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Chapter 12: Problem 69

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?

Short Answer

Expert verified
After calculating GPE at both perihelion and aphelion, we get: GPE at perihelion: \(U_{peri} = -\dfrac{(6.674\times10^{-11})(1.989\times10^{30})(5.972\times10^{24})}{(0.983 \times 1.496\times10^{11})} = -3.664\times10^{33} J\) GPE at aphelion: \(U_{aph} = -\dfrac{(6.674\times10^{-11})(1.989\times10^{30})(5.972\times10^{24})}{(1.017 \times 1.496\times10^{11})} = -3.548\times10^{33} J\) #tag_title#Step 2: Calculate the KE at perihelion and aphelion#tag_content# To calculate the KE at perihelion and aphelion, we need to find the Earth's velocities at those points. Earth's velocity can be obtained using the following formula derived from the conservation of angular momentum: \(v = \sqrt{\dfrac{GM}{r}}\) Velocity at perihelion: \(v_{peri} = \sqrt{\dfrac{(6.674\times10^{-11})(1.989\times10^{30})}{(0.983 \times 1.496\times10^{11})}} = 2.978\times10^4 m/s\) Velocity at aphelion: \(v_{aph} = \sqrt{\dfrac{(6.674\times10^{-11})(1.989\times10^{30})}{(1.017 \times 1.496\times10^{11})}} = 2.926\times10^4 m/s\) Now, using the KE formula: \(K = \dfrac{1}{2}mv^2\) KE at perihelion: \(K_{peri} = \dfrac{1}{2}(5.972\times10^{24})(2.978\times10^4)^2 = 2.643\times10^{33} J\) KE at aphelion: \(K_{aph} = \dfrac{1}{2}(5.972\times10^{24})(2.926\times10^4)^2 = 2.573\times10^{33} J\) #tag_title#Step 3: Calculate percentage change in GPE and KE#tag_content# To find the percentage change in GPE and KE, we use the following formula: \(\dfrac{(\text{final value} - \text{initial value})}{\text{initial value}} \times 100\%\) Percentage change in GPE: \(\dfrac{(-3.548\times10^{33} - (-3.664\times10^{33}))}{-3.664\times10^{33}}\times 100\% = 3.17\%\) Percentage change in KE: \(\dfrac{(2.573\times10^{33} - 2.643\times10^{33})}{2.643\times10^{33}}\times 100\% = -2.64\%\) Short answer: Between perihelion and aphelion, the Earth's gravitational potential energy changes by 3.17%, while the kinetic energy changes by -2.64%.

Step by step solution

01

Calculate the GPE at perihelion and aphelion

To calculate the GPE at perihelion and aphelion, we first need to know the distances between the Earth and the Sun at these points. The average distance between Earth and Sun is 1 astronomical unit (AU), which is approximately \(1.496\times10^8\) km. At perihelion, the Earth is about 0.983 AU from the Sun, and at aphelion, the Earth is about 1.017 AU from the Sun. Now we can calculate the GPE at both these points using the formula: \(U = - \dfrac{GMm}{r}\), where \(G = 6.674\times10^{-11}m^3kg^{-1}s^{-2}\), \(M =1.989\times10^{30}kg\) and \(m=5.972\times10^{24}kg\).

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Key Concepts

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

Perihelion and Aphelion
The terms 'perihelion' and 'aphelion' refer to the points in the Earth's orbit around the Sun where the Earth is closest and farthest away, respectively. Understanding these concepts is crucial when considering variations in gravitational potential energy (GPE) and kinetic energy as the Earth moves around the Sun.

The Earth's orbit is not a perfect circle but an ellipse, so the distance between the Earth and Sun changes throughout the year. At perihelion, the gravitational pull of the Sun is strongest due to the shorter distance, causing the Earth to move faster. By contrast, at aphelion, the gravitational force is weaker, and the Earth's velocity decreases.

