Gm1m2 / d2 = kq1q2 / d2What is the value of Q to maintain the present orbit?

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The discussion centers on calculating the charge value Q needed for electrical attraction to maintain the Moon's orbit around the Earth, instead of gravitational force. The relevant equations for gravitational force (Fg) and electrical force (Fe) are provided, along with the masses of the Earth and Moon and the radius of their orbit. The user expresses uncertainty on how to proceed after calculating the gravitational force. It is noted that if the orbit remains unchanged, the acceleration must also remain constant, implying that the forces must be equal. The thread seeks assistance in solving this physics problem.
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Homework Statement



Supppose 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?

Treat the Earth and Moon as point particles

Homework Equations



Mass of earth: 5.97x1024kg
Mass of moon: 7.35x1022kg
radius of orbit: 3.84x108m

Fg=Gm1m2 / d2
Fe= kq1q2 / d2

The Attempt at a Solution



Well I know to start by finding the force of gravity, but from there i am unsure about how to do this problem. Could someone please help?
 
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If the orbit remains the same, then so does the acceleration.
Since the masses are the same, so are the forces: Fg=Fq
 

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