*If the Earth's orbit is divided in two by the latus rectum, show that the difference in time spent in the two halves, in years, is \frac{2}{\pi}\left(e \sqrt{1-e^{2}}+\arcsin e\right) and hence for small e about twice as large as the difference computed in the example in \(\S 4.4\). (Hint: Use Cartesian co-ordinates to evaluate the required area. The identity \(\pi / 2-\arcsin \sqrt{1-e^{2}}=\arcsin e\) may be useful.)

The Earth's orbit can be divided into two halves by the latus rectum, a line segment perpendicular to the major axis of an ellipse that passes through either of the foci. We know the orbit has an eccentricity of \(e\). Let's label the two halves of the orbit as A and B. In half A, the Earth is closer to the Sun, and in half B, it is farther away. The problem asks us to calculate the difference in time spent in these two halves.

Let's first find the area of each half of the Earth's orbit. We will later use these areas to find the difference in time spent. To find the area of half A and B, we can use the polar equation of an ellipse: $$r(\theta)=\frac{p}{1+e\cos(\theta)}$$ where \(r(\theta)\) is the distance between the location of the Earth and the Sun as a function of the angle \(\theta\) and \(p\) is the latus rectum. Since the orbit is divided by the latus rectum, we will evaluate the area from \(\theta=0\) to \(\theta=\pi\) for half A and from \(\theta=\pi\) to \(\theta=2\pi\) for half B. Let Area_A be the area of half A and Area_B be the area of half B. To calculate these areas, we can use the integral formula for the area of a polar curve: $$\text{Area} = \frac{1}{2}\int_\alpha^\beta r^2(\theta) d\theta$$ where \(\alpha\) is the lower limit and \(\beta\) is the upper limit of the angle \(\theta\). Thus, we can calculate Area_A and Area_B using the following integrals: Area_A: $$\frac{1}{2}\int_0^\pi \left(\frac{p}{1+e\cos(\theta)}\right)^2 d\theta$$ Area_B: $$\frac{1}{2}\int_\pi^{2\pi} \left(\frac{p}{1+e\cos(\theta)}\right)^2 d\theta$$ Now, we need to find the difference in areas, which we will call Area_diff: $$\text{Area_diff} = \text{Area_A} - \text{Area_B}$$

The time spent in each half of the orbit will be proportional to the area of that half. Therefore, the difference in time spent in years can be found by dividing Area_diff by the total area of the Earth's orbit and multiplying by the orbital period (1 year). Time_diff: $$\frac{\text{Area_diff}}{\text{Total Area}}\times (1\,\text{year})$$ Now, we need to find an expression for Time_diff in terms of eccentricity \(e\). To do this, evaluate the above integrals for Area_A and Area_B, and thus find the Area_diff. You can then simplify the expression for Time_diff using the given identity in the hint: $$\pi / 2-\arcsin \sqrt{1-e^{2}}=\arcsin e$$ Finally, show that the difference in time spent, Time_diff, is equal to: $$\frac{2}{\pi}\left(e \sqrt{1-e^{2}}+\arcsin e\right)$$

We are now asked to compare the above result for Time_diff with the difference computed in the example in § 4.4 for small eccentricities \(e\). When \(e\) is small, the two portions of the orbit will be almost equal, and the difference in time spent will also be small. By expanding the expression for Time_diff using the small \(e\) approximation, show that the difference in time spent in the two halves of the orbit is about twice as large as the difference computed in the example in § 4.4.