# Earth Gravity Calculator

Saturday, September 1, 2018 - 20:45 | Author: wabis | Topics: Calculator, Knowlegde, Mathematics, Physics
This calculator can be used to compute the effective gravitational acceleration, the pure gravitational acceleration and the centrifugal acceleration for any point on the Reference Ellipsoid of the earth for both surface and at altitude according to the model WGS84.

## Calculation Form

Vectors are ECEF (Earth Centered, Earth Fixed) coordinates. X axes points from earth center to 0-Meridian at the equator, Y axes points from earth center to equator at 90° longitude and Z axes points from earth center to north pole.

## Legend

Reset resets all fields ​to their initial values. To reset only a specific field to its initial value, place the cursor in the field and press the ESC key.

Lat: Degree of Latitude of point P. The equator is 0, the North Pole 90 and the South Pole -90 degrees.

Alt: Altitude above sea level of point Ph.

Long: Degree of Longitude of point P. The 0-Meridian is 0, negative values are direction west, positive values are direction east.

Weight: Weight of the test mass in arbitrary units.

You may enter Lat and Long in any commonly used formats.

Note: in the following when a value is given for the point P or Ph the same value is also valid everywhere on the same latitude.

Wo: Weight of the test mass displayed on the scale at sea level at point P.

go: Effective gravitational acceleration on earth's surface at point P, see (9). This corresponds to the vectorial sum of pure gravitational acceleration goG and centrifugal acceleration aoC.

goG: Pure gravitational acceleration on earth's surface at point P, see (14). Depending on the mass M of the earth and the distance Ro from the center of the earth.

aoC: Centrifugal acceleration on earth's surface at point P, see (11). This is dependent on the rotational speed or rotation period T and the radius rφ for the latitude Lat. The radius rφ is the perpendicular distance of the point P from the earth axis. On the poles this radius is 0, on the equator it is equal to the radius of the earth at that location.

v0: Tangential velocity at the surface at point P. Depends on the rotational speed ω or the rotational period T and the radius rφ where rφ is the perpendicular distance of the point P from the earth axis.

Ro: Distance from the center of the earth to to the point P, see (6). This radius changes with position because the earth is an ellipsoid.

Wh: Weight of the test mass as displayed on the scale at altitude Alt at point Ph.

gh: Effective gravitational acceleration at altitude Alt at point Ph, see (17). This is the vectorial sum of pure gravitational acceleration ghG and centrifugal acceleration ahC.

ghG: Pure gravitational acceleration at altitude Alt at point Ph, see (21). Depending on the mass M of the earth and the distance Rh from the center of the earth, i.e. the earth radius at latitude Lat plus the altitude Alt.

ahC: Centrifugal acceleration at altitude Alt at point Ph, see (19). This is dependent on the rotational speed or rotation period T and the radius rφ for the latitude Lat. The radius rφ is the perpendicular distance of the point Ph from the earth axis.

vh: Tangential velocity at altitude Alt and at point Ph. Depends on the rotational speed ω or the rotational period T and the radius rφ where rφ is the perpendicular distance of the point Ph from the earth axis.

Rh: Distance from the center of the earth to to the point Ph at altitude Alt, see (7). This radius changes with position because the earth is an ellipsoid.

## Effective Gravitational Acceleration

On each point P on earth there is a gravitational force F acting on a mass m down to the center of the earth. Gravity can also be expressed as a gravitational acceleration g = F / m. It is prefered to use the gravitational acceleration instead of the force, because the acceleration is the same for any mass m, so m is not contained in the formula.

On a perfect non-rotating sphere the gravitational acceleration for each point on the surface of the sphere is according to Newton:

(1)
where'
 $g$ ' =' 'pure gravitational acceleration $G$ ' =' '6.674 × 10−11 m3/kg/s2 = gravitational constant $M$ ' =' '5.972 × 1024 kg = mass of the earth $R$ ' =' '6371 km = radius of the earth

Now, the earth is not a perfect sphere but an oblate ellipsoid of revolution, also called an oblate spheroid, and is rotating with the angular speed ω around its minor axis. This means that the pure gravitational acceleration is dependent on the latitude and there is another component acting against gravity: the centrifugal acceleration aC, which is also dependent on the latitude. So the formulas to compute the effective gravitational acceleration on a point on earth is much more complicated.

