Suppose a large spherical object, such as a planet, with radius Rand mass Mhas a narrow tunnel passing diametrically through it. A particle of mass m is inside the tunnel at a distance xRfrom the center. It can be shown that the net gravitational force on the particle is due entirely to the sphere of mass with radius rx; there is no net gravitational force from the mass in the spherical shell with r>x.

a. Find an expression for the gravitational force on the particle, assuming the object has uniform density. Your expression will be in terms of x, R, m, M, and any necessary constants.

b. You should have found that the gravitational force is a linear restoring force. Consequently, in the absence of air resistance, objects in the tunnel will oscillate with SHM. Suppose an intrepid astronaut exploring a 150-km-diameter, 3.5×1018kg asteroid discovers a tunnel through the center. If she jumps into the hole, how long will it take her to fall all the way through the asteroid and emerge on the other side?

Short Answer

Expert verified

(a)The gravitation on the particle is F=-GmMR3r.

(b)The time taken by the astronaut to fall all the way through the asteroid and emerge on the opposite side is localid="1650304271788" t=70min.

Step by step solution

01

The Principles.

1- Newton's Law of Universal Gravitation: The magnitude of the attractive gravitational force that an object with mass m1exerts on an object with mass m2separated by a center-to-center distance ris:

F1on2=Gm1m2r2

where G=6.67×10-11N·m2/kg2is known as the gravitational constant.

2- Hooke's Law: If any object causes a spring to stretch or compress, the spring exerts an elastic force on that object. If the object stretches the spring along the x-direction, the x-component of the force the spring exerts on the item is:

FSonO,x=-kx

where kis that the spring constant measured in newtons per meter and is a measure of the stiffness of the spring (or any elastic object), xis the distance that the object has been stretched/compressed (not the total length of the object). The elastic force exerted by the spring on the object points in a direction opposite to the direction it was stretched (or compressed), hence, the negative sing in front of kx. The object in turn exerts a force on the spring:

FOonS,x=+kx

3- The period of an oscillator in simple harmonic motion is given by

T=2πmk

Note that Tdoes not depend upon on the amplitude but only on the mass mand the force constant k.

02

The given data.

The radius of the planet is: R.

The mass of the planet is: M.

A particle of mass mis inside a narrow tunnel passing diametrically through the planet at a distance xRfrom the center.

The net gravitational force on the particle is due entirely to the sphere of mass with radius rx.

The planet has uniform density.

The diameter of the asteroid is: D=(150km)1000m1km=150×103m.

The radius of the asteroid is then: R=D2=75×103m.

The mass of the asteroid is: M=3.5×1018kg.

The asteroid has a tunnel through the center.

03

To find the required data.

In part (a), we are asked to determine an expression for the gravitational force on the particle of mass m.

In part (b), we are asked to determine the time taken by the astronaut to fall all the way through the asteroid and emerge on the other side.

04

Step 4  Expression for gravity force.

(a)Let the positive direction point outward from the center of the planet. The gravitational force on the object of mass mlocated a distance rfrom the center of the planet is found from Equation (1):

localid="1650302779928" F=-Gmmencr2

where the negative sing indicates that the gravitational force is directed inward to the center of the planet.

The mass localid="1650302784021" mencenclosed by a sphere of radius ris found in terms of the density of the planet as follows:

localid="1650302788444" menc=ρVenc

where localid="1650302792117" Vencis that the volume of the sphere of radius rwhich is equal to the density of the planet which is equal to the full mass Mof the planet divided by its volume 4/3πR3. So

menc=M43πR343πr3

=Mr3R3

Substitute for mencinto Equation (4):

F=-GmMr2r3R3

05

 To seek out the time taken.

(b)Comparing Equation(5)on the gravitational force on the object to the Equation(2)for the restoring force of a spring, we see that the object of mass mundergoes simple harmonic motion with:

k=GmMR3

which has the samedimension asthe spring constant.
The time taken by the astronaut to fall all the way through the asteroid and emerge on the otherside is half the time of the astronaut's motion:

t=T2

Substitute for Tfrom Equation (3):

t=πmk

Substitute for kfrom Equation (6):

t=π75×103m36.67×10-11N·m2/kg23.5×1018kg

=4223s

=4223s1min60sec

=70min.

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 Vaia!

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

Equation 15.25states that 12kA2=12mvmax2. What does this mean? Write a couple of sentences explaining how to interpret this equation.

It is said that Galileo discovered a basic principle of the pendulum—

that the period is independent of the amplitude—by using

his pulse to time the period of swinging lamps in the cathedral

as they swayed in the breeze. Suppose that one oscillation of a

swinging lamp takes5.5s.

a. How long is the lamp chain?

b. What maximum speed does the lamp have if its maximum

angle from vertical is 3.0?

In a science museum, a 110 kg brass pendulum bob swings at the end of a 15.0-m-long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.5 m to the side and releasing it. Because of its compact shape and smooth surface, the pendulum’s damping constant is only 0.010 kg/s. At exactly 12:00 noon, how many oscillations will the pendulum have completed and what is its amplitude?

A 757 kg student jumps off a bridge with a 12-m-long bungee cord tied to his feet. The massless bungee cord has a spring constant of 430N/m.

a. How far below the bridge is the student’s lowest point?

b. How long does it take the student to reach his lowest point? You can assume that the bungee cord exerts no force until it begins to stretch.

A block on a frictionless table is connected as shown in FIGUREP15.74to two springs having spring constants k1and k2. Show that the block's oscillation frequency is given by

f=f12+f22

where f1and f2are the frequencies at which it would oscillate if attached to spring 1or spring 2alone.

FIGURE P15.74

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.

Sign-up for free