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I don't know if you remember me or not; my name is Clark and you answered a question for my study group today. Your answer was so helpful, we wanted to ask you one other question we have had trouble with:

Determine the altitude of a lunasynchronous orbit (ignore any effects the Earth or any other factors besides those relating to the moon would have).

Thanks so much again, Clark

This problem also uses the formula for the centripetal acceleration, which is easier to use in an alternative form, namely

a_c = (ω^2)r

where ω = radians/second = angular frequency of the orbiting object and r is the altitude from the center of the moon (not its surface). The period of the orbit is

T = 2𝛑/ω

We want the period of the orbit to be equal to the period of rotation of the moon, which we'll get to in a moment. For now, we want to write down the equation which describes how the centripetal force required to an keep object in orbit is related to the gravitational force exerted by the moon;

m_obj･a_c = G･M_moon･m_obj/r^2

where

M_moon = mass of moon

m_obj = mass of object

G = 6.67259e-11 [m^3/kg/s^2] = Newton's universal gravitational constant. Canceling m_obj on both sides and making the various substitutions, we get

r = { G･M_moon･(T/2𝛑)^2 }^1/3

Now the rotational period of the moon is determined by the fact that it always faces the earth as it orbits around it, which means that the moon rotates about its own axis with the same period as it orbits around the earth, which takes about 28 days = 2,419,200 seconds. I assume you have the mass of the moon (M_moon) and so can plug it into the above formula. I haven't done it myself so I'm not sure what r is. Probably huge.

Good luck. I hope this is what you wanted.

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J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane**Education/Credentials**

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