Thread: Astronomy...Earth

1. Astronomy...Earth

For the question below, 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 it its greatest distance from the Sun and the perihelion is the shortest distance. The mean distance of a planet from the Sun is the length of the semimajor axis of the elliptical orbit.

QUESTION:

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?

2. First, let's find the eccentricity, e. Do not get this mixed up with the log e

The ratio of the perhelion to aphelion for the Sun and Earth is about 59/61.

The smallest distance is a-c and the largest is a+c.

$\frac{a-c}{a+c}=\frac{59}{61}$

$\frac{1-\frac{c}{a}}{1+\frac{c}{a}}=\frac{59}{61}$

$\frac{1-e}{1+e}=\frac{59}{61}$

Solving gives us $e=\frac{1}{60}$

From above $e=\frac{c}{a}=\frac{1}{60}$

So, $c=\frac{a}{60}$

The smallest distance is $a-c=a-\frac{a}{60}=\frac{59a}{60}=\frac{59}{60}(93)=91, 450,000 \;\ miles$

You can now find the ellipse equation which models this.

3. ok...

Originally Posted by galactus
First, let's find the eccentricity, e. Do not get this mixed up with the log e

The ratio of the perhelion to aphelion for the Sun and Earth is about 59/61.

The smallest distance is a-c and the largest is a+c.

$\frac{a-c}{a+c}=\frac{59}{61}$

$\frac{1-\frac{c}{a}}{1+\frac{c}{a}}=\frac{59}{61}$

$\frac{1-e}{1+e}=\frac{59}{61}$

Solving gives us $e=\frac{1}{60}$

From above $e=\frac{c}{a}=\frac{1}{60}$

So, $c=\frac{a}{60}$

The smallest distance is $a-c=a-\frac{a}{60}=\frac{59a}{60}=\frac{59}{60}(93)=91, 450,000 \;\ miles$

You can now find the ellipse equation which models this.
I will play around with this later.

Thanks

4. Originally Posted by magentarita
For the question below, 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 it its greatest distance from the Sun and the perihelion is the shortest distance. The mean distance of a planet from the Sun is the length of the semimajor axis of the elliptical orbit.

QUESTION:

The mean distance [= m] of Earth from the Sun is 93 million miles. If the aphelion [= a] of Earth is 94.5 million miles, what is the perihelion [= p]? Write an equation for the orbit of Earth around the Sun?
Use the mean value formula:

$\dfrac{a+p}2 = m$ That means: Solve for p:

$\dfrac{94.5 \cdot 10^6+p}2 = 93 \cdot 10^6~\implies~ p = 91.5 \cdot 10^6$

5. ok...

Originally Posted by earboth
Use the mean value formula:

$\dfrac{a+p}2 = m$ That means: Solve for p:

$\dfrac{94.5 \cdot 10^6+p}2 = 93 \cdot 10^6~\implies~ p = 91.5 \cdot 10^6$
Thanks for the extra help.

the mean distance if earth from the sun is 93million miles. if the aphelion of earth is 94.5million miles , what is the perihelion? write an equation for the orbit of the earth around the sun

Click on a term to search for related topics.