Chapter 9, Problem 41RE

To determine

Calculate the eccentricity of the orbit.

Answer to Problem 41RE

The eccentricity of the orbit is $0.0536$ .

Explanation of Solution

Given:

It is given in the question that a saturn moves in an elliptical orbit with the sun at one focus. The least distance and the greatest distance of the planet from the sun are $1.3498\times {10}^{9}and1.5035\times {10}^9$ .

Concept Used:

In this, use the concept of linear equation and the eccentricity formula $e=\frac{c}{a}$ .

Calculation: Let the largest distance is equal to a + c and the shortest distance is equal to a − c.

Now,

  $\begin{array}{l}a+c=1.5035\times {10}^9(i)\\ a-c=1.3498\times {10}^9(ii)\end{array}$

Now, add the above two equations,

  $\begin{array}{l}2a=2.854\times {10}^9\\ a=\frac{2.854\times {10}^9}{2}=1.427\times {10}^9\end{array}$

Now, use the above term to find c,

  $\begin{array}{l}c=1.5035\times {10}^9-a\\ c=1.5035\times {10}^9-1.427\times {10}^9\\ c=0.0765\times {10}^9\\ c=7.65\times {10}^7\end{array}$

Now, use the eccentricity formula,

  $\begin{array}{l}e=\frac{c}{a}\\ e=\frac{7.65\times {10}^7}{1.427\times {10}^9}\\ e=5.360\times {10}^{-2}\\ e=0.0536\end{array}$

Conclusion:

The eccentricity is $0.0536$ .