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Can you get the size of the moon using similar triangles? You have to know the distance to the moon. But presumably you can calculate it because you know its an

Can you get the size of the moon using similar triangles? You have to know the distance to the moon. But presumably you can calculate it because you know its angular velocity (it takes about 28 days to revolve around the earth). When you know that, you can calculate a distance based on the fact that the moon has a stable orbit. I can't derive a formula, but presumably it's possible.

greeneggs 5y ago
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...
marcodiego 5y ago
I also thought that, but it needs some previous knowledge about the Earth, simplifications and approximations. Let's get to calculations: First we'll consider the mass of the Moon substantially smaller than that of the Earth and that its orbit is circular. In this case, in its orbit around the Earth, the weight of the Moon is the centripetal force: W = F_cp mg = mw^2R (R being orbit Radius) g = w^2R GM/R^2 = w^2R Assuming you have the gravity g_e on the surface of the Earth and its radius R_e, we can derive GM (M being the mass of the Earth): GM/R_e^2 = g_e GM = g_e * R_e^2 Replacing GM on the previous equation, we get: g_e * R_e^2/R^2 = w^2R R = (g_e * R_e^2 / w^2)^(1/3) Note that w (angular velocity of the Moon) is easy to calculate: 28 days per rotation and gravity on the surface of the Earth is around 9.8 meters per second. Radius of the Earth can be calculated by other means. Nevertheless, we now have an approximation for the orbital radius of the Moon. Considering the Radius of the Earth, distance to the Moon from someone at the surface of the Earth is given by: D = R - R_e Now, to calculate the radius of the visible Moon disk using triangle similarity, use a small disk (with radius r) and hold it at a distance d from your eye until it becomes the same apparent size of the Moon disk. Applying similarity of triangles, the radius r_m of the Moon is given by: r_m/D = r/d r_m = D*r/d Note that we are no calculating the radius of the Moon, but the radius of the visible Moon disk. Considering D is much bigger than d, it is a good approximation. I wonder how precise this will turn out though. Personal anecdote: a gifted uncle of mine once calculated the duration of total Moon eclipse using similar methods. Distance to the Moon was approximated to 1.5 light-seconds. He got the estimated time wrong by around 5 minutes. He did it using mental calculations only.

greeneggs 5y ago

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...

marcodiego 5y ago

I also thought that, but it needs some previous knowledge about the Earth, simplifications and approximations. Let's get to calculations: First we'll consider the mass of the Moon substantially smaller than that of the Earth and that its orbit is circular. In this case, in its orbit around the Earth, the weight of the Moon is the centripetal force: W = F_cp mg = mw^2R (R being orbit Radius) g = w^2R GM/R^2 = w^2R Assuming you have the gravity g_e on the surface of the Earth and its radius R_e, we can derive GM (M being the mass of the Earth): GM/R_e^2 = g_e GM = g_e * R_e^2 Replacing GM on the previous equation, we get: g_e * R_e^2/R^2 = w^2R R = (g_e * R_e^2 / w^2)^(1/3) Note that w (angular velocity of the Moon) is easy to calculate: 28 days per rotation and gravity on the surface of the Earth is around 9.8 meters per second. Radius of the Earth can be calculated by other means. Nevertheless, we now have an approximation for the orbital radius of the Moon. Considering the Radius of the Earth, distance to the Moon from someone at the surface of the Earth is given by: D = R - R_e Now, to calculate the radius of the visible Moon disk using triangle similarity, use a small disk (with radius r) and hold it at a distance d from your eye until it becomes the same apparent size of the Moon disk. Applying similarity of triangles, the radius r_m of the Moon is given by: r_m/D = r/d r_m = D*r/d Note that we are no calculating the radius of the Moon, but the radius of the visible Moon disk. Considering D is much bigger than d, it is a good approximation. I wonder how precise this will turn out though. Personal anecdote: a gifted uncle of mine once calculated the duration of total Moon eclipse using similar methods. Distance to the Moon was approximated to 1.5 light-seconds. He got the estimated time wrong by around 5 minutes. He did it using mental calculations only.