Chapter 9.7, Problem 61E

To determine

To calculate:

Perihelion and aphelion distances.

Answer to Problem 61E

The polar equatoion of the orbit venus is $\frac{1.0820\times {10}^8}{1-0.0068\cos \theta }$ .

Perihelion distance is and Aphelion distance is $r=1.0747\times {10}^{8}Km$ and $r=1.0895\times {10}^8$ respectively.

Explanation of Solution

Given information:

Venus: $a\approx 1.0821\times {10}^{8}kilometeres,e\approx 0.0068$

Calculation:

To find the equation of the Venus orbit,

  $a\approx 1.0821\times {10}^{8}kilometeres,e\approx 0.0068$

As polar equation of the orbit of planets is

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

Where,

  $e=$ Eccentricity.

Perihelion distance (min) is $r=a\left(1-e\right)$

Aphelion distance is (max) $r=a\left(1+e\right)$

Now put the $a\approx 1.0821\times {10}^{8}kilometeres,e\approx 0.0068$ in equation $\left(1\right)$

Then,

  $\begin{array}{l}r=\frac{a\left(1-e^2\right)}{1-e\cos \theta }\\ =\frac{a\approx 1.0821\times {10}^8\left(1-0.0068\right)}{\left(1-0.068\right)\cos \theta }\\ =\frac{1.0820\times {10}^8}{1-0.0068\cos \theta }\\ =\frac{1.0820\times {10}^8}{1-0.0068\cos \theta }\end{array}$

Thus the polar equatoion of the orbit venus is $\frac{1.0820\times {10}^8}{1-0.0068\cos \theta }$ .

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

Consider the planet Venus with parameters as below:

  $a\approx 1.0821\times {10}^{8}kilometeres,e\approx 0.0068$

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

  $\begin{array}{l}r=a\left(1-e\right)\\ =1.0821\times {10}^8\left(1-0.0068\right)\\ =1.0747\times {10}^8\end{array}$

Perihelion distance is $r=1.0747\times {10}^{8}Km$

Maximum distance from the sun to Saturn is calculated by,

  $\begin{array}{l}r=a\left(1+e\right)\\ =1.0821\times {10}^8\left(1+0.0068\right)\\ =1.0895\times {10}^8\end{array}$

Aphelion distance is $r=1.0895\times {10}^8$