Problem 52 - Calculus Single Variable

Question

The astronomical unit (AU) is the mean distance of Earth to the sun. The table gives the mean distances \((D)\) in astronomical units between the sun and the six planets that were known at the time of Johannes Kepler. The table also provides the planets' periods of orbit about the sun \((T)\) in years. Plot the six points \((D, T)\). In the window \([0,10] \times[0,30],\) graph the function \(f_{q}(x)=x^{q}\) for each \(q=m / n\) with \(1 \leq m\) and \(n \leq 3 .\) For what value of \(q\) does the graph of \(f_{q}\) fit the data well? Notice that \(TD\) for \(D>1 .\) Explain why this indicates that \(q>1 .\) Which points show that \(q<2 ?\) $$ \begin{array}{|l|c|c|} \hline & \boldsymbol{T} & \boldsymbol{D} \\ \hline \text { Mercury } & 0.24 & 0.387 \\ \hline \text { Venus } & 0.615 & 0.723 \\ \hline \text { Earth } & 1.00 & 1.00 \\ \hline \text { Mars } & 1.88 & 1.524 \\ \hline \text { Jupiter } & 11.86 & 5.203 \\ \hline \text { Saturn } & 29.547 & 9.539 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The graph of \(f_q(x) = x^{3/2}\) fits the data well, indicating \(q\approx 1.5\), since \(T = D^{3/2}\).

Step 1: Plot Points

Plot the pairs \((D, T)\) for each planet on a graph with the x-axis representing the mean distance \(D\) and the y-axis representing the period \(T\). The points will be: Mercury (0.387, 0.24), Venus (0.723, 0.615), Earth (1.00, 1.00), Mars (1.524, 1.88), Jupiter (5.203, 11.86), Saturn (9.539, 29.547).

Step 2: Graph Function

Graph the function \(f_q(x) = x^q\) for each rational value of \(q = \frac{m}{n}\), where \(1 \leq m\), \(n \leq 3\). This includes different potential lines for \(q=1, \frac{4}{3}, \frac{3}{2}, 2\), etc.

Step 3: Compare Data to Function Graphs

Evaluate how well each function \(f_q(x)\) fits the plotted planet point data. Look for the curve where the plotted function closely matches the data points \((D, T)\).

Step 4: Determine if \(q>1\)

Notice that for \(D < 1\), such as Mercury and Venus, \(T < D\). For \(D > 1\), the periods such as for Mars, Jupiter, and Saturn show \(T > D\). This trend suggests \(q > 1\), since increasing \(D\) results in \(T\) increasing at a rate greater than linearly.

Step 5: Determine Upper Limit for q

Look at the data to determine when \(T \approx D^2\) isn't true. Notice that for larger values of \(D\), such as with Jupiter and Saturn, the actual \(T\) is less than what would be expected if \(T = D^2\). This indicates \(q < 2\).

Step 6: Choose Optimal q

Recognizing that Kepler's third law is approximately \(T^2 \propto D^3\), we fit the data by finding that \(T = D^{3/2}\) most accurately represents the data. This choice of \(q = \frac{3}{2}\) meets both trends defined in steps 4 and 5.

Key Concepts

Astronomical Unit

An astronomical unit, abbreviated as AU, is a unit of measurement that defines the average distance between the Earth and the Sun. It is a convenient way to measure and compare distances within our solar system because it presents these vast distances in manageable terms.
Typically, 1 AU is about 149.6 million kilometers or approximately 93 million miles. This distance is significant because it provides a baseline for comparing the distances of other planets from the Sun. For instance:

• Mercury, being closer to the Sun than Earth, is 0.387 AU away.
• Jupiter, which is much farther from the Sun than Earth, is 5.203 AU away.

Understanding the concept of an astronomical unit allows students to easily grasp the relative distances in our solar system and helps in visualizing planetary motion data like those provided in the table.

Planetary Motion

Planetary motion refers to the orbits of planets around a star, like the Sun. The concepts explored here are deeply connected to Kepler's laws, which describe how planets move in elliptical orbits and how their speeds vary based on their distance from the sun.
Johannes Kepler formulated three essential laws in the early 17th century, and his third law is particularly important in this context. Kepler's Third Law states, 'The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.' Translated into simpler terms, it means:

• The longer the distance of a planet from the Sun (like how Jupiter and Saturn are much farther away compared to Mercury and Venus), the longer the time it takes to complete one orbit around the Sun.
• It implies a predictable, mathematical relationship between a planet's distance from the Sun (D, in astronomical units) and its orbital period (T, in years).

In the exercise, the task is to observe the data comparing the distances (D) and periods (T), plotting them, and exploring functions that most accurately describe their relationship, which is an embodiment of Kepler's third law at work.

Exponential Functions

Exponential functions are mathematical expressions involving an exponent, which allows us to model growth, decay, or in this exercise, a relationship between two variables—in this case, distance and orbital period.
In the exercise, the function of interest is in the form of \( f_q(x) = x^q \), where \(q\) is a rational number. This type of function helps describe how the periods of the planets relate to their mean distances from the Sun.
The steps involved evaluating different values for \(q\) to determine which represents the data most precisely:

• By analyzing points where \(T < D\) and \(T > D\), we infer that \(q > 1\), meaning that the orbital period increases more rapidly than the distance when \(D > 1\).
• However, since we cannot describe the data where \(T \approx D^2\) (for larger planets without overestimating), it also indicates \(q < 2\). This leads us to an optimal choice of \(q = \frac{3}{2}\).

This proportionality function \(T = D^{3/2}\) closely matches Kepler's Third Law, reinforcing the relationship discovered through historical astronomical research.