<p>Originally, the concept of a reference ellipsoid was an ellipsoid that best approximated mean sea level. This concept predates satellite-based geodesy by hundreds of years. There are now better ellipsoids in terms of providing a better fit to mean sea level than the WGS84 ellipsoid. Yet the WGS84 ellipsoid is used worldwide (including the ridiculously over-precise flattening value).</p>

<p>The WGS84 reference ellipsoid is now just that, a reference ellipsoid (that is, it's the ellipsoid used as a reference by GPS and related systems). In 1993 the WGS84 governing committee voted, with one exception, to retain the 1984 defining values, despite improvements to those defining values. The one exception: The flattening in the original definition of WGS84 was a calculated value, based on the Earth's second dynamic form factor as determined from satellite orbits ($\bar C_{2,0}$ in WGS84, $J_2$ in GRS80). This calculated value became a defining value in 1993. The reason for keeping those older values was GPS. It's better to have a reference ellipsoid that is constant over time than it is to have an ever improving best fit ellipsoid.</p>

<p>Besides, not all of the changes in the best fit ellipsoid represent a reduction in measurement errors. The best fit ellipsoid truly does change over time as the Earth continues to recover from the last glaciation and as the Earth's rotation rate slows down. There are very few applications where a best fit ellipsoid truly is needed while there are many applications where a constant reference ellipsoid is the "best fit".</p>

CC BY-SA 4.0
2018-10-21T08:54:06.297Z
clever last sentence in yet another insightful answer.
uhoh
CC BY-SA 4.0
2018-07-25T02:06:32.270Z
2018-07-25T02:12:46.980Z
2018-07-25T02:12:46.980Z
David Hammen
<p>Earth's radius is usually given as 6378137 meters (<a href="https://en.wikipedia.org/wiki/Earth_radius#Global_average_radii" rel="noreferrer">equatorial radius</a>), its flattening is <strong>defined</strong> as <a href="https://en.wikipedia.org/wiki/Flattening#Numerical_values_for_planets" rel="noreferrer">1/298.257223</a> (WGS84), and thus its polar radius is <em>calculated</em> as <a href="https://en.wikipedia.org/wiki/Earth_radius#Polar_radius" rel="noreferrer">6356752 meters</a>.</p>

<p>However, the average depth of the Earth's oceans is <a href="https://www.google.com/search?q=average%20depth%20of%20all%20oceans" rel="noreferrer">3688 meters</a> (i.e. 3688 meters below mean sea level), and the average height of all land above the oceans is <a href="https://www.quora.com/What-is-the-average-elevation-of-Earth-above-the-ocean-including-land-area-below-sea-level-What-is-the-atmospheric-pressure-at-that-elevation" rel="noreferrer">840 meters</a>.</p>

<p>Assuming <a href="https://hypertextbook.com/facts/1997/EricCheng.shtml" rel="noreferrer">71% oceans and 29% dry land</a>, this gives</p>

<p>$0.71 \times (-3688~{\rm m}) ~+~ 0.29 \times 840~{\rm m} ~=~ -2374.8~{\rm m}$. In other words, 2374.8 meters below (current) mean sea level would be the "billiard-ball" Earth if all the oceans were drained<sup>1</sup>.</p>

<p>So, my question is:</p>

<p><strong>What is the defining elevation of how the Earth's radius is measured?</strong> Is it the billiard-ball elevation, current mean sea level, or something else?</p>

<p><sup>1</sup> Of course, this ignores the oblateness caused by the flattening and all the undulations described by the <em><a href="https://en.wikipedia.org/wiki/Geoid" rel="noreferrer">geoid</a></em>.</p>

CC BY-SA 4.0
2018-07-23T22:21:37.637Z
It's ironic (and kinda sad on my part) that you mentioned that site as [I, too, mentioned it in an answer regarding "sea level" on Physics SE](https://physics.stackexchange.com/a/325469/108554).
pr1268
CC BY-SA 4.0
2018-07-23T22:23:39.930Z
P.S. @BarryCarter if you want to "promote" your comment to an answer then I'd certainly accept it.  In the meanwhile I'll have to figure out what to do about my failing memory. :-\
pr1268
<geophysics><geography><earth-observation>
What physical measurement / elevation is used to define Earth's radius?
pr1268
# What physical measurement / elevation is used to define Earth's radius?

