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In SI units, this acceleration is expressed in metres per second squared (in symbols, m/s 2 or m·s −2) or equivalently in newtons per kilogram (N/kg or N·kg −1). Near Earth's surface, the acceleration due to gravity, accurate to 2 significant figures, is 9.8 m/s 2 (32 ft/s 2).
It has dimension of acceleration (L/T 2) and it is measured in units of newtons per kilogram (N/kg) or, equivalently, in meters per second squared (m/s 2). In its original concept, gravity was a force between point masses.
Newton's second law states that force equals mass multiplied by acceleration. The unit of force is the newton (N), and mass has the SI unit kilogram (kg). One newton equals one kilogram metre per second squared. Therefore, the unit metre per second squared is equivalent to newton per kilogram, N·kg −1, or N/kg. [2]
The gravitational field equation is [7] = = = | | =, where F is the gravitational force, m is the mass of the test particle, R is the radial vector of the test particle relative to the mass (or for Newton's second law of motion which is a time dependent function, a set of positions of test particles each occupying a particular point in space ...
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 ]
Assuming SI units, F is measured in newtons (N), m 1 and m 2 in kilograms (kg), r in meters (m), and the constant G is 6.674 30 (15) × 10 −11 m 3 ⋅kg −1 ⋅s −2. [12] The value of the constant G was first accurately determined from the results of the Cavendish experiment conducted by the British scientist Henry Cavendish in 1798 ...
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).
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 ...