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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!

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    $\begingroup$ 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 $\endgroup$ – uhoh May 11 '18 at 8:40
  • $\begingroup$ 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? $\endgroup$ – cms Feb 10 at 2:06
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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.



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    $\begingroup$ (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)) $\endgroup$ – Matt Jessick May 18 '18 at 23:30
  • $\begingroup$ @MattJessick ha! I forgot the 2π relating radius to circumference. Great catch, thank you! I've make an edit and added the step explicitly. $\endgroup$ – uhoh May 19 '18 at 0:05
  • $\begingroup$ 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 $\endgroup$ – Eduardo C Rojas Jun 1 '18 at 7:51
  • $\begingroup$ 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. $\endgroup$ – uhoh Jun 1 '18 at 10:46
  • $\begingroup$ 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. $\endgroup$ – uhoh Jun 1 '18 at 10:47

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