Chapter 10.3, Problem 79AYU

In Problems 79-83, use the fact that the orbit of a planet about the Sun is an ellipse, with the Sun at one focus. The aphelion of a planet is its greatest distance from the Sun, and the perihelion is its shortest distance. The mean distance of a planet from the Sun is the length of the semimajor axis of the elliptical orbit. See the illustration.

Earth The mean distance of Earth from the Sun is 93 million miles. If the aphelion of Earth is $94.5$ million miles, what is the perihelion? Write an equation for the orbit of Earth around the Sun.

To determine

To find:

a. The measurement of perihelion, if the aphelion of Earth is $94.5\text{million}$ miles.

Answer to Problem 79AYU

Solution:

a. Perihilion $=91.5$ million miles.

Explanation of Solution

Given:

Use the fact that the orbit of a planet about the Sun is an ellipse, with the Sun at one focus. The aphelion of a planet is its greatest distance from the Sun, and the perihelion is its shortest distance. The mean distance of a planet from the Sun is the length of the semi major axis of the elliptical orbit.

The mean distance of Earth from the Sun is 93 million miles.

Formula used:

Center	Major axis	Foci	Vertices	Equation
$(0,0)$	$x\text{-axis}$	$(c,0)$	$(a,0)$	$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $a>b>0 \text{and} b²=a²-c²$
$(0,0)$	$x\text{-axis}$	$(-c,0)$	$(-a,0)$

Calculation:

Mean distance $=93$ million miles, Hence $a=93$ million miles.

The length of the major axis: $2a=186$ million.

From the figure we see that,

a. $\text{Perihelion }=\text{ Length of major axis }–\text{ Apehilion}$

$\text{Perihilion }=186\text{ million }-94.5\text{ million}$

$\text{Perihilion }=91.5\text{ million miles}$

To find:

b. An equation for the orbit of Earth around.

Solution:

b. $\frac{x^2}{8649}+\frac{y^2}{8646.75}=1$ ($x\text{and}y$ in millions of miles).

Given:

Use the fact that the orbit of a planet about the Sun is an ellipse, with the Sun at one focus. The aphelion of a planet is its greatest distance from the Sun, and the perihelion is its shortest distance. The mean distance of a planet from the Sun is the length of the semi major axis of the elliptical orbit.

The mean distance of Earth from the Sun is 93 million miles.

Formula used:

Center	Major axis	Foci	Vertices	Equation
$(0,0)$	$x\text{-axis}$	$(c,0)$	$(a,0)$	$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $a>b>0 \text{and} b²=a²-c²$
$(0,0)$	$x\text{-axis}$	$(-c,0)$	$(-a,0)$

Calculation:

Mean distance $=93$ million miles, Hence $a=93$ million miles.

The length of the major axis: $2a=186$ million.

From the figure we see that,

b. Distance from center of the ellipse to focus (Sun’s position) is given by $\text{Mean}\text{distance}-\text{perihelion}$.

$c=93\text{million}-91.5\text{million}=1.5\text{million}$

To find $b$,

$b²=a²-c²$

$b²=93²-1.5²{\left(\text{in}\text{millions}\right)}^2$

$b²=8646.75{\left(\text{in}\text{millions}\right)}^2$

$b=92.99(\text{in}\text{million})\text{or}92.99\ast {10}^6$

Hence the equation is,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\frac{x^2}{{93}^2}+\frac{y^2}{92.99²}=1(\text{in}\text{millions})$

$\frac{x^2}{8649}+\frac{y^2}{8646.75}=1$ ($x\text{and}y$ in millions of miles).

To determine

To find:

b. An equation for the orbit of Earth around.

Answer to Problem 79AYU

Solution:

b. $\frac{x^2}{8649}+\frac{y^2}{8646.75}=1$ ($x\text{and}y$ in millions of miles).

Explanation of Solution

Given:

Use the fact that the orbit of a planet about the Sun is an ellipse, with the Sun at one focus. The aphelion of a planet is its greatest distance from the Sun, and the perihelion is its shortest distance. The mean distance of a planet from the Sun is the length of the semi major axis of the elliptical orbit.

The mean distance of Earth from the Sun is 93 million miles.

Formula used:

Center	Major axis	Foci	Vertices	Equation
$(0,0)$	$x\text{-axis}$	$(c,0)$	$(a,0)$	$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $a>b>0 \text{and} b²=a²-c²$
$(0,0)$	$x\text{-axis}$	$(-c,0)$	$(-a,0)$

Calculation:

Mean distance $=93$ million miles, Hence $a=93$ million miles.

The length of the major axis: $2a=186$ million.

From the figure we see that,

b. Distance from center of the ellipse to focus (Sun’s position) is given by $\text{Mean}\text{distance}-\text{perihelion}$.

$c=93\text{million}-91.5\text{million}=1.5\text{million}$

To find $b$,

$b²=a²-c²$

$b²=93²-1.5²{\left(\text{in}\text{millions}\right)}^2$

$b²=8646.75{\left(\text{in}\text{millions}\right)}^2$

$b=92.99(\text{in}\text{million})\text{or}92.99\ast {10}^6$

Hence the equation is,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\frac{x^2}{{93}^2}+\frac{y^2}{92.99²}=1(\text{in}\text{millions})$

$\frac{x^2}{8649}+\frac{y^2}{8646.75}=1$ ($x\text{and}y$ in millions of miles).