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Homework Statement
There exists a spaceship orbiting between the Moon and the Earth. Find the perfect spot where the gravity due to Earth is equal to the one caused by the Moon. I have already solved for the ratio ## \frac {R_E}{\sqrt{M_E}} = \frac {R_M}{\sqrt{M_M}} ##. ## R_E ## is the distance between the spaceship and the Earth while ##R_M## represents the distance between the ship and the Moon. Likewise ##M_M## and ##M_E## are masses. The next step in my solutions guide is ## R_E = \frac {(R_E + R_M)\sqrt{M_E}}{\sqrt{M_E}+\sqrt{M_M}} ##. I don't understand how they got to that.
Homework Equations
1) Correct answer ## = R_E = 3.46 * 10^8 ## meters
2) Mass of the Moon ##=7.347*10^22##
3) Mass of the Earth ##=5.98*10^24##
4) Distance between Moon and Earth ##= R_E + R_M = 3.844*10^8##
The Attempt at a Solution
Using ratio I solved for in "The problem statement, all..." I tried to isolate ##R_E##
##R_E = \frac{R_M\sqrt{M_E}}{\sqrt{M_M}} ##
From 4) from part 2 "Relevant equa..." I get that ##R_M = 3.844*10^8 - R_E##
Therefore ##R_E = \frac{(3.844*10^8 - R_E)\sqrt{M_E}}{\sqrt{M_M}} ##
Thus ##R_E = (3.844*10^8)(\frac{\sqrt{M_E}}{\sqrt{M_M}}) - R_E(\frac{\sqrt{M_E}}{\sqrt{M_M}}) ##
##R_E + R_E(\frac{\sqrt{M_E}}{\sqrt{M_M}}) = (3.844*10^8)(\frac{\sqrt{M_E}}{\sqrt{M_M}}) ##
##R_E(1+(\frac{\sqrt{M_E}}{\sqrt{M_M}})) = (3.844*10^8)(\frac{\sqrt{M_E}}{\sqrt{M_M}}) ##
##R_E = \frac{(3.844*10^8)(\frac{\sqrt{M_E}}{\sqrt{M_M}})}{1+(\frac{\sqrt{M_E}}{\sqrt{M_M}})}##
While I did get the correct answer, my equation is so complicated. How do I turn it into the one given by the solutions manual?
