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Chapter 4: Problem 17

The earth is attracted to an object with a force equal and opposite to the force of the earth on the object. If this is true, why is it that when you drop an object, the earth does not have an acceleration equal and opposite to that of the object?

Short Answer

Expert verified
The Earth's immense mass compared to the object results in its acceleration being negligibly small and thus imperceptible.

Step by step solution

01

Understanding Newton's Third Law

Newton's third law of motion states that for every action, there is an equal and opposite reaction. This means when an object like a ball is dropped towards the Earth, the Earth exerts a force on the ball, and the ball exerts an equal and opposite force on the Earth. These forces are equal in magnitude but act on different objects.
02

Identifying the Forces and Masses

The force exerted on both the object and the Earth due to gravity is given by the formula \[ F = G \frac{{m_1 m_2}}{{r^2}} \]where \( G \) is the gravitational constant, \( m_1 \) is the mass of the Earth, \( m_2 \) is the mass of the object, and \( r \) is the distance between their centers. Though the forces are equal, \( m_1 \) is much larger than \( m_2 \).
03

Using Newton's Second Law of Motion

Newton's second law states that the acceleration of an object is given by \[ a = \frac{F}{m} \]This means the acceleration experienced by the Earth is \[ a_{\text{Earth}} = \frac{F}{m_{\text{Earth}}} otag\]and the acceleration of the object is\[ a_{\text{object}} = \frac{F}{m_{\text{object}}} otag\].
04

Comparing Accelerations

Since \( m_{\text{Earth}} \) is much larger than \( m_{\text{object}} \), the acceleration \( a_{\text{Earth}} \) becomes negligibly small, making it imperceptible in comparison to \( a_{\text{object}} \). Therefore, the Earth's acceleration in response is not noticeable when a small object is dropped.

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Key Concepts

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

Newton's Third Law
Every action has an equal and opposite reaction. This is the essence of Newton's Third Law of Motion. It means that forces always come in pairs. When you drop a ball, the Earth exerts a gravitational pull on the ball, directing it downward. Surprisingly, the ball also pulls back on the Earth with an equal force. It's as if they're locked in a cosmic tug-of-war! However, despite the equal nature of these forces, they act on different bodies - the Earth and the ball - each responding according to its own mass.
  • This relational interaction hints at the interconnected nature of objects.
  • Forces in the real world often go unnoticed because they produce different effects on differently sized objects.
This makes the principles of Newton's Third Law both simple in theory but complex in practical observation.
Newton's Second Law
Newton's Second Law provides insight into how forces affect motion. It is mathematically defined as \( F = ma \), where \( F \) is the force applied, \( m \) is the mass, and \( a \) is the acceleration produced. This formula shows that the acceleration of an object depends directly on the net force acting upon it and inversely on its mass.
Consequently, for the Earth-object system:
  • The same gravity force acts on both Earth and the dropped object, yet yields dramatically different accelerations.
  • This difference in acceleration stems from the vast discrepancy in mass between the Earth and the ball.
Despite an equal force being applied, the consequence in motion is more visibly apparent in the object due to its smaller mass.
Gravitational Force
Gravitational force is a universal force acting between two masses. It is expressed by the equation \[ F = G \frac{m_1 m_2}{r^2} \], where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses involved, and \( r \) is the distance between their centers.
  • This formula calculates the gravitational attraction, which is mutual and equal.
  • The larger the masses, the greater the gravitational force, but often imperceptible when mired in small-scale human interactions.
The Earth's massive size results in a strong gravitational pull, although its immense mass dilutes its own response (acceleration) to this force. Therefore, understanding gravitational force explains why only the smaller object's motion is noticeable when a force is applied.
Acceleration
Acceleration is the change in velocity of an object due to a net force. Found in Newton's second law, it is described by \( a = \frac{F}{m} \). When considering the Earth and an object, this differential in mass leads to vastly different accelerations despite equivalent forces at work.
  • For a small object, like a tennis ball, the acceleration is significant because its mass is tiny compared to the Earth.
  • On the other hand, Earth's colossal mass results in an imperceptible change in velocity.
This effectively answers why, when observing objects in freefall, the object's movement is noticeable, while Earth's reverse acceleration is virtually nil. It remains an elegant demonstration of the fundamental principles dictating motion in our universe.

