Sure. From Kepler's third law, assuming that the Moon's mass is negligible (it is actually ~1.2% of Earth's mass), the semi-major axis of the moon's orbit is g

Sure. From Kepler's third law, assuming that the Moon's mass is negligible (it is actually ~1.2% of Earth's mass), the semi-major axis of the moon's orbit is given by (mu T^2 / 4 pi^2)^{1/3} [1]. Substituting T = 27.32 days and mu = G M_Earth = 3.986004418×10^14 m^3/s^2 [2], gives a distance of 3.83 x 10^8 m. Estimating the Moon's apparent angle at 31 arcminutes, its diameter is estimated to be 2 * (distance) * ArcSin[31/2 arcminutes], or 3.46 x 10^3 km. According to Google, the right answer is about 3.47 * 10^3 km. [1] https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbi... [2] https://en.wikipedia.org/wiki/Standard_gravitational_parameter https://en.wikipedia.org/wiki/Standard_gravitational_paramet... [3] https://www.wolframalpha.com/input?i=%28%28%5C%5BMu%5D+T%5E2%29%2F%284+%5C%5BPi%5D%5E2%29%29%5E%28++1%2F3%29+%2F.+%7B%5C%5BMu%5D+-%3E+3.986004418+10%5E14%2C+T+-%3E+27.32+86400%7D https://www.wolframalpha.com/input?i=%28%28%5C%5BMu%5D+T%5E2...