/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 A tunnel is dug along the diamet... [FREE SOLUTION] | 91Ó°ÊÓ

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A tunnel is dug along the diameter of the earth (radius \(R\) and mass \(M\) ). There is a particle of mass ' \(m\) ' at the centre of the tunnel. The minimum velocity given to the particle so that it just reaches to the surface of the earth is (1) \(\sqrt{\frac{G M}{R}}\) (2) \(\sqrt{\frac{G M}{2 R}}\) (3) \(\sqrt{\frac{2 G M}{R}}\) (4) it will reach with the help of a negligible velocity

Short Answer

Expert verified
The minimum velocity is \(\sqrt{\frac{2 G M}{R}}\).

Step by step solution

01

Understand the Problem

We need to find the minimum velocity required for a particle at the center of the Earth to just reach the surface. To do this, we'll consider the gravitational force acting on the particle and the energy requirements to move from the center to the radius of the Earth.
02

Consider Gravitational Potential Energy

Gravitational potential energy at distance r from the center inside the Earth is given by \[ U(r) = -\frac{G M m}{R^3} \frac{r^2}{2} \]where \(G\) is the gravitational constant, \(M\) is the Earth's mass, \(m\) is the mass of the particle, and \(R\) is the Earth's radius. At the surface \(r = R\), potential energy becomes \[ U(R) = -\frac{G M m}{R} \].
03

Apply Conservation of Energy

The conservation of mechanical energy implies that the sum of kinetic and potential energy at the center should equal the sum of kinetic and potential energy at the surface. Initially, at the center, the potential energy is \[ U(0) = 0 \].At the surface, the potential energy is \[ U(R) = -\frac{G M m}{R} \].Let the particle have initial velocity \(v\) at the center. The initial kinetic energy is \[ K_i = \frac{1}{2} m v^2 \].So, we equate initial total energy at the center to total energy at the surface, where the final kinetic energy is zero because the minimum velocity at the surface is zero.
04

Set Equation for Energy Conservation

The energy conservation equation can be set as: \[ \frac{1}{2} m v^2 + 0 = 0 + \left(-\frac{G M m}{R}\right) \].Simplifying gives: \[ \frac{1}{2} m v^2 = \frac{G M m}{R} \].
05

Solve for Minimum Velocity

Rearrange the energy conservation equation to find velocity:\[ v^2 = \frac{2 G M}{R} \].So, the minimum velocity \(v\) needed is:\[ v = \sqrt{\frac{2 G M}{R}} \].
06

Identify the Correct Answer

The correct choice that matches our derived formula is option (3) \( \sqrt{\frac{2 G M}{R}} \).

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Key Concepts

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

Mechanical Energy Conservation
When discussing physics, especially in gravitational fields, mechanical energy conservation is crucial. This principle states that the total mechanical energy in a system remains constant if only conservative forces, like gravity, act on it. Mechanical energy is the sum of kinetic energy and potential energy. In our scenario, a particle starts at the center of the Earth. As it moves outwards, its kinetic energy will decrease while its potential energy increases.

Even though energy types are changing, the system's total energy remains unchanged. So the energy possessed by the particle initially will be equal to the energy when it reaches the Earth's surface. To find that initial kinetic energy, we apply the conservation equation, balancing initial and final energy conditions. This important rule helps predict how the energy transference will take place without considering friction or air resistance.
Gravitational Force
Gravitational force is one of the fundamental forces that influence our universe. It is the force of attraction between two masses. In our problem, this force acts between the particle and the Earth. The force is stronger the closer the objects are, especially when dealing with massive objects like planets. At the Earth's center, a particle is subject to Earth's gravitational pull, drawing it towards the core.

Inside a spherical mass like the Earth, the gravitational force is not zero at all points, except at the center. The force helps dictate the potential energy that a particle experiences inside and at the surface of the Earth. This understanding is crucial to determine how much mechanical energy is needed for the particle to escape or move to the surface.
Energy Conservation Equation
The energy conservation equation is essential for solving physics problems involving forces and motion. In our context, the equation relates a particle's kinetic and potential energies at different locations. We started at the center of the Earth where potential energy is zero, and kinetic energy depends entirely on velocity.

At the Earth's surface, the velocity is zero, meaning kinetic energy vanishes—but potential energy is negative due to the gravitational influence. The equation was structured as:
  • Initial kinetic energy + Initial potential energy = Surface kinetic energy + Surface potential energy
  • \[\frac{1}{2} m v^2 + 0 = 0 - \frac{G M m}{R}\]
This equation ensures that, in the absence of non-conservative forces, the particle conserves its total energy as it moves through the Earth's gravitational field. Solving this equation gave us the velocity needed for the particle to just barely reach the Earth's surface.

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Most popular questions from this chapter

If a body is projected with a speed lesser than escape velocity, then (1) the body can reach a certain height and may fall down following a straight line path (2) the body can reach a certain height and may fall down following a parabolic path (3) the body may orbit the earth in a circular orbit (4) the body may orbit the earth in an elliptical orbit

The change in the value of \(g\) at a height \(h\) above the surtact of earth is the same as at a depth \(d\) below the earth. Whet both \(d\) and \(h\) are much smaller than the radius of earth, then which one of the following is correct? (1) \(d=\frac{h}{2}\) (2) \(d=\frac{3 h}{2}\) (3) \(d=2 h\) (4) \(d=h\)

At perihelion, the gravitational potential energy of Pluto in its orbit has (1) its maximum value (2) its minimum value (3) the same value as at every other point in the orbit (4) the value which depends on the sense of rotation

Suppose universal gravitational constant starts to decrease, then (1) length of the day on the earth will increase (2) length of the year will increase (3) the earth will follow a spiral path of decreasing radius (4) kinetic energy of the earth will decrease

Taking the earth to be a uniform sphere of radius \(6400 \mathrm{~km}\) and the value of \(g\) at the surface to be \(10 \mathrm{~m} \mathrm{~s}^{-2}\), calculate the energy needed (in \(\times 10^{10} \mathrm{~J}\) ) to raise a satellite of mass \(2000 \mathrm{~kg}\) to a height of \(800 \mathrm{~km}\) above the earth's surface and to set it into circular orbit at that altitude.

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