/*! 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 52 A body of mass \(m\) is situated... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 52

A body of mass \(m\) is situated at a distance \(4 R_{e}\) above the earth's surface, where \(R_{e}\) is the radius of earth. How much minimum energy be given to the body so that it may escape (a) \(m g R_{e}\) (b) \(2 m g R_{e}\) (c) \(\frac{m g R_{c}}{5}\) (d) \(\frac{m g R_{e}}{16}\)

Short Answer

Expert verified
(c) \(\frac{m \cdot g \cdot R_{c}}{5}\)

Step by step solution

01

Recall the formula for Escape Velocity

The formula for escape velocity is given as \(v = \sqrt{2gR}\) where \(R\) is the distance of the body from the center of Earth and \(g\) is the acceleration due to gravity on the earth's surface.
02

Compute the Gravitational Potential Energy at a height \(h\)

The Gravitational Potential Energy (U) of an object at a height \(h\) from the earth's surface is given by \(U = - \frac{Gm_{1}m_{2}}{r}\) where \(m_{1}\) is the mass of the object, \(m_{2}\) is the mass of the Earth, \(r\) is the distance from the center of the Earth to the object which equals \(R_{e} + h\). Given that \(h = 4R_{e}\), the formula becomes \(U = - \frac{Gm_{1}m_{2}}{5R_{e}}\)
03

Compute the Minimum Energy

In order for the object to escape Earth's gravitational field, the minimum energy given to the object should negate its gravitational potential energy. In other words, \(E_{min} = |U|\). Therefore, \(E_{min} = \frac{Gm_{1}m_{2}}{5R_{e}}\)
04

Substitute Variables

We know that the gravitational constant, \(G\), can be replaced by \(G = G' R_{e}^{2}\) where \(G'\) is \(g\) on Earth's surface and \(m_{2}\) can be replaced by \(\frac{m_{1} \cdot g}{g'}\). Also, \(R_{e}\) is the radius of Earth. Substituting those in the Energy equation from step 3, we have \(E_{min} = \frac{G' m_{1}^{2} \cdot g}{5R_{e}}\)
05

Simplify the Energy Equation

Hence the minimum energy needed for the object to escape Earth's gravitational field is \(E_{min} = \frac{m \cdot g \cdot R_{e}}{5}\). Therefore among the options, the closest approximation is \(\frac{m \cdot g \cdot R_{e}}{5}\) which is (c).

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Most popular questions from this chapter

The maximum and minimum distance of a comet from the sun are \(8 \times 10^{12} \mathrm{~m}\) and \(1.6 \times 10^{12} \mathrm{~m}\). If its velocity when nearest to the sun is \(60 \mathrm{~m} / \mathrm{s}\), what will be its velocity in \(\mathrm{m} / \mathrm{s}\) when it is farthest (a) 12 (b) 60 (c) 112 (d) 6

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