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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 脳 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 脳 10^-11 N路(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 脳 10^24 kg, and the mass of the Moon is approximately 7.342 脳 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 脳 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 脳 10^-11 N路(m/kg)^2) * ( (5.972 脳 10^24 kg) * (7.342 脳 10^22 kg) ) / (3.844 脳 10^8 m)^2 F 鈮� 1.982 脳 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 脳 10^20 N.

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Most popular questions from this chapter

Newton was holding an apple of mass \(100 . \mathrm{g}\) and thinking about the gravitational forces exerted on the apple by himself and by the Sun. Calculate the magnitude of the gravitational force acting on the apple due to (a) Newton, (b) the Sun, and (c) the Earth, assuming that the distance from the apple to Newton's center of mass is \(50.0 \mathrm{~cm}\) and Newton's mass is \(80.0 \mathrm{~kg}\).

A planet with a mass of \(7.00 \cdot 10^{21} \mathrm{~kg}\) is in a circular orbit around a star with a mass of \(2.00 \cdot 10^{30} \mathrm{~kg} .\) The planet has an orbital radius of \(3.00 \cdot 10^{10} \mathrm{~m}\). a) What is the linear orbital velocity of the planet? b) What is the period of the planet's orbit? c) What is the total mechanical energy of the planet?

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?

Can the expression for gravitational potential energy \(U_{\mathrm{g}}(y)=m g y\) be used to analyze high-altitude motion? Why or why not?

The astronomical unit (AU, equal to the mean radius of the Earth's orbit) is \(1.4960 \cdot 10^{11} \mathrm{~m},\) and a year is \(3.1557 \cdot 10^{7}\) s. Newton's gravitational constant is \(G=\) \(6.6743 \cdot 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2} .\) Calculate the mass of the Sun in kilograms. (Recalling or looking up the mass of the Sun does not constitute a solution to this problem.)
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