Fan and medal , HELP PLEASEEE
A satellite is in circular orbit 25,000 miles above Earth.
Write an equation for the orbit of this satellite if the origin is at the center of the Earth.
Use 8,000 as the diameter of the Earth. ??

- anonymous

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- katieb

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- anonymous

Ok so this is the equ of a circle which is :
(x-a)^2+(y-b)^2=R^2 right?

- anonymous

yeees

- anonymous

Leyla ente 3arabiye? :p

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- anonymous

Ok so let us get the coordinate of the center and as it says that the center of the satellite is of the earth soo...

- anonymous

??

- anonymous

Then the center which is taken as origin has coordinates O(0,0) so now we can write the equ. As x^2-y^2=R^2

- anonymous

Sorry x^2+y^2=R^2

- anonymous

Understood till now??

- anonymous

yes

- anonymous

but im still confused

- anonymous

Ok and the satellite is 25000 miles over the earth and the earth has a diameter of 8000 miles thus a radius of 4000miles therefore from the center pf the earth to the satellite which is the radius of the equation is 25000+4000=29000 therefore the equ. Gives us x^2+y^2=25000^2

- anonymous

Sorry 29000^2*

- anonymous

Whats confusing you?

- anonymous

I understand now , I thought that , I need more process than that

- anonymous

you know more formulas etc , but thank you so much !

- anonymous

Your welcome :) btw are you arab?

- anonymous

not I'm not lol, why ?

- wolf1728

If you need to find the amount of time the satellite takes to complete one orbit, this page should help you.
http://www.1728.org/kepler3a.htm
1.1613 days by the way

- anonymous

thank you @wolf1728

- wolf1728

thanks Leyla :-)

- anonymous

Hehe nothing forget it :p

- wolf1728

That page has the formula you will need
time = sq root[(4*PI^2 * radius^3) / (G*m)]
If you need some help with that, I'll help you

- anonymous

well I dont need the time, but thank you anyway, BUT i need help with a different question you think that you can help? @wolf1728

- wolf1728

Well, ask the question and let's see.

- anonymous

Center at (2, 4), tangent to x-axis

- wolf1728

Is that for a circle?

- anonymous

yes

- wolf1728

Okay, if the circle is tangent to the x-axis and center is at x=2 y =4 then it would need a radius of 4.

- wolf1728

Circle's equation formula is:
(x -a)² + (y -b)² = r²
where center = (a, b)

- anonymous

so 2 is a, and 4 is b ?

- wolf1728

yes and r^2 would be 16

- anonymous

thank youuuuuuuuu so much , I feel like i love u right now lol jk , but thank seriusly :)

- wolf1728

ok you are welcome :-)

- anonymous

heeyy hold on! , i dont want to bother you but i have a question,
Center in the first quadrant: tangent to x=5 the x-axis and the y-axis.
So this question is kinda the same like the previus ?

- anonymous

@wolf1728

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