/*! 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 76 You're working on the script of ... [FREE SOLUTION] | 91Ó°ÊÓ

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You're working on the script of a movie whose plot involves a hole drilled straight through Earth's center and out the other side. You're asked to determine what will happen if a person falls into the hole. You find that the gravitational acceleration inside Earth points toward Earth's center, with magnitude given approximately by \(g(r)=g_{0}\left(r / R_{\mathrm{E}}\right),\) where \(g_{0}\) is the surface value, \(r\) is the distance from Earth's center, and \(R_{\mathrm{E}}\) is Earth's radius. What do you report for the person's motion, including equations and values for any relevant parameters?

Short Answer

Expert verified
The person would oscillate back and forth through the Earth, taking a total time of \( 2 * \sqrt{\frac{2 * R_{E}}{g_{0}}} \).

Step by step solution

01

Setting up the Equation of Motion

The gravitational force on the person falling into the hole is given by \( F = m*g(r) \), which can be written as \( F = - m*g_{0}*(r / R_{\mathrm{E}}) \), taken negative as it acts towards the center. Using Newton's second law \( F = m*a \), this becomes \( m*a = - m*g_{0}*(r / R_{\mathrm{E}}) \), where \( a \) is the acceleration. By cancelling out \( m \), the equation of motion becomes \( a = - g_{0}*(r / R_{\mathrm{E}}) \).
02

Solving the Equation of Motion

This differential equation can be solved by rearranging and integrating: \( \int_{R_{E}}^{0} a dr = - \int_{0}^{t} g_{0} dt \). The left side integrates to \( \frac{- g_{0} * R_{E}}{2} \), and the right side integrates to \( - g_{0} * t \). Equating and solving for \( t \), we find \( t = \sqrt{\frac{2 * R_{E}}{g_{0}}} \). This is the time for the person to reach the center of the Earth. Because of symmetry, the total time for the person's journey (from one side of the Earth to the other) would be double, i.e. \( T = 2 * t = 2 * \sqrt{\frac{2 * R_{E}}{g_{0}}} \).
03

Describing the Person's Motion

The person falling into the hole would oscillate back and forth along the diameter of the Earth, similar to a simple harmonic motion. The total time of the journey from one side to the other side of the Earth would be \( T = 2 * \sqrt{\frac{2 * R_{E}}{g_{0}}} \).

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Most popular questions from this chapter

The equation for an ellipse is \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1 .\) Show that two-dimensional simple harmonic motion whose components have different amplitudes and are \(\pi / 2\) out of phase gives rise to elliptical motion. How are constants \(a\) and \(b\) related to the amplitudes?

The \(x\) - and \(y\) -components of an object's motion are harmonic with frequency ratio \(1.75: 1 .\) How many oscillations must each component undergo before the object returns to its initial position?

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A simple model for a variable star considers that the outer layer of the star is subject to two forces: the inward force of gravity and the outward force due to gas pressure. As a result, Newton's law for the star's outer layer reads \(m d^{2} r / d t^{2}=4 \pi r^{2} p-G M m / r^{2} .\) Here \(m\) is the mass of the outer layer, \(M\) is the total mass of the star, \(r\) is the star's radius, and \(p\) is the pressure. (a) Use this equation to show that the star's equilibrium pressure and radius are related by \(p_{0}=G M m / 4 \pi r_{0}^{4},\) where the subscript 0 represents equilibrium values. (b) As you'll learn in Chapter 18 , gas pressure and volume \(V\left(=\frac{4}{3} \pi r^{3}\right)\) are related by \(p V^{3 / 3}=p_{0} V_{0}^{5 / 3}\) (this is for an adiabatic process, a good approximation here, and the exponent \(5 / 3\) reflects the ionized gas that makes up the star). Let \(x=r-r_{0}\) be the displacement of the star's surface from equilibrium. Use the binomial approximation (Appendix A) to show that, when \(x\) is small compared with \(r,\) the righthand side of the above equation can be written \(-\left(G M m / r_{0}^{3}\right) x\) (c) since \(r\) and \(x\) differ only by a constant, the term \(d^{2}\) r/dt \(^{2}\) in the equation above can also be written \(d^{2} x / d t^{2} .\) Make this substitution, along with substituting the result of part (b) for the right- hand side, and compare your result with Equations 13.2 and 13.7 to find an expression for the oscillation period of the star. (d) What does your simple model predict for the period of the variable star Delta Cephei, with radius 44.5 times that of the Sun and mass of 4.5 Sun masses? (Your answer overestimates the actual period by a factor of about \(3,\) both because of oversimplified physics and because changes in the star's radius are too large for the assumption of a linear restoring force.)

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