Problem 57 - Algebra 1

Question

SIZE OF JUPITER Jupiter, the largest planet in our solar system, has a radius of about \(4.4 \times 10^{4}\) miles. Use the equation \(V=\frac{4}{3} \pi r^{3}\) to find Jupiter's volume.

Step-by-Step Solution

Answer

Therefore, the volume of Jupiter is approximately \(3.563 \times 10^{15}\) cubic miles.

Step 1: Understand the Given

We are given the radius of the Jupiter as \(4.4 \times 10^{4}\) miles, and required to find its volume using the formula for the volume of a sphere which is \(V=\frac{4}{3} \pi r^{3}\).

Step 2: Substitution

By substituting the given radius of the Jupiter into this formula, we obtain \(V=\frac{4}{3} \pi (4.4 \times 10^{4})^{3}\).

Step 3: Calculate the Volume

By applying the exponentiation \(r^{3}=(4.4 \times 10^{4})^{3} = 8.51 \times 10^{13}\), then multiplying by \(\frac{4}{3} \pi\), we find that the volume of the Jupiter is \(V \approx 3.563 \times 10^{15}\) cubic miles.

Key Concepts

Volume of a Sphere

Calculating the volume of a sphere is a key concept in geometry. A sphere is a perfectly symmetrical shape in three dimensions, and its volume measures the amount of space it occupies. To find the volume, we use the formula:

• \[ V = \frac{4}{3} \pi r^3 \]

Here, \(V\) represents the volume, \(\pi\) (approximately 3.14159) is a mathematical constant, and \(r\) is the radius of the sphere. It's crucial to cube the radius in the equation, which means multiplying it by itself twice.

This formula essentially tells us how the volume expands when the radius increases. Since the radius is cubed, even a small change in the radius significantly impacts the volume. In the case of Jupiter, as the radius of this massive planet is \(4.4 \times 10^4\) miles, using this formula helps us calculate its colossal volume.

Exponential Functions

Exponential functions are widely used in mathematics to represent phenomena with rapid growth or decay. An exponential function typically takes the form of \(a^x\), where \(a\) is a constant, and \(x\) is the variable raised to that constant. In the context of calculating volumes, exponential growth is seen in cubing the radius of a sphere.

When we cube the radius, as in \(r^3\), it involves exponential functions where

• \( r = 4.4 \times 10^4 \)
• \(r^3 = (4.4 \times 10^4)^3\)
• \(r^3 = 8.51 \times 10^{13}\)

This step illustrates not only multiplication but also exponentiation, showcasing how quickly numbers can expand. The larger the base number, the greater the result when cubed, highlighting the power of exponentiation in measuring astronomical objects like planets.

Mathematical Substitution

Mathematical substitution is a technique used to simplify complex equations by replacing variables with known values. This allows you to calculate specific results. In our exercise, we substitute the given radius of Jupiter into the volume formula for a sphere.

• Start by identifying the given values: here, \(r = 4.4 \times 10^4\) miles.
• Substitute \(r\) in the formula: \(V = \frac{4}{3} \pi (4.4 \times 10^4)^3\).
• Calculate \(r^3\), then multiply by \(\frac{4}{3}\pi\) to conclude the calculation.

This substitution method ensures accurate calculations, as you replace unknowns with specifics to find concrete outcomes. Understanding and applying substitution in math ensures that complex computations become manageable, making it an essential skill in technical subjects.