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Cavendish's stated aim was the "weighing of Earth", that is, determining the average density of Earth and the Earth's mass. His result, ρ 🜨 = 5.448(33) g⋅cm −3, corresponds to value of G = 6.74(4) × 10 −11 m 3 ⋅kg −1 ⋅s −2. It is surprisingly accurate, about 1% above the modern value (comparable to the claimed relative ...
Earth's density varies considerably, between less than 2700 kg/m 3 in the upper crust to as much as 13 000 kg/m 3 in the inner core. [13] The Earth's core accounts for 15% of Earth's volume but more than 30% of the mass, the mantle for 84% of the volume and close to 70% of the mass, while the crust accounts for less than 1% of the mass. [13]
Near the surface of the Earth, the acceleration due to gravity g = 9.807 m/s 2 (metres per second squared, which might be thought of as "metres per second, per second"; or 32.18 ft/s 2 as "feet per second per second") approximately. A coherent set of units for g, d, t and v is essential.
Gravity decreases with altitude as one rises above the Earth's surface because greater altitude means greater distance from the Earth's centre. All other things being equal, an increase in altitude from sea level to 9,000 metres (30,000 ft) causes a weight decrease of about 0.29%.
[citation needed] The kilogram-force is equal to the magnitude of the force exerted on one kilogram of mass in a 9.806 65 m/s 2 gravitational field (standard gravity, a conventional value approximating the average magnitude of gravity on Earth). [2] That is, it is the weight of a kilogram under standard gravity. One kilogram-force is defined as ...
Usually, the relationship between mass and weight on Earth is highly proportional; objects that are a hundred times more massive than a one-liter bottle of soda almost always weigh a hundred times more—approximately 1,000 newtons, which is the weight one would expect on Earth from an object with a mass slightly greater than 100 kilograms.
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.
It is a constant defined by standard as 9.806 65 m/s 2 (about 32.174 05 ft/s 2). This value was established by the third General Conference on Weights and Measures (1901, CR 70) and used to define the standard weight of an object as the product of its mass and this nominal acceleration .