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Gravitational force and electrical force

  1. Feb 5, 2005 #1
    Suppose that electrical attraction, rather than gravity, were responsible for holding the MOon in orbit around the Earth. If equal and opposite charges Q were placed on the Earth and the Moon, what should be the value of Q to maintain the present orbit? Given these data: Moon = [tex]7.35 \ast 10^{22} kg [/tex], Earth = [tex]5.98 \ast 10^{24} kg[/tex] , radius of orbit = [tex] 3.84 \ast 10^8 m[/tex].

    This is how I approach the problem:
    I set
    [tex] G \frac{m_{1}m_{2}}{r^2} = k \frac{Q^2}{r^2}[/tex]
    and then solve for Q

    Am I doing the right thing?
     
  2. jcsd
  3. Feb 5, 2005 #2

    dextercioby

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    Nope,u should totally disregard gravity.Wiped out.Substituted with electric force...

    HINT:What is the period of Moon's revolution around Earth ??

    Daniel.
     
  4. Feb 5, 2005 #3
    Hmmm, I don't see anything wrong with your method leolaw. :confused:
     
  5. Feb 5, 2005 #4

    dextercioby

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    Think at this way:there's suddenly no gravity and no Newton's law,just Coulomb's one...

    Another clue:do you see somthing rather suspicious about his formula??

    Daniel.
     
  6. Feb 5, 2005 #5
    Why do you need to know the period? I should use kepler's law?
     
  7. Feb 5, 2005 #6

    dextercioby

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    Not that version built with Newton's law... :wink: You can build a new one,just for the fun of it.However,you won't need the new one exactly to solve your problem...

    Just apply the second law of dynamics...

    Daniel.
     
  8. Feb 5, 2005 #7
    Second law of dynamics mean Netwon Second law? But i thought you said I should forget about newton's law before
     
  9. Feb 5, 2005 #8

    dextercioby

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    Sorry,i mislead you... :blushing: I was referring to the Law of Gravity (of Universal Attraction).

    So what's the answer & how did you get it...?

    Daniel.
     
  10. Feb 5, 2005 #9
    phew, thats a relief. I was just about to check myself into a mental institution! :smile:
     
  11. Feb 5, 2005 #10
    hahaa, this problem is even number on my textbook, so i dont have the answer.
    But i do think that i need another way to approach this problem because i should aviod the use of law of gravity.

    And by the way, why did you say there is smoething wrong with my equation before:
     
  12. Feb 5, 2005 #11

    dextercioby

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    I didn't say "wrong",i said "suspicious"... :wink: Well,by your formula,the "r" wouldn't really matter,because it would simplify,right...??

    Daniel.
     
  13. Feb 5, 2005 #12
    Yes, but now i have to make sure if my idea is right
     
  14. Feb 5, 2005 #13

    dextercioby

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    What idea??Don't keep it a secret... :tongue2:

    Daniel.
     
  15. Feb 5, 2005 #14
    if [tex] G \frac{m_{1}m_{2}}{r^2} = k \frac{Q^2}{r^2}[/tex] will get me the answer !
     
  16. Feb 5, 2005 #15

    dextercioby

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    Incidentally,yes...It will,though i initially advised to disregard Newton's gravity law and use the centripetal form of acceleration.

    Anyway,i guess you got it solved...One way or another... :smile:

    Daniel.
     
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