(e.g., longitude/latitude, elevation, and Earth’s mass and rotational speed)

I know a couple of approximations for the gravity due to Earth’s mass. The usual one I’ve been given is $g \approx 9.8 \, \mathrm{m / s^2}$; and there is also a generic equation for any two objects with mass, $F = G \frac {m_1 \cdot m_2} {r^2}$, which remains to be divided by the mass of the object of interest in order to obtain gravitational acceleration. However, I was interested (out of sheer curiosity, but also because I am taking a college-level physics course) in finding an equation that took into account other variables, such as longitude and elevation, as exemplified by some online calculators.

I did some research and found the Wikipedia article “Gravity of Earth”, with the section of interest being § Mathematical models. However, I had trouble finding an equation on this page (or on the page about Earth’s theoretical gravity) that concisely stated an accurate equation — or sequence of successively more refined equations — for the gravity experienced at a point on/above/below Earth’s surface.

Have I missed a resource? Alternatively, does someone here perhaps have the knowledge and expertise to write these equations succinctly? A couple of variables I am trying to isolate are…

  • latitude and longitude on an oblate spheroid approximation of Earth;
  • the distance away from Earth’s center of mass, or equivalently, the elevation above/below sea level;
  • the current angular velocity about Earth’s axis of rotation, which I understand undergoes minor variations that change the duration of, say, the solar year (this is relevant since the inertial frame might change w.r.t. the centripetal force of gravity); and
  • Earth’s mass, which I imagine stays more or less constant with respect to our instruments’ precision.

In the meantime, I will be reading the relevant resources linked from Physics Stack Exchange 🙂

  • $\begingroup$ What level of precision do you need? $\endgroup$ Sep 21, 2020 at 20:33
  • $\begingroup$ I gave an extensive answer to a similar but broader question at the Space Exploration SE. I don't feel like replicating that answer here. $\endgroup$ Sep 22, 2020 at 4:34

2 Answers 2


Based on your research, the answer is clearly "no". There is so simple formula, and a comprehensive discussion of Earth's gravity is well beyond the scope of a PSE answer. It is in fact, and entire field of active research.

The current standard for Earth's gravity is called "The Geoid" (EGM2008: https://earth-info.nga.mil/GandG/wgs84/gravitymod/egm2008/), and involves a multipole expansion to $l=2159$. Lookup tables for the unclassified geoid are publicly available. This represents a static tide-free equipotential surface, and not $g$. Additional tables are available for the gravitational anomaly and deflection from vertical.

The GRACE NASA mission and its follow-on (https://www.nasa.gov/mission_pages/Grace/index.html) use precision orbit metrology to measure changing mass distributions on Earth with astonishing precision.

Regarding the Earth's rotation, that is tracked by the Naval Observatory (https://www.usno.navy.mil/USNO/earth-orientation).


Compared to $F = G \frac {m_1 \cdot m_2} {r^2}$ the next level up is Mccullagh's formula.

Mccullaghs formula adds a term that is proportional to $\frac{1}{r^4}$, this term is referred to as the quadrupole term. This quadrupole term has a sine factor such that the adjustment in the plane of the equator is opposite in sign to the adjustment at the poles. This extra term accomodates that for an oblate spheroid the gravitational force is not spherically symmetrical.

Among the difficulties: celestial bodies don't have a uniform density. The core is denser than the outer layers. When the celestial body is rotating, hence has an equatorial bulge, the non-sphericity of the gravitational force also depends on how the density is a function of distance to the geometric center.

Depending on the density distribution the moment of inertia of the celestical body comes out differently. The moment of inertia of a (spinning) celestial body can be inferred from the rate of gyroscopic precession of that celestial body.

The following link is a chapter from a book on the internal dynamics of celestial bodies (particularly the Earth) In this chapter: a comprehensive derivation of Mccullagh's formula is given:

For the moment of inertia of a celestial body see esspecially slide 4.5 and 4.6 of this series of slides:


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