Show that the comet discussed at the end of \(\$ 4.4\) crosses the Earth's orbit at opposite ends of a diameter. Find the time it spends inside the Earth's orbit. (To evaluate the area required, write the equation of the orbit in Cartesian co-ordinates. See Appendix B.)

We know from the exercise statement that the comet's orbit has an equation in polar coordinates given by \(r = \frac{l}{1 + \varepsilon \cos(\theta)}\). To find the Cartesian coordinates, we will use the transformation equations: \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\) Substituting the value of \(r\) from the comet's orbit equation, we get: \(x = \frac{l\cos(\theta)}{1 + \varepsilon \cos(\theta)}\) and \(y = \frac{l\sin(\theta)}{1 + \varepsilon \cos(\theta)}\) Now, we need to eliminate \(\theta\) to obtain a single equation in \(x\) and \(y\). The easiest way to do this is to use the Pythagorean identity \(\sin^2(\theta) + \cos^2(\theta) = 1\).

Divide the equation for \(y\) by the equation for \(x\): \(\frac{y}{x} = \frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta)\) Now square both sides and use the Pythagorean identity: \(\frac{y^2}{x^2} = \tan^2(\theta) = \frac{\sin^2(\theta)}{\cos^2(\theta)} = \frac{1 - \cos^2(\theta)}{\cos^2(\theta)}\) Now substitute the expression for \(\cos(\theta)\) from the equation for \(x\): \(\frac{y^2}{x^2} = \frac{1 - (\frac{l}{x})^2}{(\frac{l}{x})^2}\) Clear the denominators and rearrange the equation in the standard form of a conic section: \((1 - \varepsilon^2)x^2 + 2 \varepsilon l y = 0\)

The equation derived in Step 2 represents an ellipse, as the square of the eccentricity is always less than 1 \((0 < \varepsilon^2 < 1)\). To show that the comet's orbit intersects the Earth's orbit at opposite ends of a diameter, we need to prove that the line \(y = 0\) (Earth's orbit) intersects the ellipse at two points diametrically opposite to each other. Substitute \(y = 0\) into the equation of the ellipse: \((1 - \varepsilon^2)x^2 = 0\) This equation has two solutions: \(x = \frac{l}{\varepsilon}\) and \(x = -\frac{l}{\varepsilon}\). These two points are diametrically opposite on the ellipse, which implies that the comet crosses the Earth's orbit at opposite ends of a diameter.

To find the time the comet spends inside Earth's orbit, we need to find the time interval between the intersection points. We will use the fact that the area swept by the radius vector is directly proportional to the time elapsed. Let \(\tau\) be the time spent by the comet inside the Earth's orbit, and \(A\) be the area of the elliptical segment corresponding to this time interval. Since the area of the ellipse can be found using the formula $$A_\text{ellipse} = \pi AB$$, where \(A\) is the length of the major axis, \(B\) is the length of the minor axis, and the entire orbital period is \(T\), the required time \(\tau\) is given by the ratio of the areas: $$\frac{\tau}{T} = \frac{A}{A_\text{ellipse}}$$ The area \(A\) of the elliptical segment is half of the ellipse's total area since the comet crosses the Earth's orbit at opposite ends of a diameter. Hence, $$\tau = \frac{1}{2}T$$ So, the time spent by the comet inside Earth's orbit is half of its orbital period.