Answer
(a) See explanations.
(b) perihelion $1.47\times10^8$ km, aphelion $1.52\times10^8$ km
Work Step by Step
(a) Step 1. Recall the equation given in Exercise 43(a) as $r=\frac{a(1-e^2)}{1-e\cdot cos\theta}$
Step 2. Use the figure given in the Exercise, the perihelion happens when $\theta=\pi$, plug-in the equation above to get $r=\frac{a(1-e^2)}{1+e}=a(1-e)$.
Step 3. Similarly, the aphelion happens when $\theta=$, plug-in the equation above to get $r=\frac{a(1-e^2)}{1-e}=a(1+e)$.
(b) Step 1. Recall the quantities given in Exercise 43(b) as $e=0.017$ and $2a=2.99\times10^8$
Step 2. The distances from the earth to the sun at perihelion is given by $d_1=a(1-e)=(1-0.017)1.495\times10^8\approx1.47\times10^8$ km
Step 3. Similarly, The distances from the earth to the sun at aphelion is given by $d_2=a(1+e)=(1+0.017)1.495\times10^8\approx1.52\times10^8$ km
