Chapter 9.7, Problem 58E

To determine

To find: the polar equation of the orbit of the planet and the perihelion and aphelion distance.

Answer to Problem 58E

The polar equation is $\frac{9.2930\times {10}^7}{1-0.0167\cos \theta }$ , perihelion is $9.1404\times {10}^7$ and aphelion is $9.4508\times {10}^7$ .

Explanation of Solution

Given:

  $\text{Mercury }a\approx 3.5983\times {10}^7\text{miles, }e\approx 0.2056$

Calculation:

Use the equation given in exercise 49. Plug in the given values.

  $\begin{array}{l}r=\frac{\left(1-e^2\right)a}{1-e\cos \theta }\\ r=\frac{\left(1-{0.2056}^2\right)\left(35.983\times {10}^6\right)}{1-0.2056\cos \theta }=\frac{3.4462\times {10}^7}{1-0.2056\cos \theta }\end{array}$

Use the formula from exercise 49 to find the perihelion.

  $\begin{array}{l}r=a\left(1-e\right)\\ r=\left(35.983\times {10}^6\right)\left(1-0.2056\right)=2.8585\times {10}^7\end{array}$

Use the formula from exercise 49 to find the aphelion.

  $\begin{array}{l}r=a\left(1+e\right)\\ r=\left(35.983\times {10}^6\right)\left(1+0.2056\right)=4.3381\times {10}^7\end{array}$

Equation: $\frac{3.4462\times {10}^7}{1-0.2056\cos \theta }$

Perihelion: $2.8585\times {10}^7$

Aphelion: $4.3381\times {10}^7$

Conclusion:

Therefore, the polar equation is $\frac{3.4462\times {10}^7}{1-0.2056\cos \theta }$ , perihelion is $2.8585\times {10}^7$ and aphelion is $4.3381\times {10}^7$ .