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Chapter 6: Problem 712

If the value of ' \(\mathrm{g}\) ' acceleration due to gravity, at earth surface is \(10 \mathrm{~ms}^{-2}\). its value in \(\mathrm{ms}^{-2}\) at the center of earth, which is assumed to be a sphere of Radius ' \(\mathrm{R}\) 'meter and uniform density is (A) 5 (B) \(10 / \mathrm{R}\) (C) \(10 / 2 \mathrm{R}\) (D) zero

Short Answer

Expert verified
The value of acceleration due to gravity at the center of Earth is \(\textbf{0 ms}^{-2}\) (option D).

Step by step solution

01

Understand the problem

First, let's understand the problem. We need to find acceleration due to gravity at the Earth's center. The key to solving this problem is to realize that within the Earth, the force of gravity varies with distance from the center. So, we need to consider how gravity affects an object at the center of the Earth, assuming that Earth has a uniform density.
02

Set up the gravitational force equation

We can use Newton's law of universal gravitation to describe the force of gravity acting on an object. It is as follows: \[ F = G \frac{m M}{r^2} \] where F is the gravitational force, G is the gravitational constant, m is the mass of the object, M is the mass of the Earth, r is the distance between the Earth's center and the object. Since we are interested in the acceleration due to gravity, we can use Newton's second law \(F = ma\) to set up the following equation: \[ a_{g} = G \frac{M}{r^2} \]
03

Calculate the mass of the Earth from its density

Considering that the Earth is a sphere with uniform density \(\rho\), we can find its mass by multiplying the density by its volume. The volume of a sphere is given by: \[ V = \frac{4}{3}\pi R^3 \] So the mass of the Earth is: \[ M = \rho V = \rho \left(\frac{4}{3}\pi R^3\right) \]
04

Calculate acceleration due to gravity at the center of Earth

Now, let's compute the value of acceleration due to gravity at the center of the Earth. Set the distance r equal to zero (since we are considering the center of Earth). Here, the gravitational force experienced by an object is not due to the entire mass of Earth, but only to the mass within the spherical shell of radius r. At the center (r=0), there is no mass inside the sphere. Hence, there is no gravitational field acting on that point. Therefore, the value of acceleration due to gravity at the center of the Earth is zero.
05

Select the correct answer

Now that we know that the acceleration due to gravity at the center of the Earth is zero, we can choose the correct answer from the given options. In this case, the answer is: (D) zero

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Most popular questions from this chapter

A particle of mass \(\mathrm{M}\) is situated at the center of a spherical shell of same mass and radius a the magnitude of gravitational potential at a point situated at (a/2) distance from the center will be (A) \([(4 \mathrm{GM}) / \mathrm{a}]\) (B) \((\mathrm{GM} / \mathrm{a})\) (C) \([(2 \mathrm{GM}) / \mathrm{a}]\) (D) \([(3 \mathrm{GM}) / \mathrm{a}]\)

The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to four times the previous value, the new time period will become \(\ldots \ldots \ldots\) hours (A) 10 (B) 120 (C) 40 (D) 80

What is the intensity of gravitational field at the center of spherical shell (A) \(\left(\mathrm{Gm} / \mathrm{r}^{2}\right)\) (B) \(\mathrm{g}\) (C) zero (D) None of these

Match the following Table-1 \(\quad\) Table-2 (A) kinetic energy (P) \([(-\mathrm{GMm}) /(2 \mathrm{r})]\) (B) Potential energy (Q) \(\sqrt{(\mathrm{GM} / \mathrm{r})}\) (C) Total energy (R) - [(GMm) / r] (D) orbital velocity (S) \([(\mathrm{GMm}) /(2 \mathrm{r})]\) Copyright (O StemEZ.com. All rights reserved.

If the density of small planet is that of the same as that of the earth while the radius of the planet is \(0.2\) times that of the earth, the gravitational acceleration on the surface of the planet is (A) \(0.2 \mathrm{~g}\) (B) \(0.4 \mathrm{~g}\) (C) \(2 \mathrm{~g}\) (D) \(4 \mathrm{~g}\)
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