Chapter 9, Problem 33CR

To determine

To prove: The eccentricity of the orbit for Saturn is 0.0565

Explanation of Solution

Given information:

The least distance and the greatest distance of the planet from the sun are $1.3526\times {10}^9$ and $1.5145\times {10}^9$ kilometers.

Formula used:

Eccentricity $\left(e\right)=\frac{c}{a}$

Proof:

The least distance of the planet from the sun is $1.3526\times {10}^9$ which means $a-c=1.3526\times {10}^9$

The greatest distance of the planet from the sun is $1.5145\times {10}^9$ which means $a+c=1.5145\times {10}^9$

Now, the given value

  $\begin{array}{l}a+c=1.5145\times {10}^9\\ a-c=1.3526\times {10}^9\\ 2a=2.8671\times {10}^9\\ a=1.43355\times {10}^9\end{array}$

Now,

  $\begin{array}{l}a+c=1.5145\times {10}^9\\ c=1.5145\times {10}^9-1.43355\times {10}^9\\ c=0.08095\times {10}^9\end{array}$

The eccentricity of the orbit can be obtained as,

  $\begin{array}{l}e=\frac{c}{a}=\frac{0.08095\times {10}^9}{1.43355\times {10}^9}\\ \approx 0.0565\end{array}$

Therefore, the eccentricity for the Saturn is $0.0565\left(\text{proved}\right)$