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Chapter 13: Problem 15

The Sun and Earth each exert a gravitational force on the Moon. What is the ratio \(F_{\text {Sun }} / F_{\text {Earth }}\) of these two forces? (The average Sun-Moon distance is equal to the Sun-Earth distance.)

Short Answer

Expert verified
The ratio \( \frac{F_{\text{Sun}}}{F_{\text{Earth}}} \approx 332,946 \) given mass effects.

Step by step solution

01

Understand the Gravitational Force Equation

The gravitational force between two objects is given by the equation \( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \), where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between them.
02

Calculate Gravitational Force by the Sun

For the gravitational force by the Sun on the Moon, we have:\[ F_{\text{Sun}} = \frac{G \cdot M_{\text{Sun}} \cdot M_{\text{Moon}}}{R_{\text{Sun-Moon}}^2} \]Since the distance and mass of the Moon will cancel out later, we do not need to calculate \( F_{\text{Sun}} \) separately.
03

Calculate Gravitational Force by the Earth

Similarly, the gravitational force by the Earth on the Moon is given by:\[ F_{\text{Earth}} = \frac{G \cdot M_{\text{Earth}} \cdot M_{\text{Moon}}}{R_{\text{Earth-Moon}}^2} \] Again, precise values are not necessary due to the subsequent ratio calculation.
04

Set Up the Ratio

To find the ratio \( \frac{F_{\text{Sun}}}{F_{\text{Earth}}} \), we express it as:\[ \frac{F_{\text{Sun}}}{F_{\text{Earth}}} = \frac{G \cdot M_{\text{Sun}} \cdot M_{\text{Moon}} / R_{\text{Sun-Moon}}^2}{G \cdot M_{\text{Earth}} \cdot M_{\text{Moon}} / R_{\text{Earth-Moon}}^2} \]This simplifies to:\[ \frac{M_{\text{Sun}}}{M_{\text{Earth}}} \cdot \left(\frac{R_{\text{Earth-Moon}}}{R_{\text{Sun-Moon}}}\right)^2 \]
05

Insert Given Distances and Simplify

Since the Sun-Moon distance equals the Sun-Earth distance and \( R_{\text{Earth-Moon}} \) is much smaller than \( R_{\text{Sun-Moon}} \), we consider the simplification. The ratio is dependent on the mass ratio alone since the distances are the same by approximation:\[ \frac{F_{\text{Sun}}}{F_{\text{Earth}}} \approx \frac{M_{\text{Sun}}}{M_{\text{Earth}}} \cdot \left(\frac{1}{r_{\text{approx}}}\right)^2 \]
06

Use Known Mass Ratios

The mass of the Sun \( M_{\text{Sun}} \approx 1.989 \times 10^{30} \text{ kg} \) and mass of Earth \( M_{\text{Earth}} \approx 5.972 \times 10^{24} \text{ kg} \). The approximate distance ratio is large:\[ \frac{M_{\text{Sun}}}{M_{\text{Earth}}} = \frac{1.989 \times 10^{30}}{5.972 \times 10^{24}} \approx 332,946 \]This simplifies the ratio, given the distances didn't need to factor in precisely due to alternative modeling.

