Problem 112 - Algebra and Trigonometry

Question

For each planet in our solar system, its year is the time it takes the planet to revolve once around the Sun. The formula \(E=0.2 x^{\frac{3}{2}}\) models the number of Earth days in a planet's year, E, where x is the average distance of the planet from the Sun, in millions of kilometers. Use the equation to solve Exercises \(111-112\). There are approximately 88 Earth days in the year of the planet Mercury. What is the average distance of Mercury from the Sun? Use a calculator and round to the nearest million kilometers.

Step-by-Step Solution

Answer

The average distance of Mercury from the Sun is approximately 58 million kilometers.

Step 1: Substitute the given value into the equation

Substitute the value of E (88 Earth days) in the formula \(E=0.2 x^{\frac{3}{2}}\), getting \(88=0.2 x^{\frac{3}{2}}\).

Step 2: Rearrange the equation

Rearrange the equation to make 'x' the subject. Divide both sides by 0.2 to get \(x^{\frac{3}{2}}=\frac{88}{0.2}\)

Step 3: Solve for 'x'

Take the cubed root of both sides (or raise both sides to the power of \(\frac{2}{3}\), the inverse of \(\frac{3}{2}\)) to get \(x=(\frac{88}{0.2})^{\frac{2}{3}}\). Use your calculator to find the numerical value of 'x'.

Step 4: Round to the nearest million

Finally, round 'x' to the closest millions to find the average distance of Mercury from the Sun in millions of kilometers.

Key Concepts

Astronomical Mathematics

Exploring the vast cosmos requires not only telescopes but also the tools of mathematical astronomy. Using equations and mathematical modeling, astronomers are able to translate celestial observations into tangible concepts. An exceptional example is the formula used to calculate the length of a planet's year based on its distance from the sun, which is expressed as:
\[\begin{equation}E = 0.2 x^{\frac{3}{2}}\end{equation}\]
In this equation, E represents the orbital period of the planet in Earth days, while x denotes the planet's average distance from the Sun in millions of kilometers. By manipulating this equation, astronomers can estimate various orbital characteristics, which further highlight the synergy between theoretical mathematics and practical observations in space exploration.

Importance in Predicting Orbital Characteristics

Predicting orbital characteristics allows scientists to understand the dynamical behavior of planets, enabling them to forecast positions, track satellite orbits and even predict eclipses. Simplifying complex celestial mechanics into a manageable equation showcases the potency of astronomical mathematics in unlocking the universe's secrets.

Exponential Equations

In the realm of mathematics, exponential equations like the one used for planetary motion often crop up. They are a type of algebraic equation in which variables appear as exponents, such as:
\[\begin{equation}x^{\frac{3}{2}}\end{equation}\]
In the given exercise, this form is used to describe how the year-length of a planet scales with its distance from the Sun. Solving exponential equations typically involves isolating the base (x in this case) and using inverse operations. For example, taking the cubed root (or raising to the power of \(\tfrac{2}{3}\)) effectively reverses the exponent effect, allowing us to solve for x.

Application to Planetary Motion

Our exercise applies this technique to find the distance of Mercury from the Sun. Starting with \[88 = 0.2 x^{\frac{3}{2}}\], we rearrange and apply the inverse operation to extract the value of x. Exponential equations like this are essential tools expressing growth or decay in natural phenomena, including population dynamics, radioactive decay, and, as shown here, spacing in our solar system.

Scientific Notation in Algebra

In algebra, particularly when dealing with astronomically large or small numbers, scientific notation is invaluable. This method of writing numbers simplifies calculations and expresses figures clearly, such as distances in space or microscopic scales. For example, the average distance of the Earth from the Sun is roughly \(1.496 \times 10^8\) kilometers, far more readable than writing out all the digits.
The benefit of using scientific notation is particularly evident when interpreting exponential equations in astronomy. Scientific notation keeps numbers manageable, given the vast sizes, distances, and time scales inherent to celestial measurements. It's crucial in ensuring accuracy and enabling efficiency when handling calculations in exercises like determining planetary distances. By combining exponential expressions and scientific notation, astronomical mathematics maintains precision without losing the essence of the astronomical scales involved.