Problem 29 - Calculus Early Transcendentals

Question

The planet Mercury travels in an elliptical orbit with eccen- tricity \(0.206 .\) Its minimum distance from the sun is \(4.6 \times 10^{7} \mathrm{km}\) . Find its maximum distance from the sun.

Step-by-Step Solution

Answer

Mercury's maximum distance from the Sun is approximately \(6.985 \times 10^{7}\) km.

Step 1: Understand Orbital Elements

The orbit of Mercury is an ellipse with the Sun at one of the foci. The eccentricity (\(e\)) of an orbit is a measure of its deviation from being circular. The minimum distance (perihelion) is given as \(r_{min} = 4.6 \times 10^{7}\, ext{km}\).

Step 2: Apply the Formula for Orbital Distances

For an elliptical orbit, the relationship between the maximum (\(r_{max}\)) and minimum (\(r_{min}\)) distance from the focus is given by the formula:\[\frac{r_{max} - r_{min}}{r_{max} + r_{min}} = e\]where \(e = 0.206\) is the eccentricity of Mercury's orbit.

Step 3: Substitute Known Values

Plug the known values into the orbital distance formula:\[\frac{r_{max} - 4.6 \times 10^{7}}{r_{max} + 4.6 \times 10^{7}} = 0.206\]

Step 4: Solve for Maximum Distance

Rearrange the equation from Step 3 to solve for \(r_{max}\):Assume \(r_{max} = x\):\[\frac{x - 4.6 \times 10^{7}}{x + 4.6 \times 10^{7}} = 0.206\]Multiply both sides by the denominator:\[x - 4.6 \times 10^{7} = 0.206 (x + 4.6 \times 10^{7})\]Expand and simplify:\[x - 4.6 \times 10^{7} = 0.206x + 0.206 \times 4.6 \times 10^{7}\]Simplifying gives:\[0.794x = 4.6 \times 10^{7} + 0.206 \times 4.6 \times 10^{7}\]Calculate:\[0.794x = 4.6 \times 10^{7} \times (1 + 0.206)\]\[x = \frac{4.6 \times 10^{7} \times 1.206}{0.794}\]Finally, calculate the numerical value for \(r_{max}\).

Step 5: Calculate the Result

Calculate:\[r_{max} = \frac{4.6 \times 10^{7} \times 1.206}{0.794}\]First, multiply: \(4.6 \times 1.206 = 5.5476\). Then,\(5.5476 \times 10^{7} = 55.476 \times 10^{6}\).Divide by \(0.794\):\[r_{max} = \frac{55.476 \times 10^{6}}{0.794} \approx 6.985 \times 10^{7} \text{km} \].Thus, the maximum distance from the Sun is approximately \(6.985 \times 10^{7} \text{ km}\).

Key Concepts

Understanding Eccentricity in Elliptical Orbits

Eccentricity is a crucial concept when dealing with elliptical orbits in astronomy. It tells us how stretched or elongated an orbit is compared to a perfect circle. An eccentricity value of 0 means a perfect circle, while values approaching 1 indicate more elongated ellipses. This elongation is essential in understanding the shape of planetary orbits.

For planets like Mercury, this eccentricity is relatively high at 0.206, meaning its orbit is quite elongated. This affects how it moves in space because the distance from the Sun is not constant. Mercury's path takes it both close to and far from the Sun as it travels along its orbit. This variation in distance is core to understanding how planetary orbits work.

In practical terms, eccentricity helps scientists calculate and predict the positions and distances of planets at various points in their orbits, helping us to better understand their motion and potential influences on their environments.

Perihelion: Minimum Distance from the Sun

The term 'perihelion' refers to the point in the orbit of a planet where it is closest to the Sun. This is a pivotal part of understanding elliptical orbits because it marks the minimum distance in the path that a planet travels. For Mercury, the perihelion distance is given as \(4.6 \times 10^{7}\) km.

This point is significant not just because it's the closest point but also because it tells us a lot about the orbital speed of the planet. Planets travel faster when they are nearer to the Sun, at perihelion, due to stronger gravitational pull. Conversely, they move slower when they are farther away.

Understanding perihelion helps astronomers make comprehensive calculations about the orbital dynamics of planets, which are used to predict phenomena such as solar transits, where Mercury passes directly between Earth and the Sun as seen from Earth.

Calculating Maximum Distance in Elliptical Orbits

Understanding how to calculate the maximum distance (or aphelion distance) of a planet from the Sun is an important exercise, especially when considering the extremes of its orbital path. This maximum distance can be determined using the orbit's eccentricity and the perihelion distance.

In the problem at hand, a formula is used:

• \( \frac{r_{max} - r_{min}}{r_{max} + r_{min}} = e \)

This formula links the maximum and minimum distances \( r_{max} \) and \( r_{min} \) with the eccentricity \( e \) of the orbit.

By rearranging this formula and using Mercury's known perihelion distance and eccentricity, one can solve for \( r_{max} \). This process involves a few algebraic steps:

• First, plug in the known values.
• Next, solve for \( r_{max} \) by isolating it on one side of the equation.
• Finally, calculate the numerical result using arithmetic.

For Mercury, this calculation resulted in a maximum distance of approximately \( 6.985 \times 10^{7} \text{ km} \). This distance gives us the furthest point along its elliptical orbit from the Sun, providing a complete picture of its orbital path.