Gravitational force and electrical force

AI Thread Summary
The discussion revolves around the hypothetical scenario of electrical forces replacing gravitational forces to keep the Moon in orbit around the Earth. Participants debate the correct approach to calculate the necessary charge (Q) on both the Earth and the Moon using Coulomb's law instead of Newton's law of gravitation. There is confusion about whether to disregard gravity entirely and how to incorporate the period of the Moon's revolution into the calculations. Ultimately, the consensus is that while gravity should be set aside, the relationship between mass and charge can still yield a solution. The conversation highlights the importance of understanding the underlying principles of both gravitational and electric forces in orbital mechanics.
leolaw
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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 = 7.35 \ast 10^{22} kg, Earth = 5.98 \ast 10^{24} kg , radius of orbit = 3.84 \ast 10^8 m.

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

Am I doing the right thing?
 
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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.
 
Hmmm, I don't see anything wrong with your method leolaw. :confused:
 
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.
 
dextercioby said:
Nope,u should totally disregard gravity.Wiped out.Substituted with electric force...

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

Daniel.
Why do you need to know the period? I should use kepler's law?
 
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.
 
dextercioby said:
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.

Second law of dynamics mean Netwon Second law? But i thought you said I should forget about Newton's law before
 
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.
 
phew, that's a relief. I was just about to check myself into a mental institution! :smile:
 
  • #10
dextercioby said:
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.
hahaa, this problem is even number on my textbook, so i don't 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:
dextercioby said:
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.
 
  • #11
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.
 
  • #12
Yes, but now i have to make sure if my idea is right
 
  • #13
What idea??Don't keep it a secret... :-p

Daniel.
 
  • #14
if G \frac{m_{1}m_{2}}{r^2} = k \frac{Q^2}{r^2} will get me the answer !
 
  • #15
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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