Question

What equal charges would have to be placed on earth and moon to neutralize their gravitaional attraction?given, mass of earth =1025 kg ,?

Answer 1

To neutralize their gravity, the force of repulsion must equal the force of attraction. 

Universal Gravitational force = (G * m1 * m2) ÷ d^2 
G = 6.67 * 10^-11 
m1 = mass of earth = 5.98 * 10^24 kg 
m2 = mass of moon = 7.36 * 10^22 kg 
d = distance from earth to moon 

Force caused by charges = (k * q1 * q2) ÷ d^2 
k = 9 * 10^9 
q1 = charge on earth 
q2 = charge on moon 
d = distance from earth to moon 

Force caused by charges = Universal Gravitational force 
(k * q1 * q2) ÷ d^2 = (G * m1 * m2) ÷ d^2 
Multiply both sides by d^2 
(k * q1 * q2) = (G * m1 * m2) 
Divide both sides by k 
q1 * q2 = (G * m1 * m2) ÷ k 

The charges are equal, so q1 * q2 = q^2 

q^2 = (G * m1 * m2) ÷ k 
q = √ (G * m1 * m2 ÷ k) 
q = √ (6.67 * 10^-11 * 5.98 * 10^24 * 7.36 * 10^22 ÷ 9 * 10^9) 
q ≈ 3.26 * 10^27 Coulombs

Answer 2

Universal Gravitational force = (G * m1 * m2) ÷ d^2
G = 6.67 * 10^-11
m1 = mass of earth = 5.98 * 10^24 kg
m2 = mass of moon = 7.36 * 10^22 kg
d = distance from earth to moon

Force caused by charges = (k * q1 * q2) ÷ d^2
k = 9 * 10^9
q1 = charge on earth
q2 = charge on moon
d = distance from earth to moon

Force caused by charges = Universal Gravitational force
(k * q1 * q2) ÷ d^2 = (G * m1 * m2) ÷ d^2
Multiply both sides by d^2
(k * q1 * q2) = (G * m1 * m2)
Divide both sides by k
q1 * q2 = (G * m1 * m2) ÷ k

The charges are equal, so q1 * q2 = q^2

q^2 = (G * m1 * m2) ÷ k
q = √ (G * m1 * m2 ÷ k)
q = √ (6.67 * 10^-11 * 5.98 * 10^24 * 7.36 * 10^22 ÷ 9 * 10^9)
q ≈ 3.26 * 10^27 Coulombs


hope this will help.