# Geosynchronous Orbit

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.
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