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Chapter 14: Problem 11

Show that, at the bottom of a vertical mine shaft dug to depth \(D\), the measured value of \(g\) will be $$g=g_{s}\left(1-\frac{D}{R}\right)$$ \(g_{s}\) being the surface value. Assume that the Earth is a uniform sphere of radius \(R\).

Short Answer

Expert verified
Therefore, the value of acceleration due to gravity at the bottom of a deep mine shaft, dug to depth \(D\), is given by \(g = g_{s}\left(1-\frac{D}{R}\right)\), where \(g_{s}\) is the surface value, \(D\) is the depth of the mine shaft, and \(R\) is the Earth's radius.

Step by step solution

01

Gravitational Field Outside the Sphere

Starting with the formula for the gravitational field outside a sphere of mass \(M\) and radius \(R\), this is given by Newton's law: \(g_{s}= \frac{GM}{R^2}\), where \(G\) is the gravitational constant.
02

Gravitational Field Inside the Sphere

Consider a point inside the Earth at a depth \(D\). According to Gauss's law for gravity, only the mass closer to the center than the point contributes to the gravity at that point. Let's consider a mass \(m\) and its distance \(r\) from the center, where \(r = R - D\). From the spherical symmetry, we have that the gravitational field at \(r\) is given by \(g = \frac{Gm}{r^2}\). Since the Earth is a uniform sphere, the mass \(m\) inside the sphere of radius \(r\) is proportional to \(r^3\), so \(m = \frac{4}{3}\pi r^3 \rho\), where \(\rho\) is the Earth's density. Substituting this into the equation, we get \(g = \frac{4\pi G \rho r}{3}\).
03

Substituting for \(r\)

We substitute for \( r = R - D \) in the equation \( g = \frac { 4\pi G \rho r } { 3 } \) to get \( g = g_{s}\left(1-\frac{D}{R}\right) \). Note that \( g_{s} = \frac{4\pi G \rho R}{3} \) is the gravitational field strength at the surface of the Earth.

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Most popular questions from this chapter

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