Astrologic Tides – Calculations 

Here we present some calculations regarding the forces mentioned in the article Astrologic Tides. There’s a bit of math involved, and if you’re not interested it’s ok to click ‘Back’ on your browser. To reach the figures presented on the article Astrological Tides we need some values first: Note: If you’re not familiar with the scientific notation, the ‘E’ followed by a number indicates that we must multiply by 10 that number of times (if the number is positive) or divide by 10 that number of times (if the number is negative). So 1E+3 means 1000 and 1E3 means 0.001. Earth’s mass: Me = 6.02E+24 Kg Note: this means 6.020.000.000.000.000.000.000.000. As you see, scientific notation can be very convenient J Earth’s radius: Re = 6.40E+06 m Moon’s mass: Mm = 7.44E+22 Kg EarthMoon distance (centre to centre): Dem = 3.84E+08 m The Gravitational constant: G = 6.67E11 N.m2/Kg2 The variation in the Moon’s gravitational pull The average gravitational acceleration the Moon creates on something on Earth is gm = G ´ Mm / Dem = 3.36E05 N/Kg To get the gravitational acceleration for something on Earth closest to the Moon and furthest from the Moon, we just need to replace the distance value of Dem by, respectively, this value minus the Earth’s radius and this value plus the Earth’s radius. The result is: gm(closest) = 3.48E05 N/Kg gm(furthest) = 3.25E05 N/Kg So the Moon’s gravitational pull varies by 2.24E –6 N/Kg between when the Moon is directly over our heads and when it is on the other side of the Earth. The units may seem a bit strange if you are not familiar with physics. The N stands for Newton, and is a unit of force. The Kg stands for Kilogram and is a measure of mass. This value gives us is a way to calculate the difference in force the Moon exerts on something on Earth if we know its mass. For example, for a 60Kg person this would mean a force variation of 2.24E –6 ´ 60 = 1.35E –4 N. One Kg of mass, here on Earth, exerts a force equal to 9.8 N due to its weight. So to get this in more familiar units, we will need to divide by 9.8, and the force difference on a person is approximately a weight of 0.01 gram, or about ten times lighter than a drop of water. The variation in centrifugal force The Earth and Moon are spinning in space like a pair of weights attached to each other, stuck together by the gravitational attraction they exert on each other. The centre point around which both turn is called the centre of mass. The distance of this point from the centre of the Earth is equal to the distance between the Earth and Moon divided by the total mass of both times the mass of the Moon. This gives us the value of Decm = 4.68E+06 m When the Moon is directly above us, we are closest to this point, at a distance equal to the Earth’s radius minus Decm, or D nearest = 1.72E+06 m (1.720 Km below our feet) Since it takes 27.3 days to go around this point, we can calculate the velocity in meters per second by converting days to seconds, calculating the perimeter and dividing the values. I’ll skip those steps and go right to the answer: Vnearest = 4.57 m/s At this speed and distance from the centre of the circle, the centrifugal acceleration is A nearest = V2 nearest / D nearest = 1.22E 5 N/Kg The procedure is identical for the case where the Moon is on the other side of the Earth and we are furthest from the centre point. All we have to do is to use the distance of Decm plus the Earth’s radius for the distance value. The answer is: A furthest = 7.86E5 N/Kg Note: We use a minus sign here. This is just because all the other accelerations so far have been pointing towards the Moon, which we arbitrarily decided would be the positive direction. This time the centrifugal force will accelerate things away from the Moon, and we invert the sign so that when we calculate the difference it gives the correct result. The difference is of 9.31E05 N/Kg, which is equivalent for a 60Kg person to the force exerted by a weight of approximately 0.6 gram, the weight of a small piece of paper. Adding it all up So the final result is that, when the Moon is overhead, an average person is being pulled by the Moon’s gravity towards it with a force 0.01 gram greater than when the Moon is on the other side of the Earth. In this case the centrifugal force from the Earth’s movement is pulling that person away from the Moon with a force that is 0.6 gram smaller than when the Moon is on the other side. The total difference is around 0.6 gram, since the Moon’s effect is so much smaller. So what causes the tides? Actually, none of this quite does it. We calculated the magnitude of the force variations, but in the extreme values (when the Moon is directly above us or precisely on the opposite side of the Earth) the forces are being applied in the same direction as Earth’s gravity, either opposing or reinforcing it. But since these forces are so much weaker (hundreds of thousands of times weaker) in these cases they have no significant effect. The effect appears when they are applied at an angle to the Earth’s gravity. In these cases things like the water in the oceans or the atmosphere can be dragged along the Earth’s surface even with only a small force being applied. Ludwig Krippahl


All the material of the authorship of the Associação Cépticos de Portugal is public domain and can be used freely for nonprofit purposes. Please refer CEPO as a source if using this material. 