# How to calculate the speed of rotation of a given point on the Earth's WGS84 ellipsoid, about its axis?

Launches often happen near the equator in order to take advantage of Earth's rotation speed. This is why more launches are towards the East than towards the West.

How can one calculate the speed of rotation of a point on the Earth's surface about the Earth's axis, given only a latitude?

While the simplest approach might be to assume the Earth is a sphere and it rotates once per day, it's rotation isn't exactly 24 hours, and the Earth is definitely not exactly a sphere!

• Hi @EdRojas, welcome to Stack Exchange! Your original question was very "homework-like", and while the answer would have been interesting to you, it didn't really meet the criteria for what a good Stack Exchange question should be like. I've edited it and included a bit more to the question. I'll try to post an answer to kph at Latitude 14.6042* for you fairly soon. You can read more about asking in the Help Center's FAQ, and also you can take the tour – uhoh May 11 '18 at 8:40
• The Earth also doesn't spin at a constant rate -- vis-a-vis the Ut1-UTC time correction. Do you care about that level of discrepancy? – cms Feb 10 '19 at 2:06

At any latitude, the Earth completes one rotation per day.

At the equator, the circumference of the Earth is about 40000 km, so the speed of rotation is 40000 km/day or 463 m/s.

If you pick a line of higher latitude and look at it on a globe, you will see that the line of latitude is smaller than the equator.

One rotation completed in a day is therefore a shorter distance; at 60 degrees latitude the line is about (cos 60 x 40000) = 20000 km long, so the speed of rotation is 20000 km/day or 230 m/s.

• thanks for the quick response. The circumference of the latitude becomes smaller as we move away from the equator. But why does it obey the cosine function ? – Soumajit Sep 4 '16 at 16:11
• Because the Earth is approximately spherical, so it's cross section is approximately circular. Draw a circle, and a radius line from the center to the edge. This is a side view cross section of the Earth. The angle made between the horizontal and the radius line is the angle of latitude. The distance from the vertical to the point where the radius meets the circle is how the cosine is defined. – Russell Borogove Sep 4 '16 at 16:17
• If you need precision, you need to factor in the shape of the earth. More info: en.wikipedia.org/wiki/Earth_ellipsoid – Cem Kalyoncu Apr 13 '17 at 8:33

According to Wikipedia's Geographic_coordinate_conversion#From_geodetic_to_ECEF_coordinates

The 3D cartesian coordinates $$X, Y, Z$$ in Earth-centered, Earth-fixed coordinates assuming an ellipsoidal shape is given by:

$$X = \left(N(\phi) + h \right) \cos\phi \cos\lambda$$

$$Y = \left(N(\phi) + h \right) \cos\phi \sin\lambda$$

$$Z = \left(\frac{b^2}{a^2} N(\phi) + h \right) \sin\phi$$

where $$\phi, \lambda, h$$ are latitude, longitude, and altitude, and $$a, b$$ are the equatorial and polar radii of the ellipsoid used, and

$$N(\phi) = \frac{a^2}{\sqrt{a^2\cos^2\phi + b^2 \sin^2\phi}}.$$

Lat, lon, alt in "GPS coordinates} is based on WGS 84 with $$a, b$$ of 6378.1370 and 6356.7523 kilometers, respectively.

Let's say I was in Manila and needed to know the speed I was moving around the Earth's axis while paying a parking ticket. According to Google Maps, the GPS coordinates are about 14.590037, 120.981361. Since we're so close to the ocean I'll estimate the altitude to be 50 meters.

Plugging all of this into the equations above, I get:

$$X, Y, Z =$$ -3178031.13, 5293031.55, 1596255.68 meters. The radius, or distance from the axis is

$$r = \sqrt{X^2 + Y^2} = \text{6173820.93 meters.}$$

Last time I checked (1970) the Earth's rotation period was 23h, 56m, 4.09s, or 86164.09 seconds. Speed is then the circumference divided by the rotation period:

$$speed = \frac{2 \pi r}{T}$$

That makes the speed 450.202 m/s or 1620.72 kph.

Thanks to @MattJessick for reminding me about the $$2 \pi$$!

Now what do we get if we do this the simpler way, assuming the Earth is a sphere. We can say the distance to the axis is just some radius times $$cos \phi =$$ 0.967752987 , but what radius? Commonly people use the equatorial radius for places near the equator, or about 6378 km, though sometimes 6371 is used for an "average".

Those give 450.09 and 449.59 m/s or 1620.34 and 1618.56 kph.

• (All very good, but you are off by a factor of 6. The spherical Earth velocity about the Earth's rotation axis at 14.6 deg latitude is more like 450 m/s. I suspect you might have used distance from Earth +Z axis as R Sin(latitude), rather than the correct R cos(latitude)) – Matt Jessick May 18 '18 at 23:30
• @MattJessick ha! I forgot the 2π relating radius to circumference. Great catch, thank you! I've make an edit and added the step explicitly. – uhoh May 19 '18 at 0:05
• Great! Thanks. But why is the answer 1,618 or 1,620KPH at Manila Philippines' Latitude so close and even faster than the speed at the equator? Equator surface speed is 1,609 KPH per this link: scientificamerican.com/article/how-fast-is-the-earth-mov – Eduardo C Rojas Jun 1 '18 at 7:51
• Hi @EduardoCRojas , I'll take look now. I think you are using an answer post to ask a question because you can not yet comment on other people's posts until you get a reputation of 50. That comes pretty quick if you participate more. Once this is cleared up, it's best i you delete this answer since it isn't an answer. Also, I noticed the similarity between your username and the OP's (original poster = question asker). If both usernames belong to the same person, there may be a way to merge and combine scores. – uhoh Jun 1 '18 at 10:46
• Also I didn't see your message until it came up in the review queue. You have to use the @ symbol (like @uhoh) to be sure someone receives a notification of your message. – uhoh Jun 1 '18 at 10:47

Linear speed in circular motion depends on angular velocity (here: 1 rotation per day, for all latitudes) and radius of the circle - distance from the axis of rotation.

$$v = \omega r$$

If $$\alpha$$ is your latitude, the radius of the circle you make with each revolution of Earth, is $$r$$. As you can see above, basic trigonometry, $${r \over R} = cos \alpha$$. The closer you are to the pole, the more each degree of latitude decreases the radius of the circle you make.

Indirect consequence is that moderate distance of the space center from the equator diminishes delta-V much less than larger distance.

Take:

French Guiana Space Centre is 5 degrees North; Cape Canaveral is 28 degrees North; Baikonur is 45 degrees North, Plesetsk is 63 degrees North.

It would superficially seem that the velocity losses between Guiana and Canaveral (23° apart) are higher than between Canaveral and Baikonur (17°) or Baikonur and Plesetsk (18°). But this is not the case -

$$\cos 5° - \cos 28° = 0.11;$$

$$\cos 28° - \cos 45° = 0.17;$$

$$\cos 45° - \cos 63° = 0.25$$

so the 17 degrees Baikonur loses to Canaveral is way worse than 23 degrees Canaveral loses to Guiana or the 58 degrees Baikonur loses to Guiana.