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An asteroid, whose mass is \(2.0 \times 10^{-4}\) times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is \(3.0\) times Earth's distance from the Sun. (a) Calculate the period of revolution of the asteroid in years. (b) What is the ratio of the kinetic energy of the asteroid to the kinetic energy of Earth?

Short Answer

Expert verified
(a) 5.20 years; (b) 3.33 x 10^-5.

Step by step solution

01

Understand the Problem

We are given that the asteroid has a mass that is a small fraction of Earth's mass and revolves around the Sun at a specified distance, in terms of Earth's distance from the Sun. We need to calculate the asteroid's orbital period and kinetic energy ratio compared to Earth.
02

Apply Kepler's Third Law for Period Calculation

According to Kepler's third law, the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. If the asteroid's orbital distance is 3 times that of Earth's, its period can be calculated using the relation: \[T_a^2 = (3 a_e)^3\]where \(T_a\) is the asteroid's period and \(a_e\) is Earth's orbital distance. As Earth's period (T_e) is 1 year (by definition):\[\left(\frac{T_a}{T_e}\right)^2 = 3^3\]This gives:\[T_a = T_e \times 3^{3/2}\]So, calculating for \(T_a\):\[T_a = 1 \times 3^{3/2} = 1 \times \sqrt{27} \approx 5.20 \text{ years}\]
03

Calculate Kinetic Energy Ratio

The kinetic energy of an object in circular orbit is given by:\[KE = \frac{1}{2} m v^2\]where \(v\) is the orbital speed. However, for objects in orbit, the kinetic energy is also:\[KE = \frac{GM_s m}{2a}\]where \(M_s\) is the mass of the Sun and \(a\) is the semi-major axis. Thus the ratio of kinetic energies (asteroid to Earth) can be simplified using their different distances \[\frac{KE_{a}}{KE_{e}} = \frac{m_{a}/a_{a}}{m_{e}/a_{e}}\]Substituting for each mass and distance given:\[\frac{KE_{a}}{KE_{e}} = \frac{10^{-4}m_e/a_{a}}{m_e/a_{e}} = \frac{10^{-4} imes 1}{3} = \frac{10^{-4}}{3}\]Or simplifying further:\[\frac{KE_{a}}{KE_{e}} \approx 3.33 \times 10^{-5}\]
04

Conclusion

The orbital period of the asteroid is approximately 5.20 years, while the ratio of the kinetic energy of the asteroid to that of the Earth is roughly 3.33x10^-5. The asteroid takes about 5.20 years to complete one orbit around the Sun, and its kinetic energy is much smaller than Earth's due to its lower mass and larger orbital radius.

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Key Concepts

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

Orbital Mechanics
Orbital mechanics is the study of the motions of celestial objects, primarily focusing on the effects of gravity. It's essentially the science of how bodies like planets and asteroids move through space under the influence of their mutual gravitational attraction. One of the key principles in orbital mechanics is Kepler's Third Law. This law connects the period of orbit to the semi-major axis, or average distance, of that orbit around the central body, such as the Sun. This relationship is essential when calculating how long it takes for an asteroid to revolve around the Sun, just like Earth and other celestial bodies.
The beauty of Kepler's law is that it simplifies our understanding of celestial motions. It allows us to predict orbital periods without needing to know the exact dynamics of the system. As shown in the solution process, if an object's orbit is larger than Earth's, its period will also be longer. By applying this law, astronomers can efficiently determine the movement and timing of bodies within our solar system.
Kinetic Energy
In the realm of astronomy and orbital mechanics, kinetic energy plays a crucial role. Kinetic energy (\( KE \)) is the energy that an object possesses due to its motion. It is given by the equation \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity of the object. However, when dealing with objects in orbit, like asteroids or planets, it can also be calculated as \( KE = \frac{GM_s m}{2a} \). Here, \( G \) is the gravitational constant, \( M_s \) is the mass of the central body (like the Sun), and \( a \) is the semi-major axis of the orbit.
This alternative formula helps to simplify calculations in orbital mechanics. It focuses on the gravitational influence rather than just speed. For the asteroid revolving around the Sun, despite having a much smaller mass, its kinetic energy is affected by both its orbital distance and the mass of the Sun. In comparing the kinetic energy of Earth and the asteroid, we notice how mass and distance play a key role, emphasizing the negligible energy in the asteroid compared to Earth.
Circular Orbits
Circular orbits are a particular type of orbit where the path of the object forms a perfect circle around the central body. This means that the gravitational force and the centripetal force required for circular motion balance perfectly at every point in the orbit, resulting in constant speeds and orbital periods. Though in reality, most orbits are elliptical, considering circular orbits simplifies many calculations.
In simpler terms, objects in circular orbits follow a predictable path and speed, making them easier to analyze than elliptical orbits. This is why circular orbits are often used in educational problems. In the example provided, the asteroid is said to revolve in a circular orbit around the Sun. This assumption paves the way to use the simpler forms of equations for calculation, like the mentioned kinetic energy equations and Kepler’s Third Law. The concept of circular orbits is essential in understanding the dynamics of celestial motion and ensuring easier computation of elements like orbital periods and speeds.

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Most popular questions from this chapter

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