Chapter 1: Problem 348
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 \((0,0) .)\)
Short Answer
Step by step solution
Understand the Problem
Recall the Equation of a Hyperbola
Calculate Semi-major Axis (a)
Calculate Semi-minor Axis (b)
Formulate the Hyperbola Equation
Confirm the Comet's Return
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Eccentricity
- 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.
Perihelion
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 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
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.