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Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed \(2.0 \times 10^{4} \mathrm{m} / \mathrm{s}\) when at a distance of \(2.5 \times 10^{11} \mathrm{m}\) from the center of the sun, what is its speed when at a distance of \(5.0 \times 10^{10} \mathrm{m} ?\)

Short Answer

Expert verified
The speed of the comet is approximately \( 3.16 \times 10^4 \mathrm{m/s} \) at a distance of \( 5.0 \times 10^{10} \mathrm{m} \).

Step by step solution

01

Understand the Conservation of Energy

The total mechanical energy of the comet, when considering its orbit, is conserved. This means that the sum of its kinetic and potential energy at one point in its orbit is equal to the sum at any other point. Mathematically, this can be expressed as \( \frac{1}{2}mv_1^2 - \frac{GMm}{r_1} = \frac{1}{2}mv_2^2 - \frac{GMm}{r_2} \), where \(v_1\) and \(v_2\) are speeds at distances \(r_1\) and \(r_2\), respectively, and \(G\) is the gravitational constant, \(M\) the mass of the sun.
02

Simplify the Equation

The mass of the comet \(m\) appears in every term, so it can be canceled out. The equation now becomes: \( \frac{1}{2}v_1^2 - \frac{GM}{r_1} = \frac{1}{2}v_2^2 - \frac{GM}{r_2} \). Rearranging gives us \( \frac{1}{2}v_1^2 + \frac{GM}{r_2} = \frac{1}{2}v_2^2 + \frac{GM}{r_1} \).
03

Isolate the Variable for Final Speed

Solve for \(v_2^2\) by rearranging the equation: \( v_2^2 = v_1^2 + 2GM \left( \frac{1}{r_2} - \frac{1}{r_1} \right) \). This equation allows for the calculation of the comet's speed at the second position using initial known values.
04

Substitute the Known Values

Insert the known values: \(v_1 = 2.0 \times 10^4 \mathrm{m/s}\), \(r_1 = 2.5 \times 10^{11} \mathrm{m}\), \(r_2 = 5.0 \times 10^{10} \mathrm{m}\), and \(G = 6.674 \times 10^{-11} \mathrm{m^3/kg \, s^2}\), \(M = 1.989 \times 10^{30} \mathrm{kg}\) (mass of the Sun):

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Energy
In the context of comets orbiting the sun, the principle of conservation of energy plays a crucial role. It states that the total mechanical energy of an object—like a comet—remains constant if only conservative forces, such as gravity, act upon it. This means that the sum of kinetic energy (energy due to motion) and potential energy (energy due to position) remains unchanged as the comet moves along its orbit.

To calculate the speed of a comet at different points in its orbit, we use the equation:
  • \( \frac{1}{2}mv_1^2 - \frac{GMm}{r_1} = \frac{1}{2}mv_2^2 - \frac{GMm}{r_2} \)
In this equation,
  • \(m\) is the mass of the comet,
  • \(v_1\) and \(v_2\) are the comet's speeds at distances \(r_1\) and \(r_2\) from the center of the sun, and
  • \(G\) is the gravitational constant and \(M\) is the mass of the sun.
In calculations involving energy conservation, observing that the mass \(m\) of the comet cancels out aids in simplifying the solution. This highlights that the speeds are solely dependent on the distances and the massive gravitational pull of the sun.
Elliptical Orbits
Elliptical orbits are a characteristic feature of planetary motion, describing the paths followed by comets around the sun. Unlike circular orbits, elliptical orbits are elongated, with the sun located at one of the foci of the ellipse. This causes variations in the speed of the comet as it travels through different regions of its orbit.

One important aspect of elliptical orbits is Kepler's First Law, which states that "The orbit of a planet (or comet) is an ellipse with the Sun at one of the two foci." This law helps explain why comets move faster when they are closer to the sun (at the perihelion) and slower when they are farther away (at the aphelion).

Understanding these orbits helps determine how gravitational forces and energy dynamics influence the motion of celestial bodies. It shows why the speed of a comet changes as it moves along its elliptical path and helps in predicting its position at given times.
Gravitational Forces
Gravitational forces are the invisible agents that govern the movement of celestial bodies such as comets in the universe. According to Newton's Law of Universal Gravitation, every two objects with mass attract each other with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula is expressed as:
  • \( F = \frac{G \, M \, m}{r^2} \)
In this formula:
  • \(F\) is the magnitude of the gravitational force,
  • \(G\) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \, \text{s}^2)\),
  • \(M\) and \(m\) are the masses of the two objects (e.g., the Sun and a comet), and
  • \(r\) is the distance between the centers of the two masses.
For comets, these gravitational forces dictate their intricate dance around the sun. The interplay of these forces and the inertia of the comet results in the observed elliptical orbits. Understanding gravitational forces allows scientists to predict the trajectory and speed changes of comets as they move within the gravitational fields of larger celestial bodies.

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Most popular questions from this chapter

Kirkwood Gaps. Hundreds of thousands of asteroids orbit the sun within the asteroid belt, which extends from about \(3 \times 10^{8} \mathrm{km}\) to about \(5 \times 10^{8} \mathrm{km}\) from the sun. (a) Find the orbital period (in years) of (i) an asteroid at the inside of the belt and (ii) an asteroid at the outside of the belt. Assume circular orbits. (b) In 1867 the American astronomer Daniel Kirkwood pointed out that several gaps exist in the asteroid belt where relatively few asteroids are found. It is now understood that these Kirkwood gaps are caused by the gravitational attraction of Jupiter, the largest planet, which orbits the sun once every 11.86 years. As an example, if an asteroid has an orbital period half that of Jupiter, or 5.93 years, on every other orbit this asteroid would be at its closest to Jupiter and feel a strong attraction toward the planet. This attraction, acting over and over on successive orbits, could sweep asteroids out of the Kirkwood gap. Use this hypothesis to determine the orbital radius for this Kirkwood gap. (c) One of several other Kirkwood gaps appears at a distance from the sun where the orbital period is 0.400 that of Jupiter. Explain why this happens, and find the orbital radius for this Kirkwood gap.

The mass of Venus is 81.5\(\%\) that of the earth, and its radius is 94.9\(\%\) that of the earth. (a) Compute the acceleration due to gravity on the surface of Venus from these data. (b) If a rock weighs 75.0 \(\mathrm{N}\) on earth, what would it weigh at the surface of Venus?

Titania, the largest moon of the planet Uranus, has \(\frac{1}{8}\) the radius of the earth and \(\frac{1}{1700}\) the mass of the earth. (a) What is the acceleration due to gravity at the surface of Titania? (b) What is the average density of Titania? (This is less the density of rock, which is one piece of evidence that Titania is made primarily of ice.)

Two satellites are in circular orbits around a planet that has radius \(9.00 \times 10^{6} \mathrm{m}\) . One satellite has mass 68.0 \(\mathrm{kg}\) , orbital radius \(5.00 \times 10^{7} \mathrm{m},\) and orbital speed 4800 \(\mathrm{m} / \mathrm{s} .\) The second satellite has mass 84.0 \(\mathrm{kg}\) and orbital radius \(3.00 \times 10^{7} \mathrm{m} .\) What is the orbital speed of this second satellite?

An object in the shape of a thin ring has radius \(a\) and mass \(M .\) A uniform sphere with mass \(m\) and radius \(R\) is placed with its center at a distance \(x\) to the right of the center of the ring, along a line through the center of the ring, and perpendicular to its plane (see Fig. E13.33). What is the gravitational force that the sphere exerts on the ring-shaped object? Show that your result reduces to the expected result when \(x\) is much larger than \(a\).

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