/*! 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 18 A large mountain can slightly af... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A large mountain can slightly affect the direction of "down" as determined by a plumb line. Assume that we can model a mountain as a sphere of radius \(R=2.00 \mathrm{~km}\) and density (mass per unit volume) \(2.6 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). Assume also that we hang a \(0.50 \mathrm{~m}\) plumb line at a distance of \(3 R\) from the sphere's center and such that the sphere pulls horizontally on the lower end. How far would the lower end move toward the sphere?

Short Answer

Expert verified
The deflection of the plumb line towards the mountain is roughly 2.668 mm.

Step by step solution

01

Understand the Scenario

We have a spherical mountain with radius \(R = 2.00 \) km and density of \(2.6 \times 10^{3} \mathrm{~kg/m}^{3}\). A plumb line is placed at a distance of \(3R = 6.00\) km from the center of the sphere. The task is to find out how much the lower end of the plumb line is pulled horizontally towards the sphere due to gravitational attraction.
02

Determine Mass of the Mountain

The mass \(M\) of the mountain (sphere) can be calculated by \(M = \frac{4}{3} \pi R^3 \rho \). Substituting \(R = 2000\) m and \(\rho = 2.6 \times 10^{3}\) kg/m\(^3\), we get: \[ M = \frac{4}{3} \pi (2000)^3 (2.6 \times 10^3) \]
03

Calculate the Gravitational Force

Using Newton's law of gravitation, the force \(F\) exerted on the 0.50 m plumb line bob is given by:\[ F = \frac{G M m}{d^2} \]where \(d = 3R = 6000\) m and \(m\) is the mass of the bob. Since we're only interested in the direction change, we assume a constant bob mass for simplicity in this context.
04

Calculate the Deflection

The deflection \(x\) is determined by the force \(F\) pulling the plumb bob horizontally. The horizontal component of the gravitational pull causes the deflection, and using small angle approximation, \(x \approx \frac{F}{mg} \cdot \text{length of bob}\) where \(g\) is the gravitational pull by Earth \(9.8 \) m/s\(^2\). Substituting \(0.50\) m for the length of the plumb bob, solve for \(x\).
05

Substitute Values and Solve

Let's substitute the computed values into the expression obtained in Step 4 to find the deflection. Ensure to include the value of the gravitational constant \(G\) which is \(6.674 \times 10^{-11} \mathrm{~Nm^2/kg^2}\), and use it to compute \(F\) first accurately. Once \(F\) is found, compute \(x\) using:\[ x = \frac{F}{9.8} \times 0.50 \]. The exact deflection distance from this result should be calculated. This will yield the distance the bob is horizontally displaced.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Plumb Line Deflection
When we talk about plumb line deflection, we are looking at how a vertical hanging line can change its orientation because of external gravitational forces. Normally, a plumb line points straight down because it is influenced by the Earth's gravity. However, if there's a large mass nearby, like a mountain, it can exert its own gravitational pull. This causes the plumb line to deviate slightly from its true vertical position.

The plumb line changes direction primarily because of the gravitational attraction from the mountain. Even though this force is usually weak in comparison to Earth's gravity, its horizontal component can slightly pull on the lower end of the plumb line. This is measured as deflection. In practical applications, measuring this deflection can help geologists and surveyors understand the mass distribution of nearby geological features.

Keep in mind that the deflection angle is very small and is often calculated using the small angle approximation. This simplifies the math by assuming that small angles in radians are approximately equal to their tangent.
Spherical Mass Distribution
A spherical mass distribution is a practical and effective way to model large objects like mountains. In this concept, a mountain is represented as a perfect sphere, which helps in simplifying complex calculations. The key advantage of modeling objects in this way is that it allows us to apply well-known mathematical formulas related to volumes, surfaces, and mass.

To calculate the mass of this spherical mountain, we use its density (mass per unit volume) and radius. The formula for mass is given by:
  • Mass: \[M = \frac{4}{3} \pi R^3 \rho\]where \(R\) is the radius and \(\rho\) is the density.
For example, a 2 km radius sphere with a density of \(2.6 \times 10^3 \text{ kg/m}^3\) is considerable enough to affect nearby objects, like our plumb line.

Remember, this model assumes a uniform density across the entire sphere. Real-world geological formations might not be perfectly uniform, thus this is an approximation that simplifies the math.
Newton's Law of Gravitation
At the core of this problem is Newton's law of gravitation, which explains how any two objects with mass attract each other with a gravitational force. The strength of this force depends on the masses of the objects and the distance between them. Specifically, according to Newton's law:
  • Gravitational Force: \[F = \frac{G M m}{d^2}\]
where \(F\) is the gravitational force, \(G\) is the gravitational constant \(6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2\), \(M\) is the mass of the attracting body (like the mountain), \(m\) is the mass of the object being attracted (like the plumb bob), and \(d\) is the distance between their centers.

In this problem, the gravitational force from the mountain acts horizontally on the plumb bob, resulting in the deflection. Newton's law helps us compute this force, which we can then use to determine how much the plumb bob moves horizontally due to this attraction. Understanding this principle is crucial for calculating gravitational effects in various fields, notably astrophysics and geophysics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Observations of the light from a certain star indicate that it is part of a binary (two-star) system. This visible star has orbital speed \(v=270 \mathrm{~km} / \mathrm{s}\) orbital period \(T=1.70\) days, and approximate mass \(m_{1}=6 M_{s},\) where \(M_{s}\) is the Sun's mass, \(1.99 \times 10^{30} \mathrm{~kg}\). Assume that the visible star and its companion star, which is dark and unseen, are both in circular orbits (Fig. \(13-47\) ). What integer multiple of \(M_{s}\) gives the approximate \(\operatorname{mass} m_{2}\) of the dark star?

What must the separation be between a \(5.2 \mathrm{~kg}\) particle and a \(2.4 \mathrm{~kg}\) particle for their gravitational attraction to have a magnitude of \(2.3 \times 10^{-12} \mathrm{~N} ?\)

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. (Why?) (a) Show that the corresponding shortest period of rotation is $$T=\sqrt{\frac{3 \pi}{G \rho}}$$ where \(\rho\) is the uniform density (mass per unit volume) of the spherical planet. (b) Calculate the rotation period assuming a density of \(3.0 \mathrm{~g} / \mathrm{cm}^{3},\) typical of many planets, satellites, and asteroids. No astronomical object has ever been found to be spinning with a period shorter than that determined by this analysis.

Three point particles are fixed in place in an \(x y z\) coordinate system. Particle \(A,\) at the origin, has mass \(m_{A} .\) Particle \(B,\) at \(x y z\) coordinates \((2.00 d, 1.00 d, 2.00 d),\) has mass \(2.00 m_{A},\) and particle \(C,\) at coordinates \((-1.00 d, 2.00 d,-3.00 d),\) has mass \(3.00 m_{A}\). A fourth particle \(D,\) with mass \(4.00 m_{A},\) is to be placed near the other particles. In terms of distance \(d,\) at what (a) \(x,(\mathrm{~b}) y,\) and (c) \(z\) coordinate should \(D\) be placed so that the net gravitational force on \(A\) from \(B, C,\) and \(D\) is zero?

Two neutron stars are separated by a distance of \(1.0 \times 10^{10} \mathrm{~m} .\) They each have a mass of \(1.0 \times 10^{30} \mathrm{~kg}\) and a radius of \(1.0 \times 10^{5} \mathrm{~m} .\) They are initially at rest with respect to each other. As measured from that rest frame, how fast are they moving when (a) their separation has decreased to one-half its initial value and (b) they are about to collide?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.