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We know that the equator has the highest centrifugal acceleration caused by the rotation of the earth (a = 0.034m / s2) but it is negligible because the vector of the acceleration of gravity is much greater (g = 9.81m / s2) and it is right in opposition to the centrifugal acceleration vector. This results in a net acceleration of g = 9,776m / s2 on the vertical axis. On the other hand, if we are located for example in the parallel 18 ° (which is where I am now) we have a centrifugal force of a = 0.032m / s2 but in this case, the centrifugal force is not parallel to the gravity vector creating an angle of 18 ° with respect to the vertical. (See figure 1) Scheme of forces acting on a body with mass m = 1Kg at the equator and the parallel 18 °

Figure 1

That means that a static pendulum with a mass 1Kg will behave according to figure 2 Scheme of forces acting on a pendulum of mass m = 1Kg in parallel 18 °

Figure 2

The resulting force scheme at the point of the ceiling where the pendulum hangs is seen in Figure 3 Diagram of forces at the point of the ceiling where the pendulum hangs

Figure 3

If the Sc force that restricts the horizontal component of the centrifugal force could be eliminated without eliminating the vertical component Sg, the pendulum would move southward with a slight acceleration but would undoubtedly move. In the case of a drone flying in place, we have that the vertical component Fg is matched to the lift force Sg. But there is no force in the horizontal component that restricts movement. Therefore the drone should move horizontally according to Figure 4 Diagram of forces of a drone flying in parallel 18 °

Figure 4

Then calculating the horizontal distance d that the drone would travel with the formula d=0.5*a*t^2 we have that in 1 minute the drone would move 17.8 meters south. In 10 minutes the drone would travel 1782m. This obviously does not happen in reality. What would be the possible reasons?

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  • $\begingroup$ To obtain the earth gravity you use this equation $m\overrightarrow{g}=\dfrac{mMG}{R}\dfrac{\overrightarrow{R}}{R}$. The earth rotation don’t affect g $\endgroup$
    – Eli
    Commented Feb 5, 2020 at 8:36
  • $\begingroup$ I don't know exactly what you mean. Does that mean that textbooks are wrong when they say that the acceleration of gravity in Ecuador is less because of centrifugal force? If it affects it in Ecuador it also affects it in other latitudes only than in other latitudes the angle formed between the vector of the centrifugal acceleration and the vector of the acceleration of gravity is different from zero $\endgroup$
    – El péndulo de Moisés
    Commented Feb 5, 2020 at 9:22
  • $\begingroup$ I used the Newton gravity law . your textbook use just $m\g$ ? If so just ignore my comment $\endgroup$
    – Eli
    Commented Feb 5, 2020 at 12:21
  • $\begingroup$ Related: physics.stackexchange.com/q/8074/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Feb 5, 2020 at 14:34

2 Answers 2

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The angle at which a plumb line hangs is, by definition vertical. Because of the centrifugal component, the local vertical when continued downward will not pass through the center of the earth. The local vertical is, however, perpendicular to the surface of a body of water. The net effect is that that the earth is slighly non-spherical. It is an oblate spheroid. The surface of a body of water is an equipotential, so there is no sideways force tending to move the water, or any drone hovering above it.

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  • $\begingroup$ Although, according to you, the plumb line is locally perpendicular to the surface, the centrifugal force will have a horizontal component (except at the equator and the poles). This means that the centrifugal acceleration component a = 0.0099m/s2 that I calculated in my example would be slightly higher. It would be in this case a = 0.01m/s2 $\endgroup$
    – El péndulo de Moisés
    Commented Feb 5, 2020 at 14:52
  • $\begingroup$ Is there any experimental evidence that the earth is an oblate spheroid? How do you know this? $\endgroup$
    – El péndulo de Moisés
    Commented Feb 5, 2020 at 15:34
  • $\begingroup$ Yes. There are many. The first was the French Geodesic mission to Lapland led by Maupertuis who measured the length of a degree of latitude at various points and so proved oblateness. Today the exact shape of the earth is measured within inches by satellite observations. $\endgroup$
    – mike stone
    Commented Feb 5, 2020 at 16:45
  • $\begingroup$ By the way, the reason that the suraface is an equipotential comes from three terms: the centrifugal term, the varying distance from the centre of the earth on the bulges, and the graviational effect of the extra material in the bulges. Do problem 3 in the following homwork set: courses.physics.illinois.edu/phys508/fa2019/508hw10.pdf to see this $\endgroup$
    – mike stone
    Commented Feb 5, 2020 at 16:49
  • $\begingroup$ When you say "The surface of a body of water is an equipotential" you are referring to the geoid. The geoid is an ideal model where the earth's surface would conform according to the combination of the gravity vector and centrifugal acceleration. Only in this model, the resulting vector would be locally perpendicular to the surface. But this does not represent reality. The real form is as you said an "oblate spheroid" with a polar radius of 6357Km and equatorial of 6378Km. Therefore there will be components of horizontal accelerations produced by the rotation of the earth. $\endgroup$
    – El péndulo de Moisés
    Commented Feb 5, 2020 at 18:17
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I believe it is to do with the fact that you are not accounting for all possible forces on the drone in your example. Yes there may be a southward component of the Centrifugal force, but is that the only possible horizontal force acting on it? The answer is no. There is e.g. air resistance stopping it from very slowly drifting away.

This problem is the same as asking the following:

Why don't I fly off a merry-go round going at a slow speed? - There is a small centrifugal force pushing me away from the centre in my rotating frame of reference, so surely this must push me off?

The answer is friction stopping me from slipping off, and contributing to the Centripetal acceleration.

This may be a good example of why friction and reaction forces are important.

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  • $\begingroup$ Yes I am considering air resistance and that is not the solution. The air resistance depends on the aerodynamic shape and the speed with which the object moves in the air mass. The drone would start moving from zero speed and slowly increase its speed. Considering the corresponding parameters for a 1Kg mini-drone, the terminal speed would be 308Km / h. $\endgroup$
    – El péndulo de Moisés
    Commented Feb 5, 2020 at 11:24
  • $\begingroup$ The air does not act like a rope that binds the objects instead helicopters and airplanes could not move forward. You can perform the experiment of flying a drone inside a rotating platform and you will see that the drone moves out of the platform due to centrifugal force (the drone must be rotating together with the platform before putting it in the air ) $\endgroup$
    – El péndulo de Moisés
    Commented Feb 5, 2020 at 11:24
  • $\begingroup$ But there may be an initial static "friction" which needs to be overcome before we can start moving. The acceleration due to centrifugal effects is small, so not enough to overcome this. $\endgroup$
    – Garf
    Commented Feb 5, 2020 at 12:20
  • $\begingroup$ I don't think the air offers an "initial static friction" against the movement. The only thing that theoretically would oppose the initial drone movement would be air molecules, which are not fixed in space, that is, they move freely because air is a gas. Every body in motion makes its way between the air molecules and will only find resistance when its speed increases. Anyway, if there is a number that quantifies the "static friction of the air" I would like to know it to compare it with the horizontal component of the centrifugal force. $\endgroup$
    – El péndulo de Moisés
    Commented Feb 5, 2020 at 14:41

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