Question

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

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

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.

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 \(g = \frac{4\pi G \rho r}{3}\).

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.

Key concepts

Gravitational Constant

The gravitational constant, represented by the symbol \( G \), plays a crucial role in the laws of gravity proposed by Isaac Newton. It is a fundamental constant used in the computation of gravitational forces between two masses. The value of \( G \) is approximately 6.674 × 10^{-11} \( \text{Nm}^2/\text{kg}^2 \).

Essentially, \( G \) helps us understand the gravitational attraction between two objects. It states how much two pieces of mass attract each other based on their distance and masses.

Gauss's Law for Gravity

Gauss's law for gravity is a powerful tool that simplifies the calculation of gravitational fields, especially within symmetrical objects like spheres.

This law assists in determining the field strength at a specific point by considering only the mass within a closed surface. For a point inside the Earth, only the mass closer to the center than the point itself contributes to the gravitational force. This accounts for the interior gravitational field being linearly proportional to the distance from the center.

• The law highlights the symmetrical nature of uniform spherical bodies in terms of their influence on gravity.
• For a point inside the Earth, the gravitational field can be calculated using \( g = \frac{Gm}{r^2} \), where \( r \) is the distance from the center.

By applying Gauss's law, calculations become more manageable, especially when dealing with complex spherical objects like planets.

Uniform Sphere Approximation

The uniform sphere approximation is a simplified model used to study gravitational fields in spherical objects like planets and stars. This assumption allows us to model the Earth as a uniform sphere, meaning its mass is evenly distributed throughout its volume.

Under this approximation, calculations become more straightforward, as it eliminates variances in density or shape. It also helps to simplify our approach to understanding the gravitational field both on the surface and inside the sphere.

Surface Gravity

Surface gravity is the acceleration due to gravity experienced by an object near or at the surface of a celestial body, such as Earth. For Earth, the surface gravity \( g_s \) can be calculated using the formula \( g_{s}= \frac{GM}{R^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the planet's radius.

This value varies with different celestial bodies depending on their mass and size.

• Surface gravity helps in determining how much an object would weigh on the surface of any celestial body.
• It is a critical factor in understanding the gravitational pull experienced by satellites and other space-bound objects.

Understanding surface gravity is pivotal in space missions, as it influences trajectories and bindings of orbiting bodies.

Depth Dependence

Depth dependence describes how gravitational field strength changes with depth inside a planet. As you go deeper, the force of gravity doesn't remain constant; instead, it decreases. This concept embodies the reality that only the mass within a radius smaller than your current depth contributes to gravitational attraction.

With depth \( D \) inside the Earth, gravity at that point can be expressed as \( g = g_{s}\left(1-\frac{D}{R}\right) \), where \( R \) is Earth's radius, and \( g_s \) is the gravitational field strength at the surface.

• Understanding depth dependence helps in fields like mining and geology.
• Gravitational pull affects everything from deep-earth drilling to crust movements.

The depth dependence of gravity brings insights into various applications and technological developments impacted by geological conditions.