Chapter 11.6, Problem 44E

a.

To determine

To show: that the perihelion distance from a planet to the sun is $a\left(1-e\right)$ and the aphelion distance is $a\left(1+e\right)$ .

Explanation of Solution

Given information:

The polar equation is $r=\frac{a\left(1-e^2\right)}{1-e\cos \theta }$ .

Proof: since, the polar equation for the elliptical orbit of the earth around the sun is $r=\frac{a\left(1-e^2\right)}{1-e\cos \theta }$ ,

so the equation can be written as,

  $\begin{array}{l}r=\frac{a\left(1-e^2\right)}{1-e\cos \theta }\\ r=\frac{a\left(1-e\right)\left(1+e\right)}{1-e\cos \theta }\end{array}$

Now, from the equation, the perihelion distance from a planet to the sun is $a\left(1-e\right)$ and the aphelion distance is $a\left(1+e\right)$ .

b.

To determine

To find: the distances from the earth to the sun at perihelion and at aphelion.

Answer to Problem 44E

The distance form the earth to the sun at perihelion and aphelion is $2.93\times {10}^8\text{km and }3.04\times {10}^8\text{km}$ .

Explanation of Solution

Given information:

The equation is $r=\frac{1.49\times {10}^8}{1-\left(0.017\right)\cos \theta }$ .

Calculation: to find the approximate polar equation, substitute in the formula,

  $\begin{array}{l}r=\frac{a\left(1-e^2\right)}{1-e\cos \theta }\\ r=\frac{\left(2.99\times {10}^8\right)\left(1-\left(0.017\right)\right)\left(1+0.017\right)}{1-\left(0.017\right)\cos \theta }\text{ [substitute]}\\ r=\frac{2.99\times {10}^8\left(0.983\right)\left(1.017\right)}{1-\left(0.017\right)\cos \theta }\text{ [simplify]}\end{array}$

Thus, the distance form the earth to the sun at perihelion and aphelion is $2.93\times {10}^8\text{km and }3.04\times {10}^8\text{km}$ .