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Chapter 7: Problem 28

The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity. (A) Will be directed towards the centre but not the same everywhere. (B) Will have the same value everywhere but not directed towards the centre. (C) Will be same everywhere in magnitude directed towards the centre. (D) Cannot be zero at any point.

Short Answer

Expert verified
The acceleration due to gravity on the surface of the Earth will be directed towards the center but not the same everywhere (A) due to uneven mass distribution in the Earth's interior.

Step by step solution

01

Understanding the Effects of Uneven Density

Since the Earth's interior is not evenly distributed, the gravitational pull caused by this uneven mass distribution will vary at different points on the surface.
02

Gravitational Force Formula

The gravitational force experienced by an object due to Earth's interior is determined by Newton's law of gravitation which states that: \[F = G \frac{m1.m2}{r^2}\] where: F : Gravitational force G : Gravitational constant, approximately \(6.674 脳 10^{-11} N路(m/kg)^2\) m1 : The mass of object 1 (in our case, a portion of the Earth's mass) m2 : The mass of object 2 (the object on the surface) r : The distance between the two objects, in our case, Earth's radius.
03

Calculate the Acceleration due to Gravity

To determine the acceleration due to gravity experienced by an object on the Earth's surface (g), we will use the gravitational force formula (F) and divide it by the mass of the object (m2): \[g = \frac{F}{m2} = G \frac{m1}{r^2}\] As the density and distribution of the Earth's mass (m1) will vary, so will the acceleration due to gravity (g), on different points on the Earth's surface.
04

Direction of Acceleration due to Gravity

The direction of the acceleration due to gravity will always be towards the center of the Earth. This is because mass attracts mass, and any object on Earth's surface will feel attraction towards the Earth's center.
05

Compare with Given Options

Let's compare our findings with the given options: (A) Will be directed towards the centre but not the same everywhere. --> Correct. As we have shown, due to the uneven mass distribution, the value of acceleration due to gravity will vary across different points on Earth's surface, but its direction will always be towards the center. (B) Will have the same value everywhere but not directed towards the centre. ---> Incorrect. The direction will be towards the center, but the value will not be the same everywhere. (C) Will be same everywhere in magnitude directed towards the centre. ---> Incorrect. While the direction is correct, the value will not be the same everywhere. (D) Cannot be zero at any point. ---> Correct. It is highly improbable for the acceleration due to gravity to be zero at any point on Earth's surface, since mass attracts mass and there will always be some gravitational pull experienced. Conclusion: The correct answer is (A), the acceleration due to gravity will be directed towards the center but not the same everywhere on the surface of the Earth.

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Key Concepts

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

Newton's Law of Gravitation
When we speak about the forces in the universe that bring every mass together, we often refer to Sir Isaac Newton's Law of Gravitation. This law is a fundamental principle that describes the gravitational attraction force between two bodies.

Simply put, Newton proposed that every particle of matter attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The equation to express this is:
\[ F = G \frac{{m_1 \cdot m_2}}{{r^2}} \]
where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between the centers of the two masses.

This universal law helps us understand everything from the orbiting of planets to the reason why objects fall to the ground here on Earth. For students, grasping Newton's law is crucial because it lays the foundation for further understanding of astrophysics, orbital mechanics, and even everyday gravity-related phenomena.
Gravitational Force
The concept of gravitational force is closely linked with Newton's law of gravitation. It is the attractive force that exists between any two masses, any two bodies, any two particles. Importantly, gravity doesn't require the two objects to be in direct contact with each other; it is a non-contact force that acts over distance.

In the context of our planet, we generally talk about the gravitational force that the Earth exerts on objects on its surface. This force is what we commonly call the force of gravity, and it gives objects weight. The gravitational force on an object on Earth is calculated with the same formula derived from Newton's law of gravitation:
\[ g = \frac{F}{m_2} = G \frac{m_1}{r^2} \]
where \( g \) is the gravitational acceleration, \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) is the mass of the Earth, \( m_2 \) is the mass of the object, and \( r \) is the radius of the Earth. Since Earth is not a perfect sphere and its mass is distributed unevenly, the gravitational force will indeed vary at different locations.
Density Variations in Earth
The density variations in Earth are a significant factor that contributes to the variations in the gravitational force experienced on its surface. Our planet is not a uniform sphere; it has a complex structure with layers of different densities, including the crust, mantle, outer core, and inner core.

These layers are made up of various materials, with each layer having a different density. Consequently, as we move from one geographical location to another, the distribution of Earth's mass changes, affecting the local gravitational force according to Newton's law of gravitation. Even local geographical features, such as mountains and valleys, can impact the gravitational force slightly due to their mass and relative density.

Therefore, students should recognize that the local topography and Earth's internal structure can cause variations in the gravitational acceleration that we experience. This concept has real-world applications, including the development of accurate GPS systems, which must take into account the gravity variations to provide precise location data.

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Most popular questions from this chapter

The period of a satellite in a circular orbit of radius \(R\) is \(T\). The period of another satellite in a circular orbit of radius \(4 R\) is (A) \(4 T\) (B) \(T / 4\) (C) \(8 T\) (D) \(T / 8\)

The change in the value of \(g\) at a height \(h\) above the surface of the earth is the same as at a depth \(d\) below the surface of earth. When both \(d\) and \(h\) are much smaller than the radius of earth, then which one of the following is correct? (A) \(d=\frac{h}{2}\) (B) \(d=\frac{3 h}{2}\) (C) \(d=2 h\) (D) \(d=h\)

Imagine a light planet revolving around a very massive star in a circular orbit of radius \(R\) with a period of revolution \(T\). If the gravitational force of attraction between the planet and the star is proportional to \(R^{-5 / 2}\), \(T^{2}\) is proportional to (A) \(R^{3}\) (B) \(R^{7 / 2}\) (C) \(R^{3 / 2}\) (D) \(R^{3.75}\)

Given that mass of the earth is \(M\) and its radius is \(R\). A body is dropped from a height equal to the radius of the earth above the surface of earth. When it reaches the ground its velocity will be (A) \(\frac{G M}{R}\) (B) \(\left[\frac{G M}{R}\right]^{1 / 2}\) (C) \(\left[\frac{2 G M}{R}\right]^{1 / 2}\) (D) \(\left[\frac{2 G M}{R}\right]\)

From a solid sphere of mass \(M\) and radius \(R\), a spherical portion of radius \(\frac{R}{2}\) is removed, as shown in Fig. 7.17. Taking gravitational potential \(V=0\) at \(r=\infty\), the potential at the centre of the cavity thus formed is \((G=\) gravitational constant \()\) (A) \(\frac{-G M}{R}\) (B) \(\frac{-2 G M}{3 R}\) (C) \(\frac{-2 G M}{R}\) (D) \(\frac{-G M}{2 R}\)
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