Question

Perihelion and Aphelion

The planets move around the sun in elliptical orbits with the sun at one focus. The point in the orbit at which the planet is closest to the sun is called perihelion, and the point at which it is farthest is called aphelion. These points are the vertices of the orbit. The earth’s distance from the sun is $\mathbf{147},\mathbf{000},\mathbf{000}$km at perihelion and $\mathbf{1}53,\mathbf{000},\mathbf{000}$km at aphelion. Find an equation for the earth’s orbit. (Place the origin at the center of the orbit with the sun on the x-axis.)

Short answer

The required equation of earth’s orbit is $\frac{x^2}{2.25\times {10}^{16}}+\frac{y^2}{2.2491\times {10}^{16}}=1$

Step-by-step solution

Step 1. Given information

Earth's distance from the sun is $147000000\text{km}$ at perihelion and $153000000\text{km}$at aphellon, origin at the center of the orbit.

Step 2. Concept used

Each fixed point is called a focus (plural: foci) of the ellipse.

In this problem, we have to find the equation for the earth's orbit satisfying the given conditions.

Step 3. Calculation

The ellipse is horizontal so we will write an equation of the ellipse as $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where the vertices are $(\pm a,0)$, the foci are at $v$, the major axis has a length of $2a$ and

$c^2=a^2-b^2$

We know that the sun is located at a focus of the ellipse, the distance from the sun to the perihelion is $147000000=1.47\times {10}^8\text{km}$and the distance from the sun to the aphelion is $153,000,000=1.53\times {10}^8\text{km}$

The length of the major axis is $2a=1.47\times {10}^8+1.53\times {10}^8$

$\begin{array}{l}2a=3\times {10}^8\\ \therefore a=1.5\times {10}^8\end{array}$

Squaring on both sides,

$a^2=2.25\times {10}^{16}$

The distance from the sun to perihelion is $a-c$ so

$\begin{array}{l}a-c=1.47\times {10}^8\\ c=a-\left(1.47\times {10}^8\right)\\ c=1.5\times {10}^8-\left(1.47\times {10}^8\right)\\ c=0.03\times {10}^8\\ c=3\times {10}^6\end{array}$

Squaring on both sides,

$c^2=9\times {10}^{12}$

Putting value of $a^2,c^2\&c^2=b^2-a^2$ we get

$\begin{array}{l}{c}^2=a^2-b^2\\ b^2=2.25\times {10}^{16}-9\times {10}^{12}\\ b^2=2.25\times {10}^{16}-0.0009\times {10}^{16}\\ b^2=2.2491\times {10}^{16}\end{array}$

So the required equation of the ellipse is

$\frac{x^2}{2.25\times {10}^{16}}+\frac{y^2}{2.2491\times {10}^{16}}=1$