Problem 54 - Calculus Volume 1

Question

The volume \(V\) of a sphere depends on the length of its radius as \(V=(4 / 3) \pi r^{3}\). Because Earth is not a perfect sphere, we can use the mean radius when measuring from the center to its surface. The mean radius is the average distance from the physical center to the surface, based on a large number of samples. Find the volume of Earth with mean radius \(6.371 \times 10^{6} \mathrm{~m}\).

Step-by-Step Solution

Answer

The Earth's volume is approximately \( 1.08321 \times 10^{21} \) cubic meters.

Step 1: Identify Given Values

Determine the known values from the problem. We are given the mean radius of Earth, which is \( r = 6.371 \times 10^{6} \) meters.

Step 2: Understand the Formula

The formula to calculate the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \). This means we need to cube the radius and multiply by \( \frac{4}{3} \pi \).

Step 3: Substitute the Radius into the Formula

Plug the mean radius of Earth into the formula: \( V = \frac{4}{3} \pi (6.371 \times 10^{6})^3 \).

Step 4: Calculate the Cubed Radius

Compute \( (6.371 \times 10^{6})^3 \). This gives \( 6.371^3 \times (10^{6})^3 = 258.5 \times 10^{18} \).

Step 5: Find the Volume

Substitute the cubed radius into the volume formula: \( V = \frac{4}{3} \pi \times 258.5 \times 10^{18} \). Calculate this to find the volume of Earth.

Step 6: Final Calculation

Calculate \( V = \frac{4}{3} \pi \times 258.5 \times 10^{18} \), which gives approximately \( 1.08321 \times 10^{21} \) cubic meters.

Key Concepts

Mean Radius

The concept of the mean radius is used to simplify the calculation of properties like volume for objects not perfectly spherical, like the Earth. Unlike a perfect sphere, the Earth is slightly flattened at the poles and bulges at the equator. Therefore, we utilize the mean radius, which represents the average radius derived from various points on the Earth's surface. This average provides a more uniform measure to use in calculations that involve spherical formulas. For Earth, this mean radius is denoted as approximately 6,371 kilometers, helping scientists and mathematicians make accurate estimations about the planet's physical attributes.

Cubing Numbers

Cubing a number involves multiplying the number by itself twice. This is mathematically represented as raising the number to the power of three, noted as \[ r^3 = r \times r \times r \]. When calculating the volume of a sphere, the radius is cubed before being used in the formula. This is a crucial step since it accounts for the three-dimensional space occupied by the sphere. In the case of Earth, cubing the mean radius \[ (6.371 \times 10^6 \, m)^3 \] involves both cubing the numerical part \(6.371^3\) and the exponential part \((10^6)^3\), resulting in a very large number that scales our calculation appropriately for the volumetric space.

Volume of Earth

Calculating the Earth's volume gives us insight into the vastness of our planet. To find a sphere's volume, which approximates the Earth, we use the formula: \[ V = \frac{4}{3} \pi r^3 \]. Plugging in the mean radius: \[ V = \frac{4}{3} \pi (6.371 \times 10^6)^3 \], we compute the huge numerical value obtained from cubing the mean radius, and multiply it by \( \frac{4}{3} \pi \). This calculation results in an impressive volume of approximately \( 1.08321 \times 10^{21} \) cubic meters. Such large numbers illustrate Earth's extensive size. This measure helps in understanding both global scalings like climate modeling and local applications like resource distribution.

Mathematical Formulas

Mathematical formulas are structured expressions that provide a precise way to perform calculations. In our exercise, the formula for the volume of a sphere \[ V = \frac{4}{3} \pi r^3 \], serves as a powerful tool for understanding planet sizes and related fields. Each component of the formula - the fraction \( \frac{4}{3} \), \( \pi \) (an irrational number approximately equal to 3.14159), and the cube of the radius - has a specific role in scaling the calculation to find the correct volume.- **Fraction \( \frac{4}{3} \):** Adjusts the formula to account for the natural proportions of a sphere, which are more extensive than those of a simpler geometric shape, like a cube.- **\( \pi \):** Reflects the relationship between a circle's diameter and its circumference, helping extend circular concepts to three-dimensional objects.- **Cubing the Radius:** Converts the linear dimension of the radius into a volumetric measure.These components come together, illustrating how formulas are constructed to solve complex problems effectively.