Chapter 9.4, Problem 3E

To determine

To find: the ratio of the length of the major and minor axis of the earth’s orbit.

Answer to Problem 3E

Ratio of the lengths is $\mathrm{1.000139.}$

Explanation of Solution

Given:

Earth semi major axis length is $1.496\times {10}^8\text{km}$ .

Eccentricity is $0.0167$ .

Concept used:

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:

Earth semi major axis length is $1.496\times {10}^8\text{km}$ ………. $\left(i\right)$

Eccentricity is $0.0167$ …………. $\left(ii\right)$

  $b=a\sqrt{1-{\epsilon }^2}$ .

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

  $\begin{array}{l}b=a\sqrt{1-{\epsilon }^2}.\\ b=1.496\times {10}^8\sqrt{1-{\left(0.0167\right)}^2}.\\ b=149,579,137.573\text{ km}\end{array}$

The major axis= $=2a=2\times 149600000=299,200,000$

The minor axis is $=2b=2\times 149,579,137.573=299,158,275.14663$

The ratio of the major and minor axis is:

  $\frac{\text{major axis}}{\text{minor axis}}=\frac{299,200,000}{299,158,275.14663}=\mathrm{1.000139.}$ .

Hence, Ratio of the lengths is $\mathrm{1.000139.}$