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anonymous

  • one year ago

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. ??

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  1. anonymous
    • one year ago
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    Ok so this is the equ of a circle which is : (x-a)^2+(y-b)^2=R^2 right?

  2. anonymous
    • one year ago
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    yeees

  3. anonymous
    • one year ago
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    Leyla ente 3arabiye? :p

  4. anonymous
    • one year ago
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    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...

  5. anonymous
    • one year ago
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    ??

  6. anonymous
    • one year ago
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    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

  7. anonymous
    • one year ago
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    Sorry x^2+y^2=R^2

  8. anonymous
    • one year ago
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    Understood till now??

  9. anonymous
    • one year ago
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    yes

  10. anonymous
    • one year ago
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    but im still confused

  11. anonymous
    • one year ago
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    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

  12. anonymous
    • one year ago
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    Sorry 29000^2*

  13. anonymous
    • one year ago
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    Whats confusing you?

  14. anonymous
    • one year ago
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    I understand now , I thought that , I need more process than that

  15. anonymous
    • one year ago
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    you know more formulas etc , but thank you so much !

  16. anonymous
    • one year ago
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    Your welcome :) btw are you arab?

  17. anonymous
    • one year ago
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    not I'm not lol, why ?

  18. wolf1728
    • one year ago
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    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

  19. anonymous
    • one year ago
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    thank you @wolf1728

  20. wolf1728
    • one year ago
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    thanks Leyla :-)

  21. anonymous
    • one year ago
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    Hehe nothing forget it :p

  22. wolf1728
    • one year ago
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    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

  23. anonymous
    • one year ago
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    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

  24. wolf1728
    • one year ago
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    Well, ask the question and let's see.

  25. anonymous
    • one year ago
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    Center at (2, 4), tangent to x-axis

  26. wolf1728
    • one year ago
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    Is that for a circle?

  27. anonymous
    • one year ago
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    yes

  28. wolf1728
    • one year ago
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    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.

  29. wolf1728
    • one year ago
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    Circle's equation formula is: (x -a)² + (y -b)² = r² where center = (a, b)

  30. anonymous
    • one year ago
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    so 2 is a, and 4 is b ?

  31. wolf1728
    • one year ago
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    yes and r^2 would be 16

  32. anonymous
    • one year ago
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    thank youuuuuuuuu so much , I feel like i love u right now lol jk , but thank seriusly :)

  33. wolf1728
    • one year ago
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    ok you are welcome :-)

  34. anonymous
    • one year ago
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    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 ?

  35. anonymous
    • one year ago
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    @wolf1728

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