Question

Some of the deepest mines in the world are in South Africa and are roughly \(3.5 \mathrm{~km}\) deep. Consider the Earth to be a uniform sphere of radius \(6370 \mathrm{~km}\). a) How deep would a mine shaft have to be for the gravitational acceleration at the bottom to be reduced by a factor of 2 from its value on the Earth's surface? b) What is the percentage difference in the gravitational acceleration at the bottom of the \(3.5-\mathrm{km}\) -deep shaft relative to that at the Earth's mean radius? That is, what is the value of \(\left(a_{\text {surf }}-a_{3,5 \mathrm{~km}}\right) / a_{\text {surf }} ?\)

Short answer

Answer: To find the depth at which the gravitational acceleration is reduced by half, first calculate the surface acceleration using the formula \(g_{surf} = -\frac{4}{3} \pi G \rho R\). Then, set up an equation to solve for the depth \(D\) such that \(g(R - D) = \frac{1}{2} g_{surf}\). Calculate the value of \(D\). To find the percentage difference in acceleration between the surface and a 3.5 km deep mine shaft, first calculate the gravitational acceleration at 3.5 km depth using \(g_{3.5 \text{ km}} = -\frac{4}{3} \pi G \rho (R - 3.5)\). Then, compute the percentage difference using the formula \(\frac{g_{surf} - g_{3.5 \text{ km}}}{g_{surf}}\). Express the result as a percentage.

Step-by-step solution

Gravitational Acceleration Formula

To calculate the gravitational acceleration inside a uniform sphere, we can use the formula: \(g(r) = - \frac{4}{3} \pi G \rho r\), where \(g(r)\) is the gravitational acceleration at radius \(r\), \(G\) is the gravitational constant, and \(\rho\) is the density of the Earth.

Find the Gravitational Acceleration at the Surface

Calculate the surface acceleration \(g_{surf}\) by using the radius of the Earth, \(R = 6370 \text{ km}\). We will also need the Earth's density \(\rho\). We can estimate \(\rho = 5500 \text{ kg/m}^3\). So, \(g_{surf} = -\frac{4}{3} \pi G \rho R\)

Determine the Depth for Half Gravitational Acceleration

We want to find the depth \(D\) such that the gravitational acceleration is half its surface value, i.e., \(g(R - D) = \frac{1}{2} g_{surf}\). To solve for \(D\), set up the equation as follows: \(\frac{1}{2} g_{surf} = -\frac{4}{3} \pi G \rho (R - D)\) Solve for \(D\) and calculate its value.

Calculate the Gravitational Acceleration at 3.5 km Depth

Next, we need to find the gravitational acceleration at the bottom of the 3.5 km deep mine shaft. Calculate \(g_{3.5 \text{ km}}\) using the formula: \(g_{3.5 \text{ km}} = -\frac{4}{3} \pi G \rho (R - 3.5)\)

Calculate the Percentage Difference in Gravitational Acceleration

Finally, we can calculate the percentage difference between the surface and 3.5 km depth accelerations as follows: \(\frac{a_{\text {surf }}-a_{3,5 \mathrm{~km}}}{a_{\text {surf }}} = \frac{g_{surf} - g_{3.5 \text{ km}}}{g_{surf}}\) Compute the value and express it as a percentage.

Key concepts

Uniform Sphere Model

When we talk about Earth as a uniform sphere, we mean imagining Earth as a large, perfect ball that has the same density throughout. Think of it like a solid rubber ball. When simplified this way, calculations for gravitational effects, like how gravity changes underground, become easier.
This model assumes that:

• Earth's mass is evenly distributed.
• There's no variation in density from the surface to the core.
• The radius of Earth remains constant.

This simplification helps in deriving formulas. In our case, it helps estimate gravitational acceleration at different depths inside Earth based on radius and density.

Earth's Density

Density, denoted as \(\rho\), is essential for calculating gravitational acceleration. It tells us how much mass is in a certain volume. For Earth, the average density used is about \(5500 \text{ kg/m}^3\).
This value helps in determining the gravitational force felt at any point inside Earth. An even distribution is assumed in the uniform sphere model, meaning each chunk of Earth has the same density.
If you were to descend into a mine, the density remains constant, simplifying calculations based on Earth's geometry and mass.

Gravitational Constant

The gravitational constant, symbolized by \(G\), is a key figure in physics, necessary to calculate gravitational forces. Its approximate value is \(6.674 \times 10^{-11} \text{ N(m/kg)}^2\).
This constant appears in Newton's law of gravitation and is vital for understanding the gravitational attraction between masses. In our formula \(g(r) = - \frac{4}{3} \pi G \rho r\), \(G\) is used alongside density and radius, determining the gravity experienced at different Earth layers.
No matter where in the universe physics is applied, \(G\) remains a consistent value, making it foundational for both terrestrial and celestial calculations.

Depth and Gravitational Variation

As we go deeper into Earth, gravitational force changes. Interestingly, inside the Earth, gravity decreases as you go lower. Using the formula \(g(r) = -\frac{4}{3} \pi G \rho r\), gravity is calculated for a given depth based on Earth's density and radius.
For a mine shaft that reaches 3.5 km, gravity slightly reduces because less mass exists below you pulling downwards. The deeper you go, the less mass is beneath, causing a decrease in felt gravity.
The exercise specifically looks at how much gravity diminishes by half at a certain depth or changes at 3.5 km deep, spotlighting how variations aren't always apparent at a surface glance but are significant in scientific undertakings.