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An Earth mass (denoted as M 🜨, M ♁ or M E, where 🜨 and ♁ are the astronomical symbols for Earth), is a unit of mass equal to the mass of the planet Earth. The current best estimate for the mass of Earth is M 🜨 = 5.9722 × 10 24 kg, with a relative uncertainty of 10 −4. [2] It is equivalent to an average density of 5515 kg/m 3.
The mass of the oceans is approximately 1.35 × 10 18 metric tons or about 1/4400 of Earth's total mass. The oceans cover an area of 361.8 million km 2 (139.7 million sq mi) with a mean depth of 3,682 m (12,080 ft), resulting in an estimated volume of 1.332 billion km 3 (320 million cu mi).
9.109 383 7139 (28) × 10 −31 kg: 3.1 ... and is strongly dependent on how those units are defined. For example, the atomic mass constant is exactly known ...
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.
The SI base unit of mass is the kilogram (kg). In physics, mass is not the same as weight, even though mass is often determined by measuring the object's weight using a spring scale, rather than balance scale comparing it directly with known masses. An object on the Moon would weigh less than it does on Earth because of the lower gravity, but ...
An overview of ranges of mass. To help compare different orders of magnitude, the following lists describe various mass levels between 10 −67 kg and 10 52 kg. The least massive thing listed here is a graviton, and the most massive thing is the observable universe.
Acceleration of gravity at the surface of the Earth M earth: kg: Mass of the Earth ... "Henry Cavendish and the Density of the Earth". The Physics Teacher. 37 (1): 34 ...
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 ...