Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 34

What is the magnitude of the free-fall acceleration of a ball (mass \(m\) ) due to the Earth's gravity at an altitude of \(2 R\), where \(R\) is the radius of the Earth?

Short Answer

Expert verified
Answer: The magnitude of the free-fall acceleration is given by the expression \(a = G\frac{M}{(3R)^2}\), where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the radius of the Earth.

Step by step solution

01

Gravitational force formula

We start by recalling the formula for the force between two objects due to gravity: \(F = G\frac{Mm}{r^2}\) where \(F\) is the gravitational force, \(G\) is the gravitational constant, \(M\) is the mass of the Earth, \(m\) is the mass of the ball, and \(r\) is the distance between the centers of the two objects. In this case, the distance \(r\) is equal to the sum of the Earth's radius and the altitude of the ball, so \(r = R + 2R = 3R\). We can now rewrite the formula as: \(F = G\frac{Mm}{(3R)^2}\)
02

Equate gravitational force to mass times acceleration

The gravitational force acting on the ball is also equal to its mass times its free-fall acceleration, which we'll denote as \(a\). So we can write: \(F = ma\)
03

Solve for acceleration

Now we can equate the expressions from steps 1 and 2, and solve for the acceleration: \(ma = G\frac{Mm}{(3R)^2}\) Divide both sides by the mass of the ball, \(m\): \(a = G\frac{M}{(3R)^2}\) That's the expression for the magnitude of the free-fall acceleration of the ball at an altitude of \(2R\).

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.

Start your free trial

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

A satellite is placed between the Earth and the Moon, along a straight line that connects their centers of mass. The satellite has an orbital period around the Earth that is the same as that of the Moon, 27.3 days. How far away from the Earth should this satellite be placed?

Careful measurements of local variations in the acceleration due to gravity can reveal the locations of oil deposits. Assume that the Earth is a uniform sphere of radius \(6370 \mathrm{~km}\) and density \(5500 . \mathrm{kg} / \mathrm{m}^{3},\) except that there is a spherical region of radius \(1.00 \mathrm{~km}\) and density \(900 . \mathrm{kg} / \mathrm{m}^{3}\) whose center is at a depth of \(2.00 \mathrm{~km} .\) Suppose you are standing on the surface of the Earth directly above the anomaly with an instrument capable of measuring the acceleration due to gravity with great precision. What is the ratio of the acceleration due to gravity that you measure compared to what you would have measured had the density been \(5500 . \mathrm{kg} / \mathrm{m}^{3}\) everywhere? (Hint: Think of this as a superposition problem involving two uniform spherical masses, one with a negative density.)

You have been sent in a small spacecraft to rendezvous with a space station that is in a circular orbit of radius \(2.5000 \cdot 10^{4} \mathrm{~km}\) from the Earth's center. Due to a mishandling of units by a technician, you find yourself in the same orbit as the station but exactly halfway around the orbit from it! You do not apply forward thrust in an attempt to chase the station; that would be fatal folly. Instead, you apply a brief braking force against the direction of your motion, to put you into an elliptical orbit, whose highest point is your present position, and whose period is half that of your present orbit. Thus, you will return to your present position when the space station has come halfway around the circle to meet you. Is the minimum radius from the Earth's center-the low point \(-\) of your new elliptical orbit greater than the radius of the Earth \((6370 \mathrm{~km})\), or have you botched your last physics problem?

The Apollo 8 mission in 1968 included a circular orbit at an altitude of \(111 \mathrm{~km}\) above the Moon's surface. What was the period of this orbit? (You need to look up the mass and radius of the Moon to answer this question!)

Three asteroids, located at points \(P_{1}, P_{2},\) and \(P_{3}\), which are not in a line, and having known masses \(m_{1}, m_{2}\), and \(m_{3}\), interact with one another through their mutual gravitational forces only; they are isolated in space and do not interact with any other bodies. Let \(\sigma\) denote the axis going through the center of mass of the three asteroids, perpendicular to the triangle \(P_{1} P_{2} P_{3} .\) What conditions should the angular velocity \(\omega\) of the system (around the axis \(\sigma\) ) and the distances $$ P_{1} P_{2}=a_{12}, \quad P_{2} P_{3}=a_{23}, \quad P_{1} P_{3}=a_{13} $$ fulfill to allow the shape and size of the triangle \(P_{1} P_{2} P_{3}\) to remain unchanged during the motion of the system? That is, under what conditions does the system rotate around the axis \(\sigma\) as a rigid body?
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Rotational Dynamics

Read Explanation

Physical Quantities And Units

Read Explanation

Electromagnetism

Read Explanation

Scientific Method Physics

Read Explanation

Atoms and Radioactivity

Read Explanation
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