The effective gravitational acceleration at any point on earth is the vector sum of the pure gravitational acceleration due to gravity plus the centrifugal acceleraion due to earth's rotation. The ellipsoid has just such a shape, so that the effective gravitational acceleration acts everywhere perpendicular to the surface of the ellipsoid. This implies that the pure gravitational acceleration generally does not act perpendicular to the surface.

Why is it that the ellispoid has a shape just right so that the effective acceleration is everywhere perpendicular to the surface?

This has to do with the fact that the earth is mostly liquid and the surface of any liquid always is level to an equipotential surface. An equipotential surface is a surface where the effective gravitational acceleration has the same magnitude everywhere. The equipotential surfaces form ellipsoidal shells around the center of the planet, while the surface of any ellipsoidal shell is always perpendicular to the effective gravitational acceleration at any point. Any local deviation of the surface of a liquid from such a shell surface would cause tangential forces at that location which produce surface streams of the liquid until the whole surface of the planet is on the same equipotential ellipsoid and no tangential forces arise.

So it is a natural process of each big rotating planet in equilibrium to form such an ellipsoidal shape.

The WGS84 ellipsoidal model of the earth is the official model used in geodesy and by GPS, called the Reference ellipsoid. On this page I use the WGS84 model of the earth and calculate all the individual accelerations involved accordingly.

## Model of the Earth, Reference Ellipsoid

In the following sections, all the formulas used in the Calculation form are listed and explained:

The earth is not a perfect sphere. It is slightly flattened at the poles and the diameter at the equator is 42.8 km greater than at the poles. Gravitation on the surface is also not uniform, but varies by mass distribution on the surface and inside the earth.

For the calculations on this page I use a Reference ellipsoid. This is a rotation ellipse with the earth axis as the rotation axis and the origin at the center of the earth. This ellipsoid has the following parameters [1]:

(2)

$a = 6{,}378{,}137\ \mathrm{m}$

Semi major axes, i.e. radius on the equator

$b = 6{,}356{,}752{.}3142\ \mathrm{m}$

Semi minor axes, i.e. radius on the poles

$R = 6{,}371{,}008{.}8\ \mathrm{m}$

Average radius of the sphere earth at about 35° latitude

$g_\mathrm{e} = 9{.}7803253359\ \mathrm{m}/\mathrm{s}^{2}$

Effective acceleration on the equator

$g_\mathrm{p} = 9{.}8321849378\ \mathrm{m}/\mathrm{s}^{2}$

Effective acceleration on the poles

$G \cdot M = 3{.}986004418 \times 10^{14}\ \mathrm{m}^{3}/\mathrm{s}^{2}$

Geocentric gravitational constant [2]

$T = 86{,}164{.}098903691\ \mathrm{s}$

Rotation period with respect to space (sidereal day) [3]

$\omega = 7{.}292115 \times 10^{-5}\ \mathrm{rad}/\mathrm{s}$

Angular speed

where'
 $G$ ' =' 'gravitational constant $M$ ' =' 'mass of the earth $\omega$ ' =' '2 π / T

The product G·M can be determined more precisely than the individual factors [4].

## Position on the Reference Ellipsoid

Ellipsoid with a degree of latitude φ and an oblique circle radius ρ

In order to calculate the accelerations at a specific point P on the reference ellipsoid, we must calculate the position from latitude φ and longitude λ. Along a latitude line φ all accelerations are the same in magnitude.

The following formula can be used to calculate a point P on the surface of the earth with h = 0 or to compute a point Ph that is at a distance h from this point perpendicular above the surface [5].

(3)
(4)
(5)
with
where'
 $P_\mathrm{r}$ ' =' 'distance of the point P measured from the earth axis $P_\mathrm{z}$ ' =' 'distance of the point P measured from the equatorial plane $\vec P$ ' =' 'position vector of point P in ECEF coordinates $h$ ' =' 'altitude measured perpendicular to the surface of the ellipsoid $\varphi$ ' =' 'degree of latitude in radian = degree · π / 180 $\lambda$ ' =' 'degree of longitude in radian = degree · π / 180 $a$ ' =' 'semi major axis of the ellipse (radius at the equator), see (2) $b$ ' =' 'semi minor axes of the ellipse (radius to the poles), see (2)

Vectors are in ECEF (Earth Centered, Earth Fixed) coordinates. X axes points from earth center to 0-Meridian at the equator, Y axes points from earth center to equator at 90° longitude and Z axes points from earth center to north pole.

