Gravitational force and electrical force

In summary, the conversation revolves around the idea of using electrical attraction to hold the Moon in orbit around the Earth instead of gravity. The value of Q is needed to maintain the present orbit, and the given data includes the masses of the Moon and the Earth, as well as the radius of the orbit. The method to approach the problem is to set an equation using Coulomb's law and solve for Q. There is also a mention of using Kepler's law and the second law of dynamics, as well as avoiding the use of Newton's law. Ultimately, the conversation ends with the confirmation that the initial equation will indeed lead to the solution.
  • #1
leolaw
85
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?
 
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  • #2
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
Hmmm, I don't see anything wrong with your method leolaw. :confused:
 
  • #4
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
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?
 
  • #6
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.
 
  • #7
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
 
  • #8
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.
 
  • #9
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... :tongue2:

Daniel.
 
  • #14
if [tex] G \frac{m_{1}m_{2}}{r^2} = k \frac{Q^2}{r^2}[/tex] 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.
 

1. What is the difference between gravitational force and electrical force?

Gravitational force is the attractive force between two objects with mass, while electrical force is the attractive or repulsive force between two objects with electric charge. Gravitational force is always attractive, while electrical force can be either attractive or repulsive.

2. How is the strength of gravitational force and electrical force determined?

The strength of gravitational force is determined by the mass of the objects and the distance between them. The greater the mass and the closer the distance, the stronger the force. The strength of electrical force is determined by the magnitude of the charges and the distance between them. The greater the charges and the closer the distance, the stronger the force.

3. Can gravitational force and electrical force cancel each other out?

Yes, in some cases, gravitational force and electrical force can cancel each other out. This can occur when there is an equal amount of positive and negative charges, or when the masses of the objects are balanced in a way that their gravitational forces cancel each other out.

4. How do gravitational force and electrical force affect the motion of objects?

Gravitational force affects the motion of objects by causing them to accelerate towards each other, while electrical force affects the motion of objects by either attracting or repelling them. Both forces can change the direction and speed of an object's motion.

5. Are gravitational force and electrical force related?

Yes, gravitational force and electrical force are both fundamental forces of nature and are related through the theory of relativity. They also have similar mathematical equations that describe their strength and behavior. However, they are distinct forces with different properties and effects on objects.

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