[SOLVED] Ellipse and Circle problem

• May 9th 2009, 08:33 PM
Stealth
[SOLVED] Ellipse and Circle problem
Hey everybody!

I have this problem and my solution was:

(x-h)/40,000 + (y-k)/10,000 = 1

Here's the problem:
The space shuttle travels in an elliptical orbit about the earth. The shuttle orbits such that it reaches a max height above the surface of the earth of 200 miles and a minimum height above the surface of 100 miles. The center of the earth is one focus of the ellipse. Assume that the center of the ellipse is at the origin and the major axis is horizontal. also assume the earth is circular. The approximate diameter of the earth 8000 miles. Write an equation for the shuttle's orbit.

The equation up there is what I got. I just seem a bit hesitant because it seemed really easy. Is it correct?

Thanks,
Stealth
• May 10th 2009, 12:27 AM
Hello Stealth
Quote:

Originally Posted by Stealth
Hey everybody!

I have this problem and my solution was:

(x-h)/40,000 + (y-k)/10,000 = 1

Here's the problem:
The space shuttle travels in an elliptical orbit about the earth. The shuttle orbits such that it reaches a max height above the surface of the earth of 200 miles and a minimum height above the surface of 100 miles. The center of the earth is one focus of the ellipse. Assume that the center of the ellipse is at the origin and the major axis is horizontal. also assume the earth is circular. The approximate diameter of the earth 8000 miles. Write an equation for the shuttle's orbit.

The equation up there is what I got. I just seem a bit hesitant because it seemed really easy. Is it correct?

Thanks,
Stealth

No, I'm afraid it isn't.

The centre of the ellipse is at the origin, so its equation will be of the form

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

The highest and lowest points, the centre and the foci all lie on the major axis. The length of the major axis, $2a$, is therefore $100+8000+200$ miles.

So $a = 4150$ and $a^2 = 17222500$

The centre of the earth is at one of the foci, and this is $50$ miles from the centre of the ellipse.

So $ae = 50$.

And $b^2 = a^2 - a^2e^2 = 4150^2 - 50^2 = 17220000$

So the equation is $\frac{x^2}{17222500}+\frac{y^2}{17220000}=1$

• May 10th 2009, 07:13 AM
Stealth
Just a few questions!

The highest and lowest points, the centre and the foci all lie on the major axis. The length of the major axis, $2a$, is therefore $100+8000+200$ miles.

Why would the lowest point also be on the major axis. I would assume it to be on the minor axis because that would be like the bsquared part of the equation.

So $a = 4150$ and $a^2 = 17222500$

The centre of the earth is at one of the foci, and this is $50$ miles from the centre of the ellipse.

Also, where did the 50 come from? I'm trying to analyze it but I don't see from where it came from.

So $ae = 50$.

And $b^2 = a^2 - a^2e^2 = 4150^2 - 50^2 = 17220000$

Also, is ae just basically the c part of the formula:
[MATH]b^2 = a^2-c^2[/MATH]?

So the equation is $\frac{x^2}{17222500}+\frac{y^2}{17220000}=1$

• May 10th 2009, 08:25 AM
Hello Stealth
Quote:

Originally Posted by Stealth
Just a few questions!

The highest and lowest points, the centre and the foci all lie on the major axis. The length of the major axis, $2a$, is therefore $100+8000+200$ miles.

Why would the lowest point also be on the major axis. I would assume it to be on the minor axis because that would be like the bsquared part of the equation.

Well it just isn't, that's all! If you draw a diagram with a circle inside an ellipse and its centre at one of the foci, the closest point the ellipse gets to the circle, and the most distant point are at the ends of the major axis. Draw it and you'll see what I mean.

Quote:

The centre of the earth is at one of the foci, and this is $50$ miles from the centre of the ellipse.

Also, where did the 50 come from? I'm trying to analyze it but I don't see from where it came from.
Again, draw a diagram. The total distance from the highest point to the lowest is 8300 miles. So the distance from the centre of the ellipse to either end of the major axis is 4150 miles. And the distance from the centre of the earth to the lowest point of the orbit is 4000 + 100 = 4100 miles. So the centre of the earth is 50 miles from the centre of the ellipse.
Quote:

Also, is ae just basically the c part of the formula:
[MATH]b^2 = a^2-c^2[/MATH]?
I don't recognise this formula. For an ellipse, we have:

$a^2e^2 = a^2 - b^2$

This is a standard formula which you should be familiar with.

• May 10th 2009, 09:15 AM
Stealth
Oh ok! I see it now!
(Nod)

• May 10th 2009, 10:33 AM
earboth
Quote:

Originally Posted by Stealth
Oh ok! I see it now!
(Nod)
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9382718801498413, "perplexity": 301.8712557312823}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988511.77/warc/CC-MAIN-20150728002308-00215-ip-10-236-191-2.ec2.internal.warc.gz"}