Chapter 8.2, Problem 58E

Solution

The eccentricity of the orbit.

It has been determined that the eccentricity of the orbit is approximately $0.9677$ .

Given:

The orbit of Halley’s comet is $36.18$ AU long and $9.12$ AU wide.

Concept used:

The eccentricity of an ellipse is given as $e=\frac{\sqrt{a^2-b^2}}{a}$ , where $a$ is the semi-major axis and $b$ is the semi-minor axis.

Calculation:

It is given that the orbit of Halley’s comet is $36.18$ AU long and $9.12$ AU wide.

Clearly, the orbit is elliptic.

Since the orbit is $36.18$ AU long, the semi-major axis is $a=\frac{36.18}{2}=18.09$ AU.

Since the orbit is $9.12$ AU wide, the semi-minor axis is $b=\frac{9.12}{2}=4.56$ AU.

Put $a=18.09$ and $b=4.56$ in $e=\frac{\sqrt{a^2-b^2}}{a}$ to get,

  $\begin{array}{c}e=\frac{\sqrt{{\left(18.09\right)}^2-{\left(4.56\right)}^2}}{18.09}\\ =\frac{\sqrt{327.2481-20.7936}}{18.09}\\ =\frac{\sqrt{306.4545}}{18.09}\\ \approx \frac{17.5058}{18.09}\end{array}$

Solving,

  $e\approx 0.9677$

This implies that the eccentricity of the orbit is approximately $0.9677$ .

Conclusion:

It has been determined that the eccentricity of the orbit is approximately $0.9677$ .