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If the Earth had a constant density ρ, the mass would be M(r) = (4/3)πρr 3 and the dependence of gravity on depth would be =. The gravity g′ at depth d is given by g′ = g(1 − d/R) where g is acceleration due to gravity on the surface of the Earth, d is depth and R is the radius of the Earth.
The result reported by Charles Hutton (1778) suggested a density of 4.5 g/cm 3 (4 + 1 / 2 times the density of water), about 20% below the modern value. [16] This immediately led to estimates on the densities and masses of the Sun , Moon and planets , sent by Hutton to Jérôme Lalande for inclusion in his planetary tables.
[1] [2] The acceleration of a body near the surface of the Earth is due to the combined effects of gravity and centrifugal acceleration from the rotation of the Earth (but the latter is small enough to be negligible for most purposes); the total (the apparent gravity) is about 0.5% greater at the poles than at the Equator. [3] [4]
The specific weight, also known as the unit weight (symbol γ, the Greek letter gamma), is a volume-specific quantity defined as the weight W divided by the volume V of a material: = / Equivalently, it may also be formulated as the product of density, ρ, and gravity acceleration, g: = Its unit of measurement in the International System of Units (SI) is newton per cubic metre (N/m 3), with ...
For instance, buoyancy's diminishing effect upon one's body weight (a relatively low-density object) is 1 ⁄ 860 that of gravity (for pure water it is about 1 ⁄ 770 that of gravity). Furthermore, variations in barometric pressure rarely affect a person's weight more than ±1 part in 30,000. [ 6 ]
For the JGM-3 model the values are: μ = 398600.440 km 3 ⋅s −2 J 2 = 1.75553 × 10 10 km 5 ⋅s −2 J 3 = −2.61913 × 10 11 km 6 ⋅s −2. For example, at a radius of 6600 km (about 200 km above Earth's surface) J 3 /(J 2 r) is about 0.002; i.e., the correction to the "J 2 force" from the "J 3 term" is in the order of 2
The first equation shows that, after one second, an object will have fallen a distance of 1/2 × 9.8 × 1 2 = 4.9 m. After two seconds it will have fallen 1/2 × 9.8 × 2 2 = 19.6 m; and so on. On the other hand, the penultimate equation becomes grossly inaccurate at great distances.
At a fixed point on the surface, the magnitude of Earth's gravity results from combined effect of gravitation and the centrifugal force from Earth's rotation. [2] [3] At different points on Earth's surface, the free fall acceleration ranges from 9.764 to 9.834 m/s 2 (32.03 to 32.26 ft/s 2), [4] depending on altitude, latitude, and longitude. A ...