Chapter 9, Problem 34CR

To determine

To find: The standard equation of Mercury’s orbit

Answer to Problem 34CR

The standard equation of the Mercury’s orbit is $\frac{x^2}{1.296\times {10}^{15}}+\frac{y^2}{1.241\times {10}^{15}}=1$

Explanation of Solution

Given information:

The eccentricity of the Mercury’s orbit is $e\approx 0.2056$

The length of the major axis is 72 million miles

Formula used:

The standard form of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

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

Calculation:

To find the standard equation of Mercury’s orbit firstly find the values of $a^2$ and $b^2$

At first find the value of $a$

The length of the major axis is 72 million miles which means $2a=7.2\times {10}^7$ so the value of $a$ will be

  $\begin{array}{l}2a=7.2\times {10}^7\\ a=\frac{7.2\times {10}^7}{2}\\ a=3.6\times {10}^7\\ a^2=1.296\times {10}^{15}\end{array}$

Now, the value of $c$ can be calculated as,

  $\begin{array}{l}c=ae\\ c=\left(3.6\times {10}^7\right)\left(0.2056\right)\\ c=7401600\end{array}$

Again, calculate the value of $b^2$

  $\begin{array}{l}{b}^2=a^2-c^2\\ b^2={\left(3.6\times {10}^7\right)}^2-{\left(7401600\right)}^2\\ b^2=1.241\times {10}^{15}\end{array}$

The standard equation can be obtained as,

  $\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \frac{x^2}{1.296\times {10}^{15}}+\frac{y^2}{1.241\times {10}^{15}}=1\end{array}$

Therefore, the standard equation of the Mercury’s orbit is $\frac{x^2}{1.296\times {10}^{15}}+\frac{y^2}{1.241\times {10}^{15}}=1$