Calculating the Required Charge for Gravitational Force

[dfetnum]

Homework Statement

The similarity of form of Newton’s law of gravitation and Coulomb’s Law caused some to speculate that the force of gravity is related to the electrostatic force. Suppose that gravitation is entirely electrical in nature—that an excess charge Q on the Moon and an equal and opposite excess charge –Q on the Earth are responsible for the gravitational force that causes the observed orbital motion of the Moon about the Earth. What is the required size of Q to reproduce the observed magnitude of the gravitational force?

Homework Equations

Fe=kQ1Q2/(r^2)
Fg=GM1M2/(r^2)

I used these values:
The gravitational constant (G) = 6.67 x 10^-11 Nm^2 / Kg^2 .
The mass of the Earth (m1) = 6.0 x 10^24 Kg .
The mass of the moon (m2) = 7.35 x 10^22 Kg .
The average distance from the center of the Earth to the center of the moon is 384,400,000 m
k=9 x 10^9 Nm^2/C^2

The Attempt at a Solution

I set Fg equal to Fe and using the above values found the value of Q to be 1.63 * 10^27 C, but it is incorrect. Any idea where I went wrong? Thanks

 

[gneill]

Hi dfetnum. Welcome to Physics Forums.

You'll have to show more of your work so that we can see what's gone wrong. How did you arrive at the number you found?

Note that if you equate your two force expressions that you can cancel out the distance (r²) variable right away.

 

[dfetnum]

Thanks!

So first I set Fe=Fg:

G*M(earth)*M(moon)/(r^2) = k*(QE)(QM)/(r^2)

radius's cancel out

G*M(earth)*M(moon) = k*(QE)(QM)

The problem stated that the charges are equal so:

G*M(earth)*M(moon) = k*2Q

Then I plugged in:

(6.67 x 10^-11 Nm^2 / Kg^2) * (6.0 x 10^24 Kg) * (7.35 x 10^22 Kg) = (9 x 10^9 Nm^2/C^2) *2Q

I used a calculator to find Q

 

[gneill]

Ah. Q*Q ≠ 2Q

 

[dfetnum]

WOWWWWWWWWWW, how could I have done that. I spent way too long on this problem.

thanks!