## Centrifugal Force at the Earth’s Surface, in the United States

### Why doesn’t Earth’s rotation fling us off in to space?

**Quick recap: **

- Gravity attracts all objects toward the Earth’s center. The acceleration due to gravity at Earth’s surface is 9.8 meters per second, per second.
- Centripetal force on Earth is the force of gravity acting on objects (us) orbiting Earth – we happen to orbit the Earth on its surface, at its exact speed of rotation, because gravity “glues” us to the surface as the Earth spins on its rotational axis.
- Centrifugal force is the unbalanced reaction of inertia to centripetal force.
- So centripetal force constantly “pulls” us toward the Earth as we orbit, while “centrifugal” force is the inertial tendency of mass (our bodies) to continue moving in a straight line (e.g. flung in to space).
- The pound (lb), in normal reference, is a unit of weight (force). The kilogram is a unit of mass.
- Force = Mass * Acceleration (F=ma)

**The math:**

- Earth’s circumference at the equator is about 40,075 km. At the equator, Earth’s surface moves at about v=464 m/s (1,044 mph).
- The Earth’s rate of rotation (w) is about 15 degrees per hour (.0042 deg/s), which of course equals one full rotation per day.
- Earth’s average radius (r=center to surface) is about 6,370 km.
- Centripetal acceleration at the equator, a = v^2/r, is about .034 m/s/s.
- The latitude, measured in degrees, of a given location starts at 0 at the equator, and ends at 90 degrees at the north or south pole. Drawing a line from the center of the Earth, this would be the angle at which the line is deflected from its nearest point on the equator.
- In the United States, the mean latitude is about 39 degrees, ranging from about 29.76 degrees in Houston, TX to about 48.23 degrees in Minot, ND.
- As you move north, the distance of the surface from its axis of rotation decreases. At the mean US latitude, the radius from the surface to the axis of rotation is only about 4,951 km
- At 39 degrees north latitude, the Earth’s linear velocity (at the surface) is about 360 m/s (811 mph)
- The resulting centripetal acceleration (a=v^2/r) is .026 m/s/s
- F=ma, so 1 kg mass (2.2 lb at Earth’s surface) * .026 m/s/s results in a centripetal force of .026 Newtons (N), or about .006 lb (about .096 oz or about 1/10 of 1 oz)

Since centrifugal force is the unbalanced reaction, the magnitude is the same, acting in the opposite direction (e.g. flinging us out toward space)

We can therefore use the formula above to calculate the centrifugal force for any arbitrary mass located in the US (at sea level, based on the mean latitude)

**The calculation:**

- Convert to kilograms: weight in lb (at the Earth’s surface) / 2.2 = mass in kg
- Multiply mass in kg by centripetal acceleration, .026 m/s/s = centripetal force in Newtons (N)
- Convert Newtons (N) to lb: force in N / 4.45 = force in lb.

Simplified, this equates to:

cf = (( lb / 2.2 ) * .026 ) /4.45

cf = ( lb * .012 ) / 4.45

cf = lb * .0027

**So any weight in lb (at Earth’s surface) times .0027 equals the centripetal (centrifugal * -1) force, on the Earth’s surface, at US mean latitude.**

**Some benchmark calculations:**

- 100 lb of mass = 0.27 lb of centrifugal force
- 150 lb of mass = 0.40 lb of centrifugal force
- 200 lb of mass = 0.54 lb of centrifugal force
- 374 lb of mass = almost exactly 1 lb of centrifugal force.

**As you can see, the inertial force acting on us, reacting to centripetal force, is a small fraction of the force of gravity on us. Thus, we don’t “fling off” in to space.**

**Cool facts:**

- So, for every 100 lb of weight, you get the benefit of about 1/4 lb of centrifugal force! If you weigh yourself on a scale, and the scale says you weigh 150 lb, you really weigh about 150.4 lb.
- Likewise, you get the benefit of about 5.4 lb per ton (2,000 lb). Chances are, that your car’s curb weight is actually 3 to 8 lb heavier than what’s listed!
- Objects in orbit are moving fast enough for the unbalanced inertia resulting from centripetal force to exactly equal the force of gravity. This means you “fall” at the same rate you “fling off”. This is why orbit is called “free fall”, because an object in orbit is constantly “falling” to Earth due to gravity, and then missing the Earth due to inertia.
- An object hovering at 10 m (about 33 feet) above the Earth’s surface, would have to travel at about 17,800 mph (26, 100 fps) to maintain a stable orbit! Flying west-to-east, that’s just under 17,000 mph relative to the ground!
- The Federal gross vehicle weight limit is 80,000 lb, and there are stiff fines for truckers who exceed the weight limit! If an empty truck plus its empty trailer is 35,000 lb, they are actually 94 lb heavier than the scale reads. If you load up 90 x 50 lb crates, that’s 45,000 lb, bringing the total to 80,000 lb, but the scale would only measure 79,784 lb — 80K minus 94 for the truck / trailer, and another 123 lb for the load itself. This is why truck scales have to be very carefully calibrated – whether they realize it or not, they are adjusting for the force of the Earth’s rotation!

Well, I did some of the numbers, anyway. I took your .026 m/s/s as a given. The calculations I made show not the derivation of the centrifugal acceleration, but rather how that centrifugal acceleration combines with the acceleration due to gravity to create a net effective acceleration.

Sorry, to contradict myself–I did end up “doing” the numbers, after all. And, of course, I made a small typo: I meant “has two sides” not “as two sides”.

Hi Justin,

You have a deluxe web site. It is truly commendable and remarkable.

I found your site, because I was wondering today about the effects of centrifugal force on perceived weight. I didn’t do the numbers, but I set up the problem much as you did. However, in my mental model I included something which it seems you neglected. Let me explain:

Acceleration is a vector. It has magnitude and direction. Vector arithmetic is not the same as normal (scalar) arithmetic. I think to do the problem correctly you need to do vector arithmetic to combine the two accelerations, as they do not act in the same direction, rather they are separated by an angle which I believe is the same as the latitude of your location–39 degrees in this case.

Adding these two vectors boils down, I think, to finding the third side of a triangle which as two sides of .026 and 9.8, and an angle of (180-39) or 141 degrees. Do you remember the law of cosines? Me neither, but I looked it up (https://www.omnicalculator.com/math/triangle-side). I’ll let you check my math, but I got a vector sum of 9.777 m/s/s. subtracting that from 9.8 gives .023 m/s/s which is still a small fraction of 9.8, but it is significantly less than .026 m/s/s which was your answer for the change in net effective acceleration.

I hope I didn’t make a math error. I think I have the right method however. Let me know what you think.