Q. 54 - Calculus

Question

Measurements indicate that the orbital eccentricity of Mars is$0.0935$  and its semimajor axis is $1.517323951$ astronomical units. 

(a) Write a Cartesian equation for the orbit of Mars. 

(b) Do $x$ and $y$ have the same meaning as in Exercise 53?

(c) Give a polar coordinate equation for the orbit of Mars, assuming that the sun is the focus of the elliptical orbit. 

Step-by-Step Solution

Answer

Part (a) The cartesian equation is $\frac{x^2}{2.230227197}+\frac{y^2}{2.28214493}=1\text{. }$

Part (b) The coordinates $x,y$have the same meaning as given for the earth's orbit.

Part (c) The polar equation is $r=\frac{1.517323951}{1+0.0935\cos \theta }$.

Part(a) Step 1. Given information.

The eccentricity is $0.0935$

Semi-major axis is $\text{ 1.517323951 }$

Part (a) Step 2. Explanation.

Now,

$$\begin{gathered} e=\frac{\sqrt{A^2-B^2}}{A} \\ A=1.517323951, e=0.0935 \\ 0.0935=\frac{\sqrt{(1.517323951{)}^2-B^2}}{1.517323951} \\ 0.141869789=\sqrt{(1.517323951{)}^2-B^2} \\ (0.141869789{)}^2={\left(\sqrt{(1.517323951{)}^2-B^2}\right)}^2 \\ 0.020127037-2.30227197=2.30227197-B^2-2.30227197 \\ {B}^2=2.28214493 \\ {A}^2=(1.517323951{)}^2=2.230227197 \end{gathered}$$

Therefore, equation becomes,

$\frac{x^2}{2.230227197}+\frac{y^2}{2.28214493}=1.$

Part (b) Step 1. Explanation.

The coordinates $x,y$have the same meaning as given for the earth's orbit. 

Part (c) Step 1. Explanation.

The polar equation will be,

$$\begin{gathered} r=\frac{eu}{1+e\cos \theta } \\ r=\frac{1.517323951}{1+0.0935\cos \theta } \end{gathered}$$