Question

Suppose we placed a positive charge Q on the Moon and an equal positive charge Q on the Earth. What value of Q would be needed to neutralize the gravitational attraction of the Moon and the Earth?

Answer 1

Answer: q=5.70 x 10^13 C

Explanation:

gravitational attraction = electrostatic repulsion GMm/d^2 = kQ^2/d^2 as you can see the d^2 cancel out. that is why lunar distance is irrelevant. G is the universal gravitational constant = 6.67 x 10^-11 m^3 / kgs^2 M is earth's mass = 5.972 × 10^24 kg m is moon's mass = 7.342×10^22 kg Q is charge on earth and moon. k is coulomb's constant = 9 x10^9 N m^2 /C^2 On solving equation for Q. Q = sqrt (GMm/k) = sqrt ( 6.67 x 10^-11 x 5.972 x 10^24 * 7.342×10^22 / 9 x10^9) = 5.70 x 10^13 C