(Asked by: username_0 on 2018-07-22)

Earth's radius is usually given as 6378137 meters ([equatorial radius](https://en.wikipedia.org/wiki/Earth_radius#Global_average_radii)), its flattening is **defined** as [1/298.257223](https://en.wikipedia.org/wiki/Flattening#Numerical_values_for_planets) (WGS84), and thus its polar radius is *calculated* as [6356752 meters](https://en.wikipedia.org/wiki/Earth_radius#Polar_radius).

However, the average depth of the Earth's oceans is [3688 meters](https://www.google.com/search?q=average%20depth%20of%20all%20oceans) (i.e. 3688 meters below mean sea level), and the average height of all land above the oceans is [840 meters](https://www.quora.com/What-is-the-average-elevation-of-Earth-above-the-ocean-including-land-area-below-sea-level-What-is-the-atmospheric-pressure-at-that-elevation).

Assuming [71% oceans and 29% dry land](https://hypertextbook.com/facts/1997/EricCheng.shtml), this gives

$0.71 \times (-3688~{\rm m}) ~+~ 0.29 \times 840~{\rm m} ~=~ -2374.8~{\rm m}$. In other words, 2374.8 meters below (current) mean sea level would be the "billiard-ball" Earth if all the oceans were drained1.

So, my question is:

**What is the defining elevation of how the Earth's radius is measured?** Is it the billiard-ball elevation, current mean sea level, or something else?

1 Of course, this ignores the oblateness caused by the flattening and all the undulations described by the *[geoid](https://en.wikipedia.org/wiki/Geoid)*.

**Tags:** geophysics, geography, earth-observation

> **Comments:**
> **username_0**: It's ironic (and kinda sad on my part) that you mentioned that site as [I, too, mentioned it in an answer regarding "sea level" on Physics SE](https://physics.stackexchange.com/a/325469/108554).  
> **username_0**: P.S. @BarryCarter if you want to "promote" your comment to an answer then I'd certainly accept it.  In the meanwhile I'll have to figure out what to do about my failing memory. :-\  

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## 1 answers

### Answer by username_1 on 2018-07-25. Score: 5 | Accepted answer

Originally, the concept of a reference ellipsoid was an ellipsoid that best approximated mean sea level. This concept predates satellite-based geodesy by hundreds of years. There are now better ellipsoids in terms of providing a better fit to mean sea level than the WGS84 ellipsoid. Yet the WGS84 ellipsoid is used worldwide (including the ridiculously over-precise flattening value).

The WGS84 reference ellipsoid is now just that, a reference ellipsoid (that is, it's the ellipsoid used as a reference by GPS and related systems). In 1993 the WGS84 governing committee voted, with one exception, to retain the 1984 defining values, despite improvements to those defining values. The one exception: The flattening in the original definition of WGS84 was a calculated value, based on the Earth's second dynamic form factor as determined from satellite orbits ($\bar C\_{2,0}$ in WGS84, $J\_2$ in GRS80). This calculated value became a defining value in 1993. The reason for keeping those older values was GPS. It's better to have a reference ellipsoid that is constant over time than it is to have an ever improving best fit ellipsoid.

Besides, not all of the changes in the best fit ellipsoid represent a reduction in measurement errors. The best fit ellipsoid truly does change over time as the Earth continues to recover from the last glaciation and as the Earth's rotation rate slows down. There are very few applications where a best fit ellipsoid truly is needed while there are many applications where a constant reference ellipsoid is the "best fit".

> **Comments:**
> **username_2**: clever last sentence in yet another insightful answer.  

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