The radius Ro at point P is the distance from the center of the earth to the point P which is the length of the vector $\vec P$:

(6)
where'
 $P_\mathrm{x}, P_\mathrm{y}, P_\mathrm{z}$ ' =' 'components of the vector $\vec P$, see (3)

The radius Rh at point Ph is the distance from the center of the earth to the point Ph which is the length of the vector $\vec P_\mathrm{h}$. You get the vector $\vec P_\mathrm{h}$ with (3) to (5) by using h0:

(7)
where'
 $P_{\mathrm{h},\mathrm{x}}, P_{\mathrm{h},\mathrm{y}}, P_{\mathrm{h},\mathrm{z}}$ ' =' 'components of the vector $\vec P_\mathrm{h}$, see (3)

## Calculations on the Surface

### Tangential Speed on the Surface

The magnitude of the tangential speed vo at the surface of the earth at position P is:

(8)
where'
 $v_\mathrm{o}$ ' =' 'tangential speed at position P $\omega$ ' =' 'angular speed, see (2) $P_\mathrm{r}$ ' =' 'distance of the point P from the axis of rotation, see (3)

### Effective Gravitational Acceleration on the Surface

In geosciences, the effective gravitational acceleration of a celestial body is composed of its pure gravitational acceleration due to the mass of the body and the centrifugal acceleration in the reference system, which rotates with the body [6] [7].

The earth is slightly flattened on the poles due to its rotation. It has the form of a rotational ellipsoid. For such an ellipsoid, according to WGS84 the effective acceleration can be calculated as follows [1]:

(9)
where'
 $g_\mathrm{o}$ ' =' 'effective acceleration at latitude φ $\varphi$ ' =' 'degree of latitude in radian = degrees · π / 180 $a$ ' =' 'semi major axis of the ellipse (radius at the equator), see (2) $b$ ' =' 'semi minor axes of the ellipse (radius to the poles), see (2) $g_\mathrm{e}$ ' =' 'effective gravitational acceleration on the equator, see (2) $g_\mathrm{p}$ ' =' 'effective gravitational acceleration at the poles, see (2)

The effective gravitational acceleration acts perpendicular to the surface of the ellipsoid at point P. It can be specified in vector form as follows:

(10)

Note: the vector $\vec g_\mathrm{o}$ with origin $\vec P$ generally does not point to the center of the earth.

### Centrifugal Acceleration on the Surface

The centrifugal acceleration on the earth's surface aoC acts perpendicular to the rotation axis, depends on the degree of latitude φ and always acts outwards, i.e. its Z component is 0. The magnitude of the centrifugal acceleration is therefore equal to its XY component.

(11)
(12)
where'
 $a_\mathrm{oC}$ ' =' 'magnitude of centrifugal acceleration at sea level $\vec a_\mathrm{oC}$ ' =' 'vector of centrifugal acceleration at sea level $\omega$ ' =' 'angular speed, see (2) $P_\mathrm{r}$ ' =' 'distance of the point P from the axis of rotation, see (3) $\varphi$ ' =' 'degree of latitude in radian = degrees · π / 180 $\lambda$ ' =' 'degree of longitude in radian = degree · π / 180

The angular velocity ω indicates how fast the earth rotates about its axis. It can be calculated from the sidereal rotation period T as follows:

 (13)

The Sidereal day is the duration of a full revolution of the earth around itself against the fixed starry sky. The mean sidereal day on the earth is almost 4 minutes shorter than the sloar day of 24 hours, see T in (2).

### Pure Gravitational Acceleration on the Surface

The pure gravitational acceleration at point P can be determined vectorially:

(14)
(15)
where'
 $\vec g_\mathrm{oG}$ ' =' 'pure gravitational acceleration at sea level $g_\mathrm{oG}$ ' =' 'magnitude of gravitational acceleration at sea level $\vec g_\mathrm{o}$ ' =' 'effective acceleration at sea level, see (9) $\vec a_\mathrm{oC}$ ' =' 'centrifugal acceleration at sea level, see (11)

## Calculations at Altitude

In order to calculate the values at altitude h, we must calculate its position Ph from the latitude φ and altitude h over sea level. The position Ph at altitude h can be calculated by the formula (3).

