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Earth Property Calculations

This page provides an explanation of CalQlata's data relating to the earth and its moon, which has been established using Isaac Newton's universal gravitational theories.

Newton's Law of Gravitational Force

Newton's theory for gravitational force is calculated thus:
F = G.m₁.m₂ / R²
F is the gravitational force of attraction {N}
G = 6.67359232004332E-11 {N.m²/kg²}⁽¹⁾
m₁ = mass of a force centre (e.g. the earth) {kg}
m₂ = mass of a satellite (e.g. the earth's moon) {kg}
R = distance between the centre of m₁ and the centre of m₂ {m}

Earth's Mass

If we apply the above formula to the earth, by removing m₂ (the mass of the earth's moon) from the equation the formula becomes:
g = G . m₁ / R² {m/s²}

At the earth's surface (e.g. sea-level): g = 9.80663139027614m/s²
From which we can accurately calculate the mass of the earth as follows:
The volumetric radius of the earth at sea-level is:
R = ³√(6356752 + 6378137²) = 6371000.68502598m
m₁ = g . R² / G
    = 9.80663139027614 x 6371000.68502598² / 6.67359232004332E-11
    = 5.9645197677618E+24kg

CalQlata's UniQon calculator has an alternative calculation method taking into account the uneven nature of the earth's composition giving 'g' a value of 9.80616 at latitude 45° and an alternative mass of 5.96629861115861E+24kg

Therefore, we can confidently predict the actual mass of the earth as:
m₁ = 5.9654091894602E+24kg ±0.01491%
however, because of the ability to match the former value for mass with Newton's laws, CalQlata shall set the earth's mass as 5.9645197677618E+24kg for all of its internal calculations

Earth's Density

The earth's surface area = 4π x 6356752 x 6378137 = 5.094938849322E+14m²

The earth's volume = 4/3π x 6371000.685³ = 1.08320726625321E+21m³

So the density of the earth can be calculated thus:
ρ₁ = 5.9645197677618E+24 / 1.0832072662532E+21 = 5506.35132682682kg/m³

Earth's Geosynchronous Orbit

Centrifugal acceleration: a = v² / R

Gravitational acceleration: g = G.m₁ / R²

Where v is the velocity of the satellite at R and g ≡ a

From the above equations:
R = G.m₁ / v²
v = 2.π.R/86400 m/s⁽⁴⁾
v² = R².(2.π/86400)² = R² x 5.288496871297E-09 m²/s²
R = G x m₁ / (R² x 5.288496871297E-09)
R = (G.m₁ / 5.288496871297E-09)

The altitude of the earth's geosynchronous orbit is 'R' minus the radius of the earth: 4.22215624808981E+07m - 6378137m = 3.584342548E+07m

The Moon's Mass

According to NASA, the average gravitational acceleration on the surface of the earth's moon is 1.624m/s², its radius is 1737494.51373m and the average distance between the centres of the earth and its moon is 3.83E+08m⁽²⁾.

By applying Newton's formula to the earth's moon: g = G . m₂ / R²
m₂ = g.R² / G
    = 1.624 x 1737494.51373² / 6.67359232004332E-11 = 7.34637741371792E+22kg

The Moon's Density

The moon's surface area: A = 4π x 1737494.51373² = 3.79364552126886E+13m²

The moon's volume is: V = 4/3π x 1737494.51373³ = 2.19714609341368E+19E+19m³

The moon has a density of: ρ₂ = m₂ / V = 3343.59987974397kg/m³

Earth's Gravitational Force on its Moon

According to CalQlata's calculations, the moon's average orbital velocity is 1021.3712417m/s. According to NASA⁽⁴⁾, the mean velocity of the moon is actually 1022m/s, which may be why it is gradually receding

Earth's gravitational acceleration at the orbital radius of its moon is calculated as follows:
From; F = m₁.v² / R we get; F/m₁ = v²/R = gm = 0.0027237143674m/s²
Alternatively, from; F = G.m₁.m₂ / R² we get;
a = F/m = G.m₂ / R² = 0.002713462204m/s²
Therefore, the mean gravitational attraction by the earth on its moon is:
gm = (0.0027237143674 + 0.002713462204)/2 = 0.0027185882857m/s²

The gravitational force between the moon and the earth is:
F = m₂ . gm = 7.34637741371792E+22 x 0.0027185882857 = 1.99717755792646E+20N
which is an average value
See Laws of Motion for accurate values

Earth's Material Content

Our estimates below are based upon the earth comprising three principle layers: the inner core, the outer core and the mantle (the plate/crust comprises less than 0.2% of the mantle thickness and so will be ignored).

As the average densities and dimensions of the earth’s internal structure is now known from our planetary spin calculations, we can estimate the amounts of iron, nickel, basalt (2900 kg/m³) and lighter materials (carbon silicon, sulphur, etc.; average density of ≈2200 kg/m³) in each layer:

ContentInner CoreOuter CoreMantle
Mass (kg)1.02534E+231.89834E+24 3.96364E+24
Radius (m)123484534348456371000.685
Average Density (kg/m³)13000117284339.2
Percentage Iron90% #80% #46.945%
Percentage Nickel5% #10% #5.907%
Percentage Silicates 5% #10% #1.450%
Percentage Basalt45.69%
Mass of Iron (kg)9.22809E+221.51867E+245.35062E+23
Mass of Nickel (kg)5.12672E+211.89834E+232.97257E+22
Mass of Silicates (kg)5.12672E+211.89834E+235.77947E+22
Mass of Basalt (kg)3.39886E+24
# estimated

Comment & Discussion

If the moon comprised the same materials as the inner planets its density should be similar (≈5,500 kg/m³). However, our moon's density (3343.6 kg/m³) is closer to that of Mars (3934.1 kg/m³), which means it may once have been another outer rocky planet that was captured by the earth when it ventured off course, probably from the Asteroid belt.


  1. Nature magazine discusses problems with the accuracy of this value
  2. NASA
  3. NASA moon facts
  4. There are ≈86400 seconds in one day {24 x 60 x 60}
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