case of Comet Halley, its speed changes as it moves through different regions of the solar system, influenced by the sun's gravity and the gravitational effects of other planets.
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Related Practice
Textbook Question

Two Jupiter-size planets are released from rest 1.0 x 10¹¹ m apart. What are their speeds as they crash together?

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Textbook Question

Three stars, each with the mass of our sun, form an equilateral triangle with sides 1.0 x 10¹² m long. (This triangle would just about fit within the orbit of Jupiter.) The triangle has to rotate, because otherwise the stars would crash together in the center. What is the period of rotation?

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Textbook Question

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Textbook Question

A satellite in a circular orbit of radius r has period T. A satellite in a nearby orbit with radius r + Δr, where Δr ≪ r, has the very slightly different period T + ΔT. Show that ΔT/T = (3/2) (Δr/r)

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Textbook Question

The solar system is 25,000 light years from the center of our Milky Way galaxy. One light year is the distance light travels in one year at a speed of 3.0 x 108 m/s. Astronomers have determined that the solar system is orbiting the center of the galaxy at a speed of 230 km/s. The gravitational force on the solar system is the net force due to all the matter inside our orbit. Most of that matter is concentrated near the center of the galaxy. Assume that the matter has a spherical distribution, like a giant star. What is the approximate mass of the galactic center?

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Textbook Question

The solar system is 25,000 light years from the center of our Milky Way galaxy. One light year is the distance light travels in one year at a speed of 3.0 x 10⁸ m/s. Astronomers have determined that the solar system is orbiting the center of the galaxy at a speed of 230 km/s. Assume that the sun is a typical star with a typical mass. If galactic matter is made up of stars, approximately how many stars are in the center of the galaxy?

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