Question

A satellite is placed between the Earth and the Moon, along a straight line that connects their centers of mass. The satellite has an orbital period around the Earth that is the same as that of the Moon, 27.3 days. How far away from the Earth should this satellite be placed?

Answer 1

well, centripetal force= gravityforce from Earth-Gravityforce from Moon

mV^2/Rs=GMem/Rs^2 -GMm*m/(Rm-Rs)^2

where Rs is the orbital distance from earth, Rm is the moon orbital distance.

Now, change to period: 2PI*Rs/Period= V

(2PI)^2*Rs/Period^2=GMe/Rs^2-GMm/(Rm-Rs)^2

Lets look at the period of the moon:

(2PI)^2 Rm/Period^2=GMmMe/Rm^2

or 1/period^2= GMmMe/Rm^3 * 1/(2PI)^2
put that into the equation...

Rs*GMmMe/Rm^3=GMe/Rs^2-GMm/(Rm-Rs)^2

Let Mm/Me = r= ratiomasses
Rs/Rm^3= (1/r)/Rs^2-r/(Rm-Rs)^2
Rs=Rm^3/rRs^2 -r Rm^3/(Rm-Rs)^2

Rs^3=Rm^3/r- rRm^3 (Rs^2/(Rm-Rs)^2)

I think I would turn to a graphical solution at this point.

double check my work. Something is bothering me about this.

Answer 2

http://www.jiskha.com/display.cgi?id=1304796180

Answer 3

The answer I posted previously is wrong. See instead

http://www.ottisoft.com/Activities/Lagrange%20point%20L1.htm