# Gravitational force and electrical force

1. Feb 5, 2005

### leolaw

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?

2. Feb 5, 2005

### dextercioby

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

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

Daniel.

3. Feb 5, 2005

### freemind

Hmmm, I don't see anything wrong with your method leolaw.

4. Feb 5, 2005

### dextercioby

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.

5. Feb 5, 2005

### leolaw

Why do you need to know the period? I should use kepler's law?

6. Feb 5, 2005

### dextercioby

Not that version built with Newton's law... 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.

7. Feb 5, 2005

### leolaw

Second law of dynamics mean Netwon Second law? But i thought you said I should forget about newton's law before

8. Feb 5, 2005

### dextercioby

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

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

Daniel.

9. Feb 5, 2005

### freemind

phew, thats a relief. I was just about to check myself into a mental institution!

10. Feb 5, 2005

### leolaw

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:

11. Feb 5, 2005

### dextercioby

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

Daniel.

12. Feb 5, 2005

### leolaw

Yes, but now i have to make sure if my idea is right

13. Feb 5, 2005

### dextercioby

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

Daniel.

14. Feb 5, 2005

### leolaw

if $$G \frac{m_{1}m_{2}}{r^2} = k \frac{Q^2}{r^2}$$ will get me the answer !

15. Feb 5, 2005

### dextercioby

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...

Daniel.