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Key Concepts

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

Gravitational Constant
The gravitational constant, commonly denoted as \( G \), is a fundamental constant that appears in the equation for gravitational force between two masses. Its approximate value is \( 6.674 \times 10^{-11} \; \text{N} \cdot \text{m}^2/\text{kg}^2 \). This constant helps us quantify and calculate the force of attraction between two objects in space. It is part of the formula: \[F = \frac{G \cdot m_1 \cdot m_2}{r^2}\]
  • \( F \) is the gravitational force
  • \( m_1 \) and \( m_2 \) are the masses of the two objects
  • \( r \) is the distance between the centers of the two masses
The gravitational constant \( G \) helps describe why objects with mass attract each other and is essential to understand when exploring gravitational interactions in the universe. This concept is key in calculating the forces between larger celestial bodies like the Sun, Earth, and Moon.
Mass of Sun
The mass of the Sun is a crucial element in calculating its gravitational influence on other bodies. The Sun's enormous mass plays a significant role in its ability to exert a strong gravitational force on planets as well as on the Moon in our exercise. The approximate mass of the Sun is:\[M_{\text{Sun}} \approx 1.989 \times 10^{30} \text{ kg}\]Understanding the mass of the Sun helps us appreciate why it has such a dominant gravitational impact in the solar system. Its mass is tremendously larger than that of Earth, affecting the gravitational forces and movements around it. For instance, when we calculate the force exerted by the Sun on the Moon using the gravitational force equation, this sizeable mass becomes a key factor.
Mass of Earth
The Earth's mass is considerably smaller than the Sun's, which results in a gravitational force that is lesser in comparison. When evaluating gravitational interactions such as those involving the Moon, it's important to recognize this difference. Earth's mass is:\[M_{\text{Earth}} \approx 5.972 \times 10^{24} \text{ kg}\]This value is vital for understanding gravitational force comparisons, like the problem of Sun versus Earth鈥檚 influence on the Moon. The significant disparity in mass between Earth and the Sun explains why the gravitational force exerted by the Earth on the Moon is less than that exerted by the Sun. Recognizing these mass values allows us to set up the ratios needed to understand the gravitational dynamics.
Distance Ratio
In gravitational calculations, the distance between two objects plays a critical role in determining the force magnitude. Although the problem states the average Sun-Moon distance is equal to the Sun-Earth distance, the key aspect lies in comparing the upper approximation via distance ratio.The formula for gravitational force includes \( r^2 \), meaning any change in distance can drastically influence the resulting force. In our exercise, we use the concept:\[\frac{R_{\text{Earth-Moon}}}{R_{\text{Sun-Moon}}}\]This distance ratio may simplify when the distances are nearly identical or considerably smaller compared to other parameters. In practical terms, the Earth's gravitational pull on the Moon considers a smaller \( R_{\text{Earth-Moon}} \), influencing only small variations. This conceptual understanding aids in grasping why certain gravitational ratios might only depend on mass differences and across vast distances, your calculations often rely on large-scale mean approximations.

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Most popular questions from this chapter

What multiple of the energy needed to escape from Earth gives the energy needed to escape from (a) a moon of mass \(1.472 \times 10^{23} \mathrm{~kg}\) and radius \(3.480 \times 10^{6} \mathrm{~m}\) and (b) Jupiter?

Three dimensions. Three point particles are fixed in place in an \(x y z\) coordinate system. Particle \(A\), at the origin, has mass \(m_{A}\). Particle \(B\), at \(x y z\) coordinates \((2.00 d, 1.00 d, 2.00 d)\), has mass \(2.00 \mathrm{~m}_{A}\), and particle \(C\), at coordinates \((-1.00 d, 2.00 d,-3.00 d)\), has mass \(3.00 m_{A} \cdot A\) fourth particle \(D\), with mass \(4.00 \mathrm{~m}_{A}\), is to be placed near the other particles. In terms of distance \(d\), at what (a) \(x\), (b) \(y\), and (c) z coordinate should \(D\) be placed so that the net gravitational force on \(A\) from \(B, C\),and \(D\) is zero?

In 1993 the spacecraft Galileo sent an image (Fig. 13-32) of asteroid 243 Ida and a tiny orbiting moon (now known as Dactyl), the first confirmed example of an asteroid-moon system. In the image, the moon, which is \(1.5 \mathrm{~km}\) wide, is \(100 \mathrm{~km}\) from the center of the asteroid, which is \(55 \mathrm{~km}\) long. Assume the moon's orbit is circular with a period of \(27 \mathrm{~h}\). (a) What is the mass of the asteroid? (b) The volume of the asteroid, measured from the Galileo images, is \(14100 \mathrm{~km}^{3}\). What is the density (mass per unit volume) of the asteroid?

The Sun's center is at one focus of Earth's orbit. How far from this focus is the other focus, (a) in meters and (b) in terms of the solar radius, \(6.96 \times 10^{8} \mathrm{~m}\) ? The eccentricity is \(0.0167\), and the semimajor axis is \(1.50 \times 10^{11} \mathrm{~m}\).

One way to attack a satellite in Earth orbit is to launch a swarm of pellets in the same orbit as the satellite but in the opposite direction. Suppose a satellite in a circular orbit \(500 \mathrm{~km}\) above Earth's surface collides with a pellet having mass \(5.0 \mathrm{~g}\). (a) What is the kinetic energy of the pellet in the reference frame of the satellite just before the collision? (b) What is the ratio of this kinetic energy to the kinetic energy of a \(5.0 \mathrm{~g}\) bullet from a modern army rifle with a muzzle speed of \(950 \mathrm{~m} / \mathrm{s}\) ?
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