Question

How much equal charge should be placed on the Earth and the Moon so that the electrical repulsion balances the gravitational force of1.98 times 10^20 N? Treat the Earth and Moon as point charges a distance 3.84 times 10^8 m apart. Express your answer to three significant figures and include appropriate units.

Answer 1

Given that, Gravitational force, F_G = 1.98 times 10^20 N rightarrow This force balances the electrical repulsion (F_e) therefore F_E = F_G rightarrow Earth and moon are treated as point charges at a distance, r = 3.84 times 10^8 m therefore kq_1q_2/r^2 = GMm/r^2 (F_E = F_G) kq^2/r^2 = 1.98 times 10^20 [as equal charge is placed on earth & moon] implies 9 times 10^9 times q^2/(3.84 times 10^8)^2 = 1.98 times 10^20 implies q^2 = 1.98 times 10^20 times (3.84 times 10^8)^2/9 times 10^9 = 29.19 times 10^36/9 times 10^9 q^2 = 3.244 times 10^27 = 32.44 times 10^26 therefore q = 5.695 times 10^13C therefore The equal charge that should be placed on the Earth and the Moon so that F_E = F_G is given by q = 5.695 times 10^13 coulombs