The mass of the Moon is 7.35x10^22 kg. At some point between Earth and the Moon, the force of Earth's gravitational attraction on an object is cancelled by the Moon's force of gravitational attraction. If the distance between Earth and the Mon (centre to centre) is 3.84x10^5 km, calculate where this will occur, relative to Earth.
Fg = (G)(m1)(m2) / r^2
The Attempt at a Solution
FgEARTH = (G)(m1)(mE) / r^2
FgMOON = (G)(m1)(mM) / r^2
dM + dE = 3.84x10^5 km = 3.84x10^8 m
(G)(m1)(mE) / dE^2 = (G)(m1)(mM) / dM^2
(5.98x10^24) / dE^2 = (7.35x10^22) / (3.84x10^8 - dE)^2
(7.35x10^22) / (5.98x10^24) = [(3.84x10^8 - dE)^2](dE^2)
0.01229 = (1.5x10^17 - 3.84x10^8dE + dE^2)(dE^2)
I dont know where to go from here? can i move the 0.01229 to the other side and at the same time move the dE^2 to the other side (so that it gets cancelled out)??