**Question:** If Earth spins 1000 miles per hour, why does it take 24 hours for a day and not just a few seconds? — DSF via gmail

**Answer: **When somebody claims “Earth spins at 1000 mph” they’re referring to a point on the equator where the rotation speed is greatest. In fact, the speed of a point on the equator is 1038 mph. You get that speed by dividing the circumference of Earth (24,901 miles) by the 24 hours it takes Earth to rotate once.

However, if you’re at a location other than the equator the speed will be less. For example, at a higher latitude, say 30°, the distance you get carried in one day by Earth’s rotation is not the equatorial circumference. It’s only 21,507 miles around in 24 hours, so that works out to a speed of 896 mph. At 60° latitude the speed is further reduced to 518 mph.

And if you were located at either pole, you wouldn’t be “moving” at all … you’d just be rotating in place once every 24 hours so your linear speed would be 0 mph. Check out the graphic for more details.

When talking about the speed of a rotating object it’s easy to make unintentionally misleading statements. Physics textbooks get it right, but the popular media often doesn’t. The proper way to talk about “rotation speed” is to express it in units of *angular speed* (ω), i.e., “degrees or radians / hour”, or “degrees or radians / second”. From that number, and the dimensions of the rotating object, it’s possible to calculate the speed at any point on its surface. That’s what I did in the graphic.

Earth rotates once every day, so that means its angular speed is: ω = Δθ / Δt = 360° / 24 hr = 15°/hr (the symbol “θ” is used for angles). This allows you to calculate the speed of a point on the surface by using the usual v = Δd / Δt, where:

- Δd is the circumference of the circular path = 2πR

- Δt is the time it takes for one rotation = 360° / ω).

There are several consequences of this global speed distribution. First, it affects rocket launches. Although rockets can be launched from any point on Earth’s surface, and in any direction required for the mission, it turns out that launching toward the east from a point near the equator allows for larger payloads.

Achieving low Earth orbit (LEO) requires a speed of 7.8 km/s relative to the Earth’s center. Launching to the east from the equator means the rocket is already moving at 0.5 km/s (1038 mph) in that direction. So the ΔV provided by the fuel need only be 7.3 km/s. Less fuel weight means more payload weight. Launching to the west changes that calculus and, all things being equal, would require a ΔV of 8.3 km/s. This is why most launch sites are as close to the equator as possible within geographic and border constraints.

Another significant consequence of the speed distribution is the *Coriolis Force* — the east or west deflection of objects moving to the north or south. This includes ocean currents and air flow in the atmosphere, and that in turn influences the development and movement of weather systems. It also affects ICBMs and long-range artillery shells, and has to be compensated for when aiming.

One other (minor) effect of this speed distribution is the action of *centrifugal force*. Because of this force you’ll weigh less at the equator and more at the poles. Not by much — the difference amounts to only 0.35%. So don’t bother moving to the equator as part of your weight loss regimen 🙂

Next Week in Sky Lights ⇒ The Coriolis Force