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Chapter 13: Problem 66

Assume a planet is a uniform sphere of radius \(R\) that (somehow) has a narrow radial tunnel through its center (Fig. 13-7). Also assume we can position an apple anywhere along the tunnel or outside the sphere. Let \(F_{R}\) be the magnitude of the gravitational force on the apple when it is located at the planet's surface. How far from the surface is there a point where the magnitude is \(\frac{1}{3} F_{R}\) if we move the apple (a) away from the planet and (b) into the tunnel?

Short Answer

Expert verified
For outside: \(d = (\sqrt{3} - 1)R\); for inside: \(s = \frac{2}{3}R\).

Step by step solution

01

Understand the Problem

We need to find the distance from the planet surface where the gravitational force on an apple is \(\frac{1}{3}F_R\) in two situations: (a) when the apple is moved outside the planet and (b) when the apple is moved inside the tunnel that passes through the planet.
02

Apply Gravitational Force Equation for Outside

For a point outside the planet, the gravitational force is given by \[ F = \frac{G M m}{(R + d)^2} \]where \(d\) is the distance from the surface, \(G\) is the gravitational constant, \(M\) is the planet's mass, and \(m\) is the apple's mass. Set this force to \(\frac{1}{3}F_R = \frac{1}{3}\frac{G M m}{R^2}\) to find \(d\).
03

Solve for Distance Outside the Planet

Equating the forces, we have: \[ \frac{G M m}{(R + d)^2} = \frac{1}{3}\frac{G M m}{R^2} \]Canceling out common terms and solving for \(d\),\[ (R + d)^2 = 3R^2 \]Taking the square root of both sides,\[ R + d = \sqrt{3}R \]\[ d = (\sqrt{3} - 1)R \].
04

Apply Gravitational Force Equation for Inside

For a point inside the planet, the gravitational force depends on only the mass enclosed by the radius from the center to the point object鈥檚 location. Hence,\[ F = \frac{G (\frac{4}{3}\pi\rho )(R - s)^3 m}{(R - s)^2} \]where \(s\) is the distance inside the tunnel. \(\rho\) is planet density. Set this force equal to \(\frac{1}{3}F_R\).
05

Solve for Distance Inside the Planet

Substitute the densities giving,\[ \frac{G M}{R^3}(R - s) = \frac{1}{3}\frac{G M}{R^2} \]Solving for \(s\),\[ R - s = \frac{1}{3}R \]\[ s = \frac{2}{3}R \].

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Key Concepts

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

Uniform Sphere
Imagine a planet shaped as a perfect ball, or more formally, a uniform sphere. This means the planet is symmetrical all around, with its mass evenly distributed throughout. This concept is crucial because it simplifies calculations of gravitational forces, allowing us to treat the planet as a point mass when considering forces outside it. The symmetry of a uniform sphere ensures that from any point outside, the gravitational effect is as though all mass were concentrated at the center. This makes calculations involving gravitational forces straightforward by using spherical symmetry principles.
Radial Tunnel
Now, consider if you could drill a tunnel from one side of the planet directly through its center to the other side鈥攖his is what we call a radial tunnel. Such a scenario is hypothetical but interesting because it invites us to think about how gravity behaves inside this tunnel. As an object moves closer to the center of the planet inside this tunnel, it experiences gravitational forces differently than it would on the surface. By understanding how gravity varies inside such a tunnel, we can explore fascinating possibilities about force balance and stable equilibria within a planet.
Newton's Law of Universal Gravitation
Sir Isaac Newton gifted us with a universal law that describes the gravitational attraction between two bodies. This law states that every point mass attracts every other point mass by a force directed along the line intersecting both points. The force is proportional to the product of their masses and inversely proportional to the square of the distance between them. This is expressed as \( F = \frac{G M m}{r^2} \), where \( G \) is the gravitational constant. Employing this formula, we can predict the gravitational pull on an object at various distances, both outside and within the planet.
Mass Distribution
Mass distribution within a planet is fundamental to calculating forces accurately鈥攅specially within a radial tunnel. A planet being a uniform sphere implies its mass is evenly spread throughout its volume. However, when considering forces inside the planet, only the mass enclosed within a given radius affects the force on an object. This scenario sees a decrease in gravitational force as you move inward because less mass is affecting the object directly. Understanding this to anticipate how and why gravitational forces change can make you appreciate how beautifully Newtonian physics models our universe.

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Most popular questions from this chapter

In a certain binary-star system, each star has the same mass as our Sun, and they revolve about their center of mass. The distance between them is the same as the distance between Earth and the Sun. What is their period of revolution in years?

A uniform solid sphere of radius \(R\) produces a gravitational acceleration of \(a_{g}\) on its surface. At what distance from the sphere's center are there points (a) inside and (b) outside the sphere where the gravitational acceleration is \(a_{g} / 2\) ?

Mile-high building. In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constructed. Ignoring Earth's rotation, find the change in your weight if you were to ride an elevator from the street level, where you weigh \(630 \mathrm{~N}\), to the top of the building.

A \(30 \mathrm{~kg}\) satellite has a circular orbit with a period of \(2.0 \mathrm{~h}\) and a radius of \(9.0 \times 10^{6} \mathrm{~m}\) around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is \(6.0 \mathrm{~m} / \mathrm{s}^{2}\), what is the radius of the planet?

What are (a) the speed and (b) the period of a \(400 \mathrm{~kg}\) satellite in an approximately circular orbit \(900 \mathrm{~km}\) above the surface of Earth? Suppose the satellite loses mechanical energy at the average rate of \(1.4 \times 10^{5} \mathrm{~J}\) per orbital revolution. Adopting the reasonable approximation that the satellite's orbit becomes a "circle of slowly diminishing radius," determine the satellite's (c) altitude, (d) speed, and (e) period at the end of its 1500 th revolution. (f) What is the magnitude of the average retarding force on the satellite? Is angular momentum around Earth's center conserved for \((\mathrm{g})\) the satellite and \((\mathrm{h})\) the satellite-Earth system (assuming that system is isolated)?
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