Problem 348 - Calculus Volume 3

Question

The C/1980 E1 comet was observed in 1980. Given an eccentricity of 1.057 and a perihelion (point of closest approach to the Sun) of \(3.364 \mathrm{AU},\) find the Cartesian equations describing the comet's trajectory. Are we guaranteed to see this comet again? (Hint: Consider the Sun at point \((0,0) .)\)

Step-by-Step Solution

Answer

The Cartesian equation is \(\frac{x^2}{3484.26} - \frac{y^2}{405.62} = 1\). The comet will not return as it is on a hyperbolic path.

Step 1: Understand the Problem

We need to find the Cartesian equation of a comet's hyperbolic trajectory around the Sun given its eccentricity and perihelion. Eccentricity (\(e > 1\)) indicates a hyperbolic orbit, meaning the comet may not return. The Sun is at the origin \((0,0)\).

Step 2: Recall the Equation of a Hyperbola

The standard form equation of a hyperbola centered at the origin with the Sun at focus is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). We need to find the values for \(a\) and \(b\) using given information.

Step 3: Calculate Semi-major Axis (a)

The relationship between semi-major axis \(a\), perihelion \(q\), and eccentricity \(e\) is given by \(q = a(1 - e)\), where \(q = 3.364 \ AU\) and \(e = 1.057\). Solve for \(a\):\[a = \frac{q}{1 - e} = \frac{3.364}{1 - 1.057} = \frac{3.364}{-0.057} \approx -59.03 \ AU\]

Step 4: Calculate Semi-minor Axis (b)

Use the relationship between semi-major axis \(a\), semi-minor axis \(b\), and eccentricity \(e\) in hyperbolic orbits: \(b = a \sqrt{e^2 - 1}\). Substitute \(a\) and \(e\):\[b = (-59.03) \times \sqrt{1.057^2 - 1} \approx -59.03 \times \sqrt{0.116449} \approx -59.03 \times 0.3413 \approx -20.14 \ AU\]

Step 5: Formulate the Hyperbola Equation

Substitute \(a\) and \(b\) into the hyperbola equation:\[\frac{x^2}{(-59.03)^2} - \frac{y^2}{(-20.14)^2} = 1\]Simplify:\[\frac{x^2}{3484.26} - \frac{y^2}{405.62} = 1\]

Step 6: Confirm the Comet's Return

Since the eccentricity \(e = 1.057 > 1\), the comet's trajectory is hyperbolic. This indicates it is on an escape trajectory and is not gravitationally bound to the Sun, meaning the comet will not return.

Key Concepts

Eccentricity

Eccentricity is a measure that describes how much an orbit deviates from being circular. In celestial mechanics, eccentricity ( \(e\) ) helps determine the shape of the orbit.

• If \(e = 0\) , the orbit is a perfect circle.
• If \(0 < e < 1\) , the orbit is elliptical.
• If \(e = 1\) , the orbit is parabolic.
• If \(e > 1\) , the orbit is hyperbolic.

In this particular problem, the given eccentricity is \(e = 1.057\) , which is greater than 1. This hyperbolic eccentricity indicates that the comet C/1980 E1 is on an escape trajectory. It implies that it is not gravitationally bound to the Sun and, therefore, is not expected to return once it passes by. The hyperbolic trajectory means the comet was likely a visitor from outside the solar system.

Perihelion

The perihelion is the point in the orbit of a celestial body (such as a comet or planet) where it is closest to the Sun. It plays a crucial role in determining various parameters of the orbit, including the semi-major axis.
For the comet C/1980 E1, the perihelion is given as \(3.364 \, \mathrm{AU}\) (Astronomical Units). This is the closest distance between the comet and the Sun during its orbit.
Knowing the perihelion distance provides insights into how close the object will get to the Sun. In this case, because the comet has a high eccentricity and is following a hyperbolic trajectory, the perihelion represents just one point along its path as it swings close to the Sun before continuing out into space, not to return.

Semi-major Axis

The semi-major axis is one half of the longest diameter of an ellipse. It plays a significant role in defining the size of the orbit.
The relationship between the semi-major axis \(a\) , perihelion \(q\) , and eccentricity \(e\) for hyperbolic trajectories is given by: \[ q = a(1 - e) \]Solving for \(a\) with \(q = 3.364 \, \mathrm{AU}\) and \(e = 1.057\) , we get: \[ a = \frac{3.364}{1 - 1.057} = \frac{3.364}{-0.057} \approx -59.03 \, \mathrm{AU} \]The negative value of \(a\) reflects the nature of the hyperbolic orbit, indicating the path extends infinitely away from the Sun. Unlike elliptical orbits, semi-major axis in hyperbolic orbits doesn't signify a physical length but rather a way to quantify the trajectory's escape characteristics.

Semi-minor Axis

The semi-minor axis is the shortest radius along an ellipse or hyperbola. Although not present in non-elliptical paths, it helps in the mathematical representation of these trajectories.
In hyperbolic trajectories, the semi-minor axis \(b\) is related to the semi-major axis \(a\) and eccentricity \(e\) by the equation: \[ b = a \sqrt{e^2 - 1} \]Using the previously calculated \(a = -59.03 \, \mathrm{AU}\) and \(e = 1.057\), we find: \[ b \approx -59.03 \times \sqrt{0.116449} \approx -59.03 \times 0.3413 \approx -20.14 \, \mathrm{AU} \]Again, the negative sign simply underscores the characteristics of a hyperbolic orbit, pointing to the fact that unlike circles and ellipses, hyperbolas do not enclose an area but extend infinitely.