Importance of Perihelion and Aphelion

These points hold significance not only in calculating energy changes but also in understanding Earth's seasonal variations. For example, perihelion occurs in early January, around the same time as the northern hemisphere's winter, which may seem counterintuitive since being closer to the Sun would suggest warmer temperatures. However, seasons are primarily determined by the tilt of the Earth's axis rather than its distance from the Sun.
Astronomical Unit
When discussing distances in space, particularly within our solar system, we often use the term 'astronomical unit' (AU) for convenience. One AU is defined as the average distance between the Earth and the Sun, approximately 149.6 million kilometers (about 93 million miles). It provides a standard measurement that makes it easier to convey and comprehend the vast distances involved in astronomy.

Knowing that perihelion is about 0.983 AU and aphelion is about 1.017 AU helps us quantify the variation in distance in a more relatable way. Moreover, using the AU simplifies calculations when comparing distances across different points in Earth's orbit or when measuring distances to other astronomical bodies within our solar system.

Standardizing Space Measurements

By using the AU as a reference, we achieve a clearer understanding of how far objects are relative to each other, such as planets' positions or the extent of cometary orbits. It acts as a bridge between the unfathomable scale of space and more familiar units of measurement.
Kinetic Energy of Earth
The kinetic energy (KE) of the Earth is the energy it possesses due to its motion as it revolves around the Sun. This energy can be expressed using the equation KE = (1/2)mv^2, where 'm' is the mass of the Earth and 'v' is its orbital velocity. As the Earth travels from perihelion to aphelion, its velocity changes due to the variations in the gravitational force exerted by the Sun.

At perihelion, the Earth's kinetic energy reaches its peak because it's traveling at its fastest rate along the orbital path. Conversely, the kinetic energy is at its minimum at aphelion since the Earth slows down due to the greater distance from the Sun.

Energy Conservation in Orbital Motion

It's important to note that while the kinetic energy changes, the total energy of Earth's orbit, which is the sum of its kinetic and potential energy, remains constant if we ignore any external forces or energy loss mechanisms. This is an illustration of the conservation of mechanical energy in celestial mechanics. As the Earth moves in its elliptical path, its kinetic and potential energies continuously change in such a way that when one form of energy increases, the other decreases accordingly.

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Most popular questions from this chapter

The Moon causes tides because the gravitational force it exerts differs between the side of the Earth nearest the Moon and that farthest from the Moon. Find the difference in the accelerations toward the Moon of objects on the nearest and farthest sides of the Earth.

You have been sent in a small spacecraft to rendezvous with a space station that is in a circular orbit of radius \(2.5000 \cdot 10^{4} \mathrm{~km}\) from the Earth's center. Due to a mishandling of units by a technician, you find yourself in the same orbit as the station but exactly halfway around the orbit from it! You do not apply forward thrust in an attempt to chase the station; that would be fatal folly. Instead, you apply a brief braking force against the direction of your motion, to put you into an elliptical orbit, whose highest point is your present position, and whose period is half that of your present orbit. Thus, you will return to your present position when the space station has come halfway around the circle to meet you. Is the minimum radius from the Earth's center-the low point \(-\) of your new elliptical orbit greater than the radius of the Earth \((6370 \mathrm{~km})\), or have you botched your last physics problem?

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?

After a spacewalk, a 1.00 -kg tool is left \(50.0 \mathrm{~m}\) from the center of gravity of a 20.0 -metric ton space station, orbiting along with it. How much closer to the space station will the tool drift in an hour due to the gravitational attraction of the space station?

Halley's comet orbits the Sun with a period of 76.2 yr. a) Find the semimajor axis of the orbit of Halley's comet in astronomical units ( \(1 \mathrm{AU}\) is equal to the semimajor axis of the Earth's orbit). b) If Halley's comet is \(0.56 \mathrm{AU}\) from the Sun at perihelion, what is its maximum distance from the Sun, and what is the eccentricity of its orbit?
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