Chapter 9.4, Problem 2E

To determine

To find: the planet with the least circular orbit, length of major and minor axis, ratio of these lengths and sketch the graph.

Answer to Problem 2E

The major axis is $115,818,100$ , The minor axis is $113,343,046$ and Ratio of the lengths is $1.0218$ .

Explanation of Solution

Given:

Mercury semi major axis length is $57,909,050\text{km}$ .

Eccentricity is $0.205630$ .

Aphelion is $69,816,900\text{ km}$ .

Perihelion is $49,001,200\text{ km}$ .

Concept used:

According mathematic measurement:

Mercury semi major axis length is $57,909,050\text{km}$ .

Eccentricity is $0.205630$ .

Aphelion is $69,816,900\text{ km}$ .

Perihelion is $49,001,200\text{ km}$ .

The major and minor axis of the orbit in elliptical shapes can be solved through the formula:

  $\epsilon =\frac{\sqrt{a^2-b^2}}{a}$ .

Calculation:

The formula to find the major and minor axis of the orbit in elliptical shapes is:

  $\begin{array}{l}\epsilon =\frac{\sqrt{a^2-b^2}}{a}=\sqrt{1-{\left(\frac{b}{a}\right)}^2.}\\ b=a\sqrt{1-{\epsilon }^2}.\end{array}$

As, my mathematical measurement:

Mercury semi major axis length is $57,909,050\text{km}$ $=2a=2\times 57,909,050=115,818,100$ .......... $\left(i\right)$

Eccentricity is $0.205630$ ….......... $\left(ii\right)$

Aphelion is $69,816,900\text{ km}$ .

Perihelion is $49,001,200\text{ km}$ .

  $b=a\sqrt{1-{\epsilon }^2}$ .

Putting the value from $\left(i\right)$ and $\left(ii\right)$ :

  $\begin{array}{l}b=a\sqrt{1-{\epsilon }^2}.\\ b=57,909,050\sqrt{1-{0.305630}^2}.\\ b=56,671,523\text{ km}\end{array}$

The major axis= $=2a=2\times 57,909,050=115,818,100$

The minor axis is $=2b=2\times 56,671,523=113,343,046.$

The ratio of the major and minor axis is:

  $\frac{\text{major axis}}{\text{minor axis}}=\frac{115,818,100}{113,343,046}=1.0218$ .

Ratio of the lengths is $1.0218$ .

The graph of the mercury is:

  

Hence, the major axis is $115,818,100$ , The minor axis is $113,343,046$ and Ratio of the lengths is $1.0218$ .