How Much Charge Would the Moon and Earth Need to Replace Gravity?

AI Thread Summary
The discussion centers on calculating the amount of charge needed on the Moon and Earth to replicate gravitational attraction through electrostatic force. The formula used is Fe = ke |q1||q2|/r^2, where gravitational pull is equated to electrostatic force. The participant initially calculated a charge value of approximately -0.64766 but expressed uncertainty about the problem. Guidance was provided to equate the gravitational force with the electrostatic force and to use the known radius and Coulomb's constant to find the charges. The conversation concludes with the participant gaining clarity on the approach to solve the problem.
itzxmikee
Messages
10
Reaction score
0
1. The Moon and Earth are bound together by gravity. If, instead, the force of attraction were the result of each having a charge of the same magnitude but opposite in sign, find the quantity of charge that would have to be placed on each to produce the required force.



2. Fe = ke |q1||q2|/r^2



3. 9.81 = ke |x||-x|/r^2
solving for x, I ended up getting -.64766

Not sure if I understand the problem correctly or not. Please help steer in the right direction. Thanks
 
Physics news on Phys.org
I think you need to equate the gravitational pull of the Earth on the moon with F_e. Since you know the radius from the center of the Earth to the center of the moon, and you know k_e, you can find the two "q"s by a simple square root operation, making one positive and the other negative.
 
I got it..thanks
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top