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The distances from the Sun at perihelion and aphelion for Pluto are \(4410 \cdot 10^{6} \mathrm{~km}\) and \(7360 \cdot 10^{6} \mathrm{~km},\) respectively. What is the ratio of Pluto's orbital speed around the Sun at perihelion to that at aphelion?

Short Answer

Expert verified
Answer: The ratio of Pluto's orbital speed around the Sun at perihelion to that at aphelion is approximately 1.667.

Step by step solution

01

Determine the distances at perihelion and aphelion

Given that the distances from the Sun at perihelion and aphelion for Pluto are \(4410 \cdot 10^{6} \mathrm{~km}\) and \(7360 \cdot 10^{6} \mathrm{~km},\) respectively. Convert these distances to meters by multiplying by 1000. \(r_{peri} = 4410 \cdot 10^{6} \mathrm{~km} \times 1000 = 4410 \cdot 10^{9} \mathrm{~m}\) \(r_{aph} = 7360 \cdot 10^{6} \mathrm{~km} \times 1000 = 7360 \cdot 10^{9} \mathrm{~m}\)
02

Apply Kepler's second law of planetary motion

Kepler's second law states that a line connecting a planet and the Sun sweeps out equal areas during equal intervals of time. Mathematically, we can express this as: \(r_{peri} \cdot v_{peri} = r_{aph} \cdot v_{aph}\) Where \(r_{peri}\) and \(r_{aph}\) are the distances from the Sun at perihelion and aphelion, and \(v_{peri}\) and \(v_{aph}\) are the orbital speeds at perihelion and aphelion respectively.
03

Solve for the ratio of the orbital speeds

We need to find the ratio of \(v_{peri}\) to \(v_{aph}\): \(\frac{v_{peri}}{v_{aph}} = \frac{r_{aph} \cdot v_{aph}}{r_{peri} \cdot v_{aph}}\) Since we want the ratio of the orbital speeds, the \(v_{aph}\) in the numerator and denominator cancels out: \(\frac{v_{peri}}{v_{aph}} = \frac{r_{aph}}{r_{peri}}\) Now, substitute the values of \(r_{peri}\) and \(r_{aph}\): \(\frac{v_{peri}}{v_{aph}} = \frac{7360 \cdot 10^{9} \mathrm{~m}}{4410 \cdot 10^{9} \mathrm{~m}}\) Finally, calculate the ratio: \(\frac{v_{peri}}{v_{aph}} = 1.667\) Therefore, the ratio of Pluto's orbital speed around the Sun at perihelion to that at aphelion is approximately \(1.667\).

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Key Concepts

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

Orbital Speed
Orbital speed is a fascinating concept that describes how fast a celestial body travels along its orbit around a larger body, like Pluto around the Sun. Imagine a planet zipping around a racetrack; the speed at which it travels depends on its position along the track. The orbital speed is calculated by taking the distance covered over a specific time period. However, this speed isn't constant for objects like planets; it varies at different points in their elliptical orbits.

Kepler's second law of planetary motion provides insights into this variation. It states that a line connecting a planet and the Sun sweeps out equal areas during equal time periods. This means that the orbital speed must increase when a planet is closer to the Sun (perihelion), and decrease when it is farther away (aphelion). This speed change ensures that the area swept over remains constant over time.

In terms of mathematics, we use Kepler's second law by expressing the relationship between speeds and distances at perihelion and aphelion:
  • \(r_{peri} \cdot v_{peri} = r_{aph} \cdot v_{aph}\)
This formula helps us understand why the ratio of orbital speeds can be calculated and applied to different celestial bodies, like Pluto.
Perihelion and Aphelion
Perihelion and aphelion are key points in the elliptical orbits of planets around the Sun. Understanding these terms gives us a clearer picture of the dynamics involved in a planet's journey through space.

**Perihelion** refers to the point in a planet's orbit where it is closest to the Sun. At this position, not only is the planet closest to the Sun, but it also travels at its fastest speed due to the gravitational pull being stronger.
**Aphelion**, on the other hand, is when the planet is farthest away from the Sun. Here, the gravitational pull is weaker, and the planet moves slower than at perihelion.

These changing speeds at perihelion and aphelion help explain the variation of planetary speeds, which is crucial for calculating orbital paths and predicting planetary motion. The distance from the Sun at perihelion, for instance, can be used to determine orbital speed as observed in the exercises involving Pluto. Therefore, recognizing perihelion and aphelion helps us understand significant aspects of a planet’s speed and energy along its orbit.
Planetary Motion
The concept of planetary motion dates back to the 17th century and Johannes Kepler, who formulated laws that provide a framework for understanding planetary orbits. These laws describe how planets like Pluto move through space, which is crucial for astronomy and space exploration.

Among these laws, Kepler's second law, often called the "law of equal areas," reveals much about how planets maintain their revolutions around the Sun at varying speeds. This law indicates that planets move in elliptical orbits with the Sun at one of the two foci. This means planets do not orbit in perfect circles but rather in flattened circles, known as ellipses.

Moreover, Kepler's laws allow scientists to predict planetary positions at any given time, which is vital for missions such as sending spacecraft to other planets or calculating satellite paths.
  • Elliptical orbits are characterized by major and minor axes, determining the shape of the orbit.
  • Understanding these orbits allows for more accurate predictions and technology usage in space exploration.
In summary, knowledge of planetary motion helps us not only perceive how celestial bodies behave in space but also how they interact with each other, contributing to the broader understanding of our solar system and beyond.

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Most popular questions from this chapter

a) By what percentage does the gravitational potential energy of the Earth change between perihelion and aphelion? (Assume the Earth's potential energy would be zero if it is moved to a very large distance away from the Sun.) b) By what percentage does the kinetic energy of the Earth change between perihelion and aphelion?

Express algebraically the ratio of the gravitational force on the Moon due to the Earth to the gravitational force on the Moon due to the Sun. Why, since the ratio is so small, doesn't the Sun pull the Moon away from the Earth?

Two 30.0 -kg masses are held at opposite corners of a square of sides \(20.0 \mathrm{~cm} .\) If one of the masses is released and allowed to fall toward the other mass, what is the acceleration of the first mass just as it is released? Assume that the only force acting on the mass is the gravitational force of the other mass. a) \(1.5 \cdot 10^{-8} \mathrm{~m} / \mathrm{s}^{2}\) b) \(2.5 \cdot 10^{-8} \mathrm{~m} / \mathrm{s}^{2}\) c) \(7.5 \cdot 10^{-8} \mathrm{~m} / \mathrm{s}^{2}\) d) \(3.7 \cdot 10^{-8} \mathrm{~m} / \mathrm{s}^{2}\)

Satellites in low orbit around the Earth lose energy from colliding with the gases of the upper atmosphere, causing them to slowly spiral inward. What happens to their kinetic energy as they fall inward?

Even though the Moon does not have an atmosphere, the trajectory of a projectile near its surface is only approximately a parabola. This is because the acceleration due to gravity near the surface of the Moon is only approximately constant. Describe as precisely as you can the actual shape of a projectile's path on the Moon, even one that travels a long distance over the surface of the Moon.

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