Geosynchronous Orbit Satellite Altitude Calculated from Moon

[KillerZ]

Homework Statement

NASA would like to place a satellite in orbit around the moon such that the satellite always remains in the same position over the lunar surface. What is the satellite's altitude?

Homework Equations

T^{2} = \left(\frac{4\pi^{2}}{GM}\right)r^{3}

The Attempt at a Solution

I think I have this right:

r_{geo} = R_{e} + h_{geo} = \left[\left(\frac{GM}{4\pi^{2}}\right)T^{2}\right]^{1/3}

r_{geo} = R_{e} + h_{geo} = \left[\left(\frac{(6.67x10^{-11})(5.98x10^{24})}{4\pi^{2}}\right)(2358720)^{2}\right]^{1/3}

= 383065776.5m

h_{geo} = r_{geo} - R_{e} = 383065776.5m - 6.67x10^{6}m

= 376695776.5m from earth.

=>3.84x10^{8}m - 376695776.5m

= 7304223.492m

= 7.30x10^{6}m from moon

I calculated it from the Earth then subtracted that from the distance between the Earth and the moon to get the distance from moon. I used the Earth because the moon is in synchronous orbit with the Earth and for T I used the moons period around the Earth to keep the satellite with it.

 

[gneill]

You want the satellite to be in orbit around the Moon. I don't understand why you're involving the Earth. I don't see the Moon's mass used anywhere.

 

[Phrak]

You would want your satellite at any of the Earth-Moon Lagrange points.