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Chapter 12: Problem 22

Compare the magnitudes of the gravitational force that the Earth exerts on the Moon and the gravitational force that the Moon exerts on the Earth. Which is larger?

Short Answer

Expert verified
Answer: The magnitudes of the gravitational forces that the Earth and the Moon exert on each other are equal, with a value of approximately 1.982 × 10^20 N.

Step by step solution

01

Recall Newton's Law of Universal Gravitation

Newton's Law of Universal Gravitation states that any two objects with mass attract each other with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between their centers. The formula for the gravitational force (F) between two masses (m1 and m2), separated by a distance (d), can be written as: F = G * (m1 * m2) / d^2 where G is the gravitational constant, approximately 6.67430 × 10^-11 N·(m/kg)^2.
02

Identify the masses of the Earth and the Moon

In order to use Newton's Law, we need to know the masses of the Earth (m1) and the Moon (m2). The mass of the Earth is approximately 5.972 × 10^24 kg, and the mass of the Moon is approximately 7.342 × 10^22 kg.
03

Identify the distance between the Earth and the Moon

We also need to know the distance (d) between the centers of the Earth and the Moon. On average, the distance between the Earth and the Moon is 3.844 × 10^8 meters. This value will be used for d.
04

Calculate the gravitational force between the Earth and the Moon

Now we have all the needed values to calculate the gravitational force (F) exerted by the Earth on the Moon, as well as the force exerted by the Moon on the Earth. As previously mentioned, the forces are equal because they result from the same interaction between the Earth and the Moon. We will calculate the force using the values identified for the mass of the Earth (m1), the mass of the Moon (m2), and the average distance between them (d), and the gravitational constant (G): F = G * (m1 * m2) / d^2 F = (6.67430 × 10^-11 N·(m/kg)^2) * ( (5.972 × 10^24 kg) * (7.342 × 10^22 kg) ) / (3.844 × 10^8 m)^2 F ≈ 1.982 × 10^20 N
05

Compare the magnitudes of the gravitational forces

Since the gravitational force is a mutual force that acts on both the Earth and the Moon, the magnitude of the force that the Earth exerts on the Moon is equal to the magnitude of the force that the Moon exerts on the Earth. Therefore, neither of these magnitudes is larger; they are equal. The gravitational force between the Earth and the Moon is approximately 1.982 × 10^20 N.

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Most popular questions from this chapter

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) \(3.7 \cdot 10^{-8} \mathrm{~m} / \mathrm{s}^{2}\)

The Apollo 8 mission in 1968 included a circular orbit at an altitude of \(111 \mathrm{~km}\) above the Moon's surface. What was the period of this orbit? (You need to look up the mass and radius of the Moon to answer this question!)

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?

A 1000.-kg communications satellite is released from a space shuttle to initially orbit the Earth at a radius of \(7.00 \cdot 10^{6} \mathrm{~m}\). After being deployed, the satellite's rockets are fired to put it into a higher altitude orbit of radius \(5.00 \cdot 10^{7} \mathrm{~m} .\) What is the minimum mechanical energy supplied by the rockets to effect this change in orbit?

Consider the Sun to be at the origin of an \(x y\) coordinate system. A telescope spots an asteroid in the \(x y\) -plane at a position given by \(\left(2.0 \cdot 10^{11} \mathrm{~m}, 3.0 \cdot 10^{11} \mathrm{~m}\right)\) with a velocity given by \(\left(-9.0 \cdot 10^{3} \mathrm{~m} / \mathrm{s},-7.0 \cdot 10^{3} \mathrm{~m} / \mathrm{s}\right) .\) What will the asteroid's speed and distance from the Sun be at closest approach?
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