Ok so this is the equ of a circle which is : (x-a)^2+(y-b)^2=R^2 right?
Leyla ente 3arabiye? :p
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...
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
Understood till now??
but im still confused
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
Whats confusing you?
I understand now , I thought that , I need more process than that
you know more formulas etc , but thank you so much !
Your welcome :) btw are you arab?
not I'm not lol, why ?
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
thank you @wolf1728
thanks Leyla :-)
Hehe nothing forget it :p
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
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
Well, ask the question and let's see.
Center at (2, 4), tangent to x-axis
Is that for a circle?
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.
Circle's equation formula is: (x -a)² + (y -b)² = r² where center = (a, b)
so 2 is a, and 4 is b ?
yes and r^2 would be 16
thank youuuuuuuuu so much , I feel like i love u right now lol jk , but thank seriusly :)
ok you are welcome :-)
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 ?