The value of g is altered by approximately Δg≈−2gΔrrE\Delta g\thickapprox-2g\frac{\Delta r}{r_{E}}Δg≈−2grEΔr at a height ∆r above the Earth’s surface, where rE is the radius of the Earth, as long as ∆r ≪ rE. What is the meaning of the minus sign in this relation?

Ch. 06 - Gravitation and Newton's Synthesis

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

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Giancoli Douglas 5th edition

Ch. 06 - Gravitation and Newton's Synthesis

Problem 25b

Chapter 6, Problem 25b

The value of g is altered by approximately $$\Delta$ g$\thickapprox$-2g$\frac{\Delta r}{r_{E}$}$ at a height ∆r above the Earth’s surface, where r_E is the radius of the Earth, as long as ∆r ≪ r_E. What is the meaning of the minus sign in this relation?

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The given relation, ∆g ≈ -2g ∆r/r_E, describes how the acceleration due to gravity (g) changes with a small height (∆r) above the Earth's surface, assuming ∆r is much smaller than the Earth's radius (r_E). The minus sign in this equation indicates the direction of the change in g relative to the increase in height.

To understand the minus sign, recall that the acceleration due to gravity decreases as we move farther from the Earth's surface. This is because gravity weakens with distance according to the inverse-square law.

In this context, the minus sign signifies that as ∆r (the height above the Earth's surface) increases, the value of ∆g (the change in gravity) becomes negative, meaning g decreases.

Mathematically, the negative sign reflects the fact that the derivative of g with respect to r (distance from the Earth's center) is negative. This aligns with the physical reality that gravity diminishes with increasing distance.

Thus, the minus sign in the relation ∆g ≈ -2g ∆r/r_E conveys the inverse relationship between the change in gravity and the increase in height above the Earth's surface.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gravitational Acceleration (g)

Gravitational acceleration, denoted as 'g', is the acceleration experienced by an object due to the gravitational force exerted by a massive body, such as Earth. Near the Earth's surface, this value is approximately 9.81 m/s². The value of 'g' decreases with increasing height above the Earth's surface, which is described by the formula provided in the question.

Height Above Earth's Surface (∆r)

The term ∆r represents a small change in height above the Earth's surface. In the context of the question, it indicates how much higher an object is compared to the Earth's surface. The relationship between ∆g and ∆r shows that as one moves away from the Earth, the gravitational acceleration decreases, which is significant for understanding how gravity varies with altitude.

Negative Sign in the Relation

The negative sign in the relation ∆g ≈ -2g ∆r/r_E indicates that the change in gravitational acceleration (∆g) is a decrease as height (∆r) increases. This means that as you move away from the Earth's surface, the gravitational pull becomes weaker, which is a fundamental concept in gravitational physics. The negative sign emphasizes the inverse relationship between height and gravitational acceleration.

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Textbook Question

Use the binomial expansion ${(1 \pm x)}^{n} = 1 \pm n x + \frac{n (n - 1)}{2} x^{2} \pm \dots$ to show that the value of g is altered by approximately $Δ g \approx - 2 g \frac{Δ r}{r_{E}}$ at a height ∆r above the Earth’s surface, where r_E is the radius of the Earth, as long as ∆r ≪ r_E.

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Textbook Question

An inclined plane, fixed to the inside of an elevator, makes a 38° angle with the floor. A mass m slides on the plane without friction. What is its acceleration relative to the plane if the elevator moves upward at constant speed?

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Textbook Question

Four masses are arranged as shown in Fig. 6–28. Determine the x and y components of the gravitational force on the mass at the origin (m). Write the force in vector notation (î, ĵ).

The value of g is altered by approximately Δg≈−2gΔrrE\(\Delta\) g\(\thickapprox\)-2g\(\frac{\Delta r}{r_{E}\)}Δg≈−2grEΔr at a height ∆r above the Earth’s surface, where rE is the radius of the Earth, as long as ∆r ≪ rE. What is the meaning of the minus sign in this relation?

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Key Concepts

Gravitational Acceleration (g)

Height Above Earth's Surface (∆r)

Negative Sign in the Relation

The value of g is altered by approximately $\(\Delta\) g\(\thickapprox\)-2g\(\frac{\Delta r}{r_{E}\)}$ at a height ∆r above the Earth’s surface, where r_E is the radius of the Earth, as long as ∆r ≪ r_E. What is the meaning of the minus sign in this relation?