Chapter 9.7, Problem 60E

To determine

To calculate:

Perihelion and aphelion distances.

Answer to Problem 60E

Polar equation is $\frac{1.4228\times {10}^9}{1-0.0542\cos \theta }$ .

Perihelion distance and Aphelion distance is $r=1.3497\times {10}^{9}Km$ and $r=1.5043\times {10}^{9}km$ respectively.

Explanation of Solution

Given information:

Saturn: $a\approx 1.4267\times {10}^{9}kilometeres,e\approx 0.0539$

Calculation:

Consider the planet travelling in elliptical orbits with the sun at one of its focus.

Lets assume that the focus is at the pole, major axis lies on the polar axis and the length of the major axis is $2a$ .

Then the polar equation of the orbit is given by,

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

Where,

  $e=$ Eccentricity.

Figure is given as below:

  

Maximum distance also called as perihelion distance from the sun to the planet is $r=a\left(1-e\right)$ and max distance (aphelion distance) from the sun to the planet is $r=a\left(1+e\right)$ .

Consider the planet Saturn with parameters as below:

  $a\approx 1.4267\times {10}^{9}km,e\approx 0.0539$

So substitute the values in equation $\left(1\right)$ .

  $\begin{array}{l}r=\frac{a\left(1-e^2\right)}{1-e\cos \theta }\\ =\frac{1.427\times {10}^9\left(1-{\left(0.0542\right)}^2\right)}{1-\left(0.0542\right)\cos \theta }\\ =\frac{1.427\times {10}^9\times 0.997}{1-\left(0.0542\right)\cos \theta }\\ =\frac{1.4228\times {10}^9}{1-0.0542\cos \theta }\end{array}$

So minimum distance from the sun to Saturn is calculated by,

  $\begin{array}{l}r=a\left(1-e\right)\\ =1.427\times {10}^9\left(1-0.0542\right)\\ =1.427\times {10}^9\left(0.9458\right)\\ =1.3497\times {10}^9\end{array}$

Perihelion distance is $r=1.3497\times {10}^{9}km$

Maximum distance from the sun to Saturn is calculated by,

  $\begin{array}{l}r=a\left(1+e\right)\\ =1.427\times {10}^9\left(1+0.0542\right)\\ =1.427\times {10}^9\left(1.0542\right)\\ =1.5043\times {10}^9\end{array}$

Aphelion distance is $r=1.5043\times {10}^{9}km$