### Tangential Speed at Altitude

The magnitude of the tangential speed vh at altitude h at position Ph is:

(16)
where'
 $v_\mathrm{h}$ ' =' 'tangential speed at position Ph $\omega$ ' =' 'angular speed, see (2) $P_\mathrm{r}(h)$ ' =' 'distance of the point Ph above the point P from the axis of rotation, see (3)

### Effective Gravitational Acceleration at Altitude

The effective gravitational acceleration at altitude h according to WGS84 can be calculated as follows [1]:

(17)
with
where'
 $g_\mathrm{h}$ ' =' 'effective gravitational acceleration according to WGS84 at a distance h from the ellipsoid $g_\mathrm{o}$ ' =' 'effective gravitational acceleration according to WGS84 at point P on the ellipsoid, see (9) $a$ ' =' 'semi major axis of the ellipse (radius at the equator), see (2) $\omega$ ' =' 'angular velocity of the earth's rotation, see (2) $\varphi$ ' =' 'degree of latitude in radian = degrees · π / 180 $h$ ' =' 'altitude above the surface of the sea in m = feets · 0.3048 m/ft

The effective acceleration according to WGS84 acts perpendicular to the surface of the ellipsoid. The vector representation is therefore:

(18)
where'
 $\vec g_\mathrm{h}$ ' =' 'effective acceleration vector $g_\mathrm{h}$ ' =' 'effective acceleration magnitude $\varphi$ ' =' 'degree of latitude in radian = degrees · π / 180 $\lambda$ ' =' 'degree of longitude in radian = degree · π / 180

### Centrifugal Acceleration at Altitude

The centrifugal acceleration due to the rotation of the earth at point Ph is:

(19)
(20)
where'
 $a_\mathrm{hC}$ ' =' 'magnitude of centrifugal acceleration due to earths rotation in altitude h $\vec a_\mathrm{hC}$ ' =' 'vector of centrifugal acceleration due to earths rotation in altitude h $\omega$ ' =' 'angular speed, see (2) $P_\mathrm{r}(h,\varphi)$ ' =' 'distance of the point Ph from the rotational axis, see (3) $\varphi$ ' =' 'degree of latitude in radian = degrees · π / 180 $\lambda$ ' =' 'degree of longitude in radian = degree · π / 180

### Pure Gravitational acceleration at Altitude

The pure gravitational acceleration $g_\mathrm{hG}$ at altitude is obtained by subtracting the centrifugal acceleration $a_\mathrm{hC}$ due to earths rotation from the effective gravitational acceleration $g_\mathrm{h}$ at altitude h calculated according to WGS84.

(21)
where'
 $\vec g_\mathrm{hG}$ ' =' 'pure gravitational acceleration at altitude $\vec g_\mathrm{h}$ ' =' 'effective gravitational acceleration according to WGS84 for altitude h, see (FallF84) $\vec a_\mathrm{hC}$ ' =' 'centrifugal acceleration with respect to the rotational axis of the earth at the Point Ph, see (ZentriF84)

The magnitude of the pure gravitational acceleration is:

(22)

## Sources

[1]
Technical Report, TR 8350.2, 3rd edition; January 2000, WGS84
http://earth-info.nga.mil/GandG/publications/tr8350.2/wgs84fin.pdf
World Geodetic System 1984; Wikipedia
https://de.wikipedia.org/wiki/World%5FGeodetic%5FSystem%5F1984
Earth; Wikipedia(en)
https://en.wikipedia.org/wiki/Earth
World Geodetic System; Wikipedia
https://de.wikipedia.org/wiki/World%5FGeodetic%5FSystem
Referenzellipsoid; Wikipedia
https://de.wikipedia.org/wiki/Referenzellipsoid
Schwerefeld; Wikipedia
https://de.wikipedia.org/wiki/Schwerefeld
Gravitation und Schwere; Das GOCE-Projektbüro Deutschland am Institut für Astronomische und Physikalische Geodäsie (IAPG) der TU München
http://www.goce-projektbuero.de/7863--~goce~Goce~Produkte~Level_2_Produkt~gravitation_und_schwere.html

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