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Chapter 12: Problem 26

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.

Short Answer

Expert verified
Answer: The approximate difference in the accelerations towards the Moon of objects on the nearest and farthest sides of the Earth is 1.1 x 10^-6 m/s^2.

Step by step solution

01

Identify given information

We are given the following information: 1. Mass of Earth (M_e): 5.97 x 10^24 kg 2. Mass of the Moon (M_m): 7.34 x 10^22 kg 3. Distance between the centers of Earth and Moon (D): 3.84 x 10^8 m 4. Radius of Earth (R): 6.37 x 10^6 m
02

Write down Newton's law of universal gravitation

Newton's law of universal gravitation states that the gravitational force (F) between two objects with masses M1 and M2, and separated by a distance D, is given by: F = G * (M1 * M2) / D^2 where G is the gravitational constant (G = 6.674 x 10^-11 N(m/kg)^2).
03

Calculate the gravitational force on the nearest side of the Earth

The distance between the center of the Earth and the nearest side of the Earth is (D - R). Therefore, we can calculate the gravitational force (F_n) on the nearest side by plugging this distance into the equation from Step 2: F_n = G * (M_e * M_m) / (D - R)^2
04

Calculate the gravitational force on the farthest side of the Earth

The distance between the center of the Earth and the farthest side of the Earth is (D + R). Therefore, we can calculate the gravitational force (F_f) on the farthest side by using this distance in the equation from Step 2: F_f = G * (M_e * M_m) / (D + R)^2
05

Calculate the difference in gravitational forces

The difference in the gravitational forces (ΔF) can be found by subtracting the force on the farthest side from the force on the nearest side: ΔF = F_n - F_f
06

Compute the difference in accelerations

Since acceleration (a) is equal to the gravitational force (F) divided by the mass of the object (M), we can find the difference in the accelerations (Δa) by dividing the difference in gravitational forces (ΔF) by the mass of the objects (M_e): Δa = ΔF / M_e After calculating the values, you will find that the difference in the accelerations toward the Moon of objects on the nearest and farthest sides of the Earth is approximately 1.1 x 10^-6 m/s^2.

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Most popular questions from this chapter

Careful measurements of local variations in the acceleration due to gravity can reveal the locations of oil deposits. Assume that the Earth is a uniform sphere of radius \(6370 \mathrm{~km}\) and density \(5500 . \mathrm{kg} / \mathrm{m}^{3},\) except that there is a spherical region of radius \(1.00 \mathrm{~km}\) and density \(900 . \mathrm{kg} / \mathrm{m}^{3}\) whose center is at a depth of \(2.00 \mathrm{~km} .\) Suppose you are standing on the surface of the Earth directly above the anomaly with an instrument capable of measuring the acceleration due to gravity with great precision. What is the ratio of the acceleration due to gravity that you measure compared to what you would have measured had the density been \(5500 . \mathrm{kg} / \mathrm{m}^{3}\) everywhere? (Hint: Think of this as a superposition problem involving two uniform spherical masses, one with a negative density.)

Consider a 0.300 -kg apple (a) attached to a tree and (b) falling. Does the apple exert a gravitational force on the Earth? If so, what is the magnitude of this force?

The more powerful the gravitational force of a planet, the greater its escape speed, \(v,\) and the greater the gravitational acceleration, \(g\), at its surface. However, in Table 12.1 , the value for \(v\) is much greater for Uranus than for Earth - but \(g\) is smaller on Uranus than on Earth! How can this be?

Determine the minimum amount of energy that a projectile of mass \(100.0 \mathrm{~kg}\) must gain to reach a circular orbit \(10.00 \mathrm{~km}\) above the Earth's surface if launched from (a) the North Pole or from (b) the Equator (keep answers to four significant figures). Do not be concerned about the direction of the launch or of the final orbit. Is there an advantage or disadvantage to launching from the Equator? If so, how significant is the difference? Do not neglect the rotation of the Earth when calculating the initial energies.

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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