Our weight and our position on Earth

How to control our weight with Physics

Generated by Deepai.org (prompt by author)

S., (she’s my wife), like many women, feels the need to control her weight daily. It’s not wrong, clearly: even a man should pay attention; being overweight is not good and not only for one’s silhouette. Let’s say that my ideal scale could have a sensitivity of around 3 or 4 kilos, while that of S. has an upper limit of 100 grams…

That being said, under Cloe’s watchful eye, the cat that shares her house with us, I decided to verify the effect of the Earth’s rotation on the weight of a person of mass m, calculated as mg.

Even if Cloe doesn’t like it, I choose the most intuitive path, considering a reference system fixed with a point on Earth's surface (Cloe always prefers inertial systems).

Let’s imagine a man on a scale at some point on the Equator: he is stationary on Earth’s surface, so no Coriolis force. I only have to add a (fictitious) centrifugal force, as the reference rotates with the Earth: this force has a direction opposite to the system’s acceleration and the same value. The forces acting on him are:

1. The weight, toward the Earth’s center, downward
2. The reaction from the surface, upward
3. The centrifugal force, upward.

By Author

The three vectors are parallel to each other, and they are not scaled in the figure.

“Of course they aren’t! — Cloe says — I expect the centrifugal force to be rather small, compared to the weight. Anyway, the normal reaction is the one that I’m interested in, since it’s measured by the scale. At the Equator, the centrifugal force has its maximum effect; everyone knows that. But how much less is the weight?”

The calculation is simple: remember that the centrifugal force is the same as the centripetal one:

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while the weight on the surface is mg. The equatorial radius is about 6378 km; as per the rotation speed, we know that the sidereal day is 23 hours, 56 minutes, 4 seconds.

“That’s right. As we need the true rotation time versus something fixed — Cloe whispers, a paw on the paper — But you have to transform this into seconds: as you know, I can’t put my paws on the calculator…”

Sure: 23h 56m 4s = 86 164 s, this is the time it takes for one revolution, that is, the period T of the Earth’s rotational motion. If we use the angular velocity ω:

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“Yes, but I like to see the usual speed, please multiply by the radius…”

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which is the same value computed time ago. Then the centrifugal acceleration for an equatorial location is:

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That is about 0.3% of g.

“So, if I make the difference between the two accelerations, I get the apparent acceleration, which, when multiplied by the mass, gives me the weight. For me, with a mass of 3 kg…”.

Well, I always prefer to use physical laws; the man is standing still, so I have to use a static equation: the sum of all forces must be zero. If N is the reaction, P the weight, and Fc the centrifugal force:

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By solving for N, that is, the measure from the scale:

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“…but I would like to know which is the value shown by the scale, not newtons, things for physicists…”.

Ok, the scale is calibrated to provide a value in kilos, although it is measuring a force, and it does it by dividing the force by g, showing the apparent mass:

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“Meeoow… Nothing! But I like this: the ratio shows the percentage change. And for a person, let’s say 70 kg?”

We only have to insert this value: it sums to 69.8 kg. The difference is always very little, but S. would be happy to weigh less.

“Yeah, but don’t tell her, because it could be worse: if the scale shows 70 kg, it means that the true mass is greater!”

You’re right, Chloe, this could be seen oppositely…

“Forget it… So, only at a pole, the weight is mg, the highest, because there is no centrifugal acceleration.”

Yeah! The reaction equals the weight!

“So, if we have the maximum weight at the pole and the minimum one at the Equator, I think that in a zone between the two a cosine has to pop up somewhere”.

Imagine what we’ll do? We’ll find the effect at an intermediate latitude!

“Yes, but please, make a drawing. At a latitude in the middle, the centripetal force… oops, centrifugal (!) given your system… is always perpendicular to the polar rotation axis, but it forms an angle with the reaction: the forces are no longer parallel”.

Medium latitude schema (by Author)

Exactly that! The value of the centrifugal force is changed because the radius is now different:

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and its value along the reaction and weight direction is

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As before, the sum of the forces along this direction must sum to zero, so the reaction is now given by:

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As you were saying before, there is a cosine, but it is squared! The apparent mass now becomes:

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A 70 kg mass on the scale appears to be 69.9 kg.

“Okay, if I send you to get 500 g of fish for my dinner, this doesn’t change my life…”

I’d say not: the sensitivity of the scale wouldn’t even see the difference.

“But… — Chloe has a doubtful expression (who says cats don’t have expressions?) — there is something I can’t explain. In the previous picture, a body on the surface couldn’t stay still! The centrifugal force has a component perpendicular to the reaction that is not balanced by anything: the body should move toward the Equator. But I don’t feel such a force. What’s missing?”

Well done, Chloe! That’s called physical intuition! (Chloe looks at me with an “anyone would have noticed it” expression). The horizontal component is unbalanced; this is why the Earth is not a true sphere: every point on its surface is affected by the gravitational and centrifugal forces. The result is that all points are pushed toward the Equator: the effect is stronger at lower latitudes and weaker at higher ones. The Earth became a geoid, which is approximated by an ellipsoid, to have mathematical formulas to use.

“It’s clear! The direction of the reaction, normal to the surface, doesn’t point to the center O of the Earth, but to a different point R on the polar axis. This direction intersects the equatorial plane in C, so the reaction and the weight have different directions. For this reason, the weight has a component along the x-axis, and this equals the component of the centrifugal force, and the body in P can finally remain still!”

That’s right! The only problem is that the mathematical things get a little tricky…

“I can understand: starting from the angle defining the latitude, is it the one in C or the one in O? In my opinion, the latitude should be related to the zenith, that is, the normal direction.”

Scientists solved the question by defining two kinds of latitude:

1. The geodetic latitude 𝜆, the one used on maps and GPS navigators, that is, the angle between the normal to the surface and the equatorial plane (the one you chose).
2. The geocentric latitude 𝜑, the angle formed by the radius OP.

The formulas become complex; if the ellipsoid is represented by the equation

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(a is the equatorial radius, while c is the polar one, with a > c) we could demonstrate that the two angles are related:

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Always with analytical calculus, we could obtain that the segment RP is

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where e = 0.006723 is the eccentricity of the Earth. By using the right triangle QPR, the radius used in the centrifugal force is

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That means a correction of 0,17%. So the previous equation for the reaction is now:

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and the apparent mass is

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This leads to a difference in the second decimal digit: the apparent mass at a latitude of 45° is 69.83 kg.

“So we justified that the body doesn’t move because the Earth is squashed to the right degree, to balance the horizontal components of the weight and the centrifugal force… But wait, there’s one more thing: the value of g depends on the distance from the surface to the Earth’s center, but this radius now is different, depending on the latitude.”

It’s true, indeed; the value of g is different at different 𝜆, following the universal gravitation formula:

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So we should replace the g value everywhere. If you are interested, you could try this site.

“Well, this is just calculator stuff. I just need to understand the effects. But at the end, I’d like to see the force diagram if we choose an inertial system, fixed to the center of the Earth.”

The gravitational and the reaction forces would be the same, but in this system, the man is not at rest, and we should use Newton’s law: the sum of the forces is equal to the acceleration times the mass of the body, and the acceleration is the centripetal one:

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This is the same equation obtained at the start.

“But you see how much nicer it is to talk about centripetal force and use Newton’s second law! I would have done it like this.” Chloe rubs herself against the warm lamp.

But with the rotating system, we use the very same law, only in a static situation.

“Yes, but with the inertial system, you have the perspective of the stars, which I like better than something looking more like a carousel…”

I wasn’t expecting such a philosophical conclusion; she heads off into the other room, tail held high, presumably for a piece of sleep.