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Chapter 7: Problem 72

Average density of the earth (A) does not depend on \(g\). (B) is a complex function of \(g\). (C) is directly proportional to \(g\). (D) is inversely proportional to \(g\).

Short Answer

Expert verified
The average density of the Earth, \(\rho\), is related to the acceleration due to gravity, \(g\), through the formula \(\rho = \frac{3g}{4 \pi G R}\), which also involves the gravitational constant \(G\) and the radius of Earth \(R\). Therefore, the average density of the Earth is a complex function of \(g\) (Option B).

Step by step solution

01

Recall the formula for gravitational force

We begin by recalling the formula for gravitational force between two objects: \[F = G \frac{m_1 m_2}{r^2}\] where \(F\) is the gravitational force, \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of two objects, and \(r\) is the distance between their centers.
02

Relate gravitational force to the acceleration due to gravity

The gravitational force acting on an object near the Earth's surface can also be expressed as: \[F = mg\] where \(m\) is the mass of the object and \(g\) is the acceleration due to gravity. From the expressions in Steps 1 and 2, we can write: \[mg = G \frac{mM}{R^2}\] where \(M\) is the mass of Earth and \(R\) is the radius of Earth. Notice that the mass \(m\) cancels out: \[g = G \frac{M}{R^2}\]
03

Express the mass of Earth in terms of its density and volume

The mass of Earth can be expressed in terms of its average density, \(\rho\), and volume, \(V\): \[M = \rho V\] The volume of Earth can be expressed in terms of its radius, \(R\): \[V = \frac{4}{3} \pi R^3\] Substituting the expression for \(V\) into the equation for \(M\), we obtain: \[M = \rho \frac{4}{3} \pi R^3\]
04

Determine the relation between the average density and \(g\)

Now, we will substitute our expression for \(M\) from Step 3 into our equation relating \(g\) and \(M\) from Step 2: \[g = G \frac{\rho \frac{4}{3} \pi R^3}{R^2}\] Simplifying this expression, we get: \[g = \rho G \frac{4}{3} \pi R\] Finally, we solve for \(\rho\): \[\rho = \frac{3g}{4 \pi G R}\] Now we can see the relationship between the average density of the Earth, \(\rho\), and the acceleration due to gravity, \(g\). Based on this relationship, we can conclude that the average density of the Earth (A) does not depend on \(g\) only, but also involves the gravitational constant \(G\) and the radius of Earth \(R\). Therefore, the correct answer is (B) is a complex function of \(g\).

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