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Most popular questions from this chapter

Find the angle between the following two vectors: $$ \begin{array}{l} \hat{\mathbf{x}}+2 \hat{\mathbf{y}}+3 \hat{\mathbf{z}} \\ 4 \hat{\mathbf{x}}+5 \hat{\mathbf{y}}+6 \hat{\mathbf{z}} \end{array} $$ \mathrm{\\{} ~ h w h i n t ~ \\{ h w h i n t : a n g l e b e t w e e n \\} ( a n s w e r ~ c h e c k ~ a v a i l a b l e ~ a t ~ l i g h t a n d m a t t e r . c o m ) ~

(a) We observe that the amplitude of a certain free oscillation decreases from \(A_{0}\) to \(A_{0} / Z\) after \(n\) oscillations. Find its \(Q\). (answer check available at lightandmatter.com) (b) The figure is from Shape memory in Spider draglines, Emile, Le Floch, and Vollrath, Nature 440:621 (2006). Panel 1 shows an electron microscope's image of a thread of spider silk. In 2, a spider is hanging from such a thread. From an evolutionary point of view, it's probably a bad thing for the spider if it twists back and forth while hanging like this. (We're referring to a back-and- forth rotation about the axis of the thread, not a swinging motion like a pendulum.) The authors speculate that such a vibration could make the spider easier for predators to see, and it also seems to me that it would be a bad thing just because the spider wouldn't be able to control its orientation and do what it was trying to do. Panel 3 shows a graph of such an oscillation, which the authors measured using a video camera and a computer, with a \(0.1 \mathrm{~g}\) mass hung from it in place of a spider. Compared to human-made fibers such as kevlar or copper wire, the spider thread has an unusual set of properties: 1\. It has a low \(Q\), so the vibrations damp out quickly. 2\. It doesn't become brittle with repeated twisting as a copper wire would. 3\. When twisted, it tends to settle in to a new equilibrium angle, rather than insisting on returning to its original angle. You can see this in panel 2, because although the experimenters initially twisted the wire by 35 degrees, the thread only performed oscillations with an amplitude much smaller than \(\pm 35\) degrees, settling down to a new equilibrium at 27 degrees. 4\. Over much longer time scales (hours), the thread eventually resets itself to its original equilbrium angle (shown as zero degrees on the graph). (The graph reproduced here only shows the motion over a much shorter time scale.) Some humanmade materials have this "memory" property as well, but they typically need to be heated in order to make them go back to their original shapes. Focusing on property number 1 , estimate the \(Q\) of spider silk from the graph.(answer check available at lightandmatter.com)

A bird is initially flying horizontally east at \(21.1 \mathrm{~m} / \mathrm{s}\), but one second later it has changed direction so that it is flying horizontally and \(7^{\circ}\) north of east, at the same speed. What are the magnitude and direction of its acceleration vector during that one second time interval? (Assume its acceleration was roughly constant.) (answer check available at lightandmatter.com)

A batter hits a baseball at speed \(v\), at an angle \(\theta\) above horizontal. (a) Find an equation for the range (horizontal distance to where the ball falls), \(R\), in terms of the relevant variables. Neglect air friction and the height of the ball above the ground when it is hit. \hwans\\{hwans:baseballrange\\} (b) Interpret your equation in the cases of \(\theta=0\) and \(\theta=90^{\circ}\). (c) Find the angle that gives the maximum range. Wwans\\{hwans:baseballrange\\}

A bullet leaves the barrel of a gun with a kinetic energy of \(90 \mathrm{~J}\). The gun barrel is \(50 \mathrm{~cm}\) long. The gun has a mass of \(4 \mathrm{~kg}\), the bullet \(10 \mathrm{~g}\). (a) Find the bullet's final velocity. (answer check available at lightandmatter.com) (b) Find the bullet's final momentum. (answer check available at lightandmatter.com) (c) Find the momentum of the recoiling gun. (d) Find the kinetic energy of the recoiling gun, and explain why the recoiling gun does not kill the shooter. (answer check available at lightandmatter.com)
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