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Chapter 7: Problem 60

A particle of mass \(m\) is dropped from the earth surface into a tunnel dug through a diameter of the earth. The velocity with which it cross the centre of the earth is \(\sqrt{n g R}\), then the value of \(n\) is? Assume the earth to be of uniform density. Express your answer in terms of radius \(R\) of the earth and the acceleration due to gravity \(g\) at the surface of the earth.

Short Answer

Expert verified
The value of \(n\) is 2.

Step by step solution

01

Expression for Gravitational Potential Energy

We have to find the gravitational potential energy at the surface of earth . We know that, Gravitational potential energy \(U = mgh\), where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity, and \(h\) is the height from ground which is equal to the radius of the earth \(R\) in this case. So, \(U = m g R\).
02

Expression for Final Kinetic Energy

The kinetic energy of particle when it crosses the center of the earth is given by the formula, \(K.E. = \frac{1}{2} m v^2\), where \(m\) is the mass of particle and \(v\) is the velocity. Here, the velocity \(v =\sqrt{n g R}\). Substituting \(v\) in the kinetic energy equation, we have \(K.E.= \frac{1}{2} m (\sqrt{n g R})^2\), which simplifies to \(K.E.= \frac{1}{2} m n g R\).
03

Applying the Principle of Conservation of Energy

According to the conservation of energy, the total energy of the system remains constant. So, in this case the Gravitational potential energy at the earth's surface is converted into kinetic energy at the center of the earth, we can set these equal to each other: \(m g R = \frac{1}{2} m n g R\).
04

Solving for n

Cancel out the common terms on both sides to solve for \(n\): \(1 = \frac{1}{2} n\). This gives us \(n = 2\).

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