Question

The planets in our solar system do not travel in circular paths. Rather, their orbits are elliptical. The Sun is located at a focus of the ellipse.
The perihelion is the point in a planet’s orbit that is closest to the Sun. So, it is the endpoint of the major axis that is closest to the Sun.

The aphelion is the point in the planet’s orbit that is furthest from the Sun. So, it is the endpoint of the major axis that is furthest from the Sun.

The closest Mercury comes to the Sun is about 46 million miles. The farthest Mercury travels from the Sun is about 70 million miles.


1. What is the distance between the perihelion and the aphelion?


2. What is the distance from the center of Mercury’s elliptical orbit and the Sun?


3. Write the equation of the elliptical orbit of Mercury, where the major axis runs horizontally. Allow a and b to be measured in millions of miles. Use the origin as the center of the

Answer 1

Answer:

(1) 83.764 million miles

(2) 52.766 million miles

(3) $\frac{x^2}{4900}+\frac{y^2}{2116}=1.$

Step-by-step explanation:

Let the origin C(0,0) be the center of the elliptical path as shown in the figure, where the location of the sun is at one of the two foci, say f.

The standard equation of the ellipse having the center at the origin is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\;\cdots(i)$

where $a$ and $b$ are the semi-axes of the ellipse along the x-axis and y-axis respectively.

Let the points P and A represent the points of perihelion (nearest to the sun) and the aphelion (farthest to the sun) of the closest planet Mercury.

Given that,

CP=46 million miles and

CA=70 million miles.

So, $CP=b$ is the semi-minor axis and $CA=a$ is the semi-major axis.

Let the distances on the axes are in millions of miles. So, the coordinates of the point P and A are $P(0,46)$ and $A(70,0)$ respectively.

(1) From the distance formula, the distance between the perihelion and the aphelion is

$PA=\sqrt{(0-70)^2+(46-0)^2}=83.764$ million miles.

(2) Location of the Sun is at focus, $f$, of the elliptical path.

From the standard relation, the distance of the focus from the center of the ellipse, c, is

$c=ae\;\cdots(ii)$

where $a$ and $e$ are the semi-major axis and the eccentricity of the ellipse.

The eccentricity of the ellipse is

$e=\sqrt{1-\frac{b^2}{a^2}}$

$\Rightarrow e=\sqrt{1-\frac{46^2}{70^2}}=0.7538$.

Hence, from the equation (i) the distance of the Sun from the center of the elliptical path of the Mercury is

$c=70\times0.7538=52.766$ million miles.

(3) From the equation (i), the equation of the elliptical orbit of Mercury is

$\frac{x^2}{70^2}+\frac{46^2}{b^2}=1$

$\Rightarrow \frac{x^2}{4900}+\frac{y^2}{